Rd Sharma Xi 2018 Solutions for Class 11 Commerce Math Chapter 32 Statistics are provided here with simple step-by-step explanations. These solutions for Statistics are extremely popular among Class 11 Commerce students for Math Statistics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2018 Book of Class 11 Commerce Math Chapter 32 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2018 Solutions. All Rd Sharma Xi 2018 Solutions for class Class 11 Commerce Math are prepared by experts and are 100% accurate.

#### Question 1:

Calculate the mean deviation from the median of the following frequency distribution:

 Heights in inches 58 59 60 61 62 63 64 65 66 No. of students 15 20 32 35 35 22 20 10 8

In order to find the mean deviation from the median, we will first have to calculate the median.
M is the value of $x$ corresponding to the cumulative frequency just greater than or equal to $N}{2}$.

 ${x}_{i}$ fi Cumulative Frequency ${f}_{i}\left|{d}_{i}\right|$ 58 15 15 3 45 59 20 35 2 40 60 32 67 1 32 61 35 102 0 0 62 35 137 1 35 63 22 159 2 44 64 20 179 3 60 65 10 189 4 40 66 8 197 5 40 $N=\Sigma {f}_{i}=197$ $\sum _{i=1}^{n}{f}_{i}\left|{d}_{i}\right|=336$

Here,
$\phantom{\rule{0ex}{0ex}}\frac{N}{2}=\frac{197}{2}=98.5$

The cumulative frequency just greater than 98.5 is 102. The corresponding value of x is 61.
Therefore, the median is 61.

#### Question 2:

The number of telephone calls received at an exchange in 245 successive one-minute intervals are shown in the following frequency distribution:

 Number of calls 0 1 2 3 4 5 6 7 Frequency 14 21 25 43 51 40 39 12
Compute the mean deviation about median.

We will first calculate the median.

 ${x}_{i}$ fi Cumulative Frequency $\left|{d}_{i}\right|=\left|{x}_{i}-4\right|$ ${f}_{i}\left|{d}_{i}\right|$ 0 14 14 4 56 1 21 35 3 63 2 25 60 2 50 3 43 103 1 43 4 51 154 0 0 5 40 194 1 40 6 39 233 2 78 7 12 245 3 36 $N=\Sigma {f}_{i}=245$ $\sum _{i=1}^{n}{f}_{i}\left|{d}_{i}\right|=366$

Here,
$\phantom{\rule{0ex}{0ex}}\frac{N}{2}=\frac{245}{2}=122.5$

The cumulative frequency just greater than 122.5 is 154 and the corresponding value of x is 4.

∴

#### Question 3:

Calculate the mean deviation about the median of the following frequency distribution:

 xi 5 7 9 11 13 15 17 fi 2 4 6 8 10 12 8

We will first calculate the median for the data.

 ${x}_{i}$ fi Cumulative Frequency ${f}_{i}\left|{d}_{i}\right|$ 5 2 2 8 16 7 4 6 6 24 9 6 12 4 24 11 8 20 2 16 13 10 30 0 0 15 12 42 2 24 17 8 50 4 32 $N=\Sigma {f}_{i}=50$ $\sum _{i=1}^{n}{f}_{i}\left|{d}_{i}\right|=136$

Here,
$\frac{N}{2}=\frac{50}{2}=25$
The cumulative frequency just greater than 25 is 30 and the corresponding value of x is 13.

#### Question 4:

Find the mean deviation from the mean for the following data:
(i)

 xi 5 7 9 10 12 15 fi 8 6 2 2 2 6

(ii)
 xi 5 10 15 20 25 fi 7 4 6 3 5

(iii)
 xi 10 30 50 70 90 fi 4 24 28 16 8

(iv)
 Size 20 21 22 23 24 Frequency 6 4 5 1 4

[NCERT EXEMPLAR]
(v)
 Size 1 3 5 7 9 11 13 15 Frequency 3 3 4 14 7 4 3 4

[NCERT EXEMPLAR]

i)

 xi fi fixi $\left|{x}_{i}-\overline{x}\right|$ ${f}_{i}\left|{x}_{i}-9\right|$ 5 8 40 4 32 7 6 42 2 12 9 2 18 0 0 10 2 20 1 2 12 2 24 3 6 15 6 90 6 36 $N=\Sigma {f}_{i}=26$ $\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-9\right|=88$
$\overline{)x}=\frac{\underset{i=1}{\overset{n}{\sum {f}_{i}}}{x}_{i}}{N}=\frac{234}{26}=9$
$M.D.=\frac{1}{N}\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-\overline{)x}\right|=\frac{1}{26}×88=3.39$

ii)
 xi fi fixi $\left|{x}_{i}-\overline{x}\right|$ ${f}_{i}\left|{x}_{i}-14\right|$ 5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 $N=25$ $\sum _{i=1}^{n}{f}_{i}{x}_{i}=350$ $\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-14\right|=158$
$\overline{)x}=\frac{\underset{i=1}{\overset{n}{\sum {f}_{i}}}{x}_{i}}{N}=\frac{350}{25}=14$
$MD=\frac{1}{N}\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-\overline{)x}\right|=\frac{1}{25}×158=6.32$

iii)
 xi fi fi​xi $\left|{x}_{i}-\overline{x}\right|$ ${f}_{i}\left|{x}_{i}-50\right|$ 10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 $N=80$ $\sum _{i=1}^{n}{f}_{i}{x}_{i}=4000$ $\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-50\right|=1280$
$\overline{)x}=\frac{\underset{i=1}{\overset{n}{\sum {f}_{i}}}{x}_{i}}{N}=\frac{4000}{80}=50$
$MD=\frac{1}{N}\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-\overline{)x}\right|=\frac{1}{80}×1280=16$

(iv)
 Size(xi) Frequency (fi) fi​xi $\left|{x}_{i}-\overline{x}\right|\phantom{\rule{0ex}{0ex}}=\left|{x}_{i}-21.65\right|\phantom{\rule{0ex}{0ex}}$ ${f}_{i}\left|{x}_{i}-\overline{)x}\right|\phantom{\rule{0ex}{0ex}}={f}_{i}\left|{x}_{i}-21.65\right|$ 20 6 120 1.65 9.9 21 4 84 0.65 2.6 22 5 110 0.35 1.75 23 1 23 1.35 1.35 24 4 96 2.35 9.4 $N=20$ $\sum _{i=1}^{n}{f}_{i}{x}_{i}=433$ $\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-\overline{)x}\right|=25$
$\overline{)x}=\frac{\underset{i=1}{\overset{n}{\sum {f}_{i}}}{x}_{i}}{N}=\frac{433}{20}=21.65$
$MD=\frac{1}{N}\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-\overline{)x}\right|=\frac{1}{20}×25=1.25$

(v)
 Size(xi) Frequency (fi) fi​xi $\left|{x}_{i}-\overline{x}\right|\phantom{\rule{0ex}{0ex}}=\left|{x}_{i}-8\right|$ ${f}_{i}\left|{x}_{i}-\overline{)x}\right|\phantom{\rule{0ex}{0ex}}={f}_{i}\left|{x}_{i}-8\right|$ 1 3 3 7 21 3 3 9 5 15 5 4 20 3 12 7 14 98 1 14 9 7 63 1 7 11 4 44 3 12 13 3 39 5 15 15 4 60 7 28 $N=42$ $\sum _{i=1}^{n}{f}_{i}{x}_{i}=336$ $\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-\overline{)x}\right|=124$
$\overline{)x}=\frac{\underset{i=1}{\overset{n}{\sum {f}_{i}}}{x}_{i}}{N}=\frac{336}{42}=8$
$MD=\frac{1}{N}\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-\overline{)x}\right|=\frac{1}{42}×124=2.95$

#### Question 5:

Find the mean deviation from the median for the following data:
(i)

 xi 15 21 27 30 35 fi 3 5 6 7 8

(ii)
 xi 74 89 42 54 91 94 35 fi 20 12 2 4 5 3 4

​i)

 xi fi Cumulative Frequency $\left|{x}_{i}-30\right|$ ${f}_{i}\left|{x}_{i}-30\right|$ 15 3 3 15 45 21 5 8 9 45 27 6 14 3 18 30 7 21 0 0 35 8 29 5 40 $N=\Sigma {f}_{i}=29$ $\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-30\right|=148$

Here, $\phantom{\rule{0ex}{0ex}}\frac{N}{2}=\frac{29}{2}=14.5$

The cumulative frequency greater than 14.5 is 21 and the corresponding value of x is 30.

$MD=\frac{1}{N}\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-M\right|=\frac{1}{29}×148=5.10$

ii)
 xi fi Cumulative Frequency $\left|{x}_{i}-74\right|$ ${f}_{i}\left|{x}_{i}-74\right|$ 35 4 4 39 156 42 2 6 32 64 54 4 10 20 80 74 20 30 0 0 89 12 42 15 180 91 5 47 17 85 94 3 50 20 60 $N=\Sigma {f}_{i}=50$ $\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-74\right|=625$

Here, $\frac{N}{2}=\frac{50}{2}=25$

The cumulative frequency greater than 25 is 30 and the corresponding value of x is 74.

$MD=\frac{1}{N}\sum _{i=1}^{n}{f}_{i}\left|{x}_{i}-M\right|=\frac{1}{50}×775=12.5$

#### Question 1:

Compute the mean deviation from the median of the following distribution:

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 5 10 20 5 10

 Class Frequency ${f}_{i}$ Cumulative frequency Mid-values ${x}_{i}$ $\left|{d}_{i}\right|=\left|{x}_{i}-25\right|$ ${f}_{i}\left|{d}_{i}\right|$ 0−10 5 5 5 20 100 10−20 10 15 15 10 100 20−30 20 35 25 0 0 30−40 5 40 35 10 50 40−50 10 50 45 20 200 $N=\underset{\mathrm{i}=1}{\overset{5}{\sum {f}_{\mathit{i}}}}=50\phantom{\rule{0ex}{0ex}}$ $\underset{i=1}{\overset{5}{\sum {f}_{i}\left|{d}_{i}\right|}}=450\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The cumulative frequency greater than $\frac{N}{2}=25$  is 35 and the corresponding class is 20−30.
Therefore, the median class is  20−30.

#### Question 2:

Find the mean deviation from the mean for the following data:
(i)

 Classes 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 Frequencies 4 8 9 10 7 5 4 3

(ii)
 Classes 95-105 105-115 115-125 125-135 135-145 145-155 Frequencies 9 13 16 26 30 12

(iii)
 Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequencies 6 8 14 16 4 2

(i)  We will compute the mean deviation from the mean in the following way:

 Classes ${f}_{i}$ Midpoints ${x}_{i}$ ${f}_{i}{x}_{i}$ = 0−100 4 50 200 308 1232 100−200 8 150 1200 208 1664 200−300 9 250 2250 108 972 300−400 10 350 3500 8 80 400−500 7 450 3150 92 644 500−600 5 550 2750 192 960 600−700 4 650 2600 292 1168 700−800 3 750 2250 392 1176 $\sum _{i=1}^{8}{f}_{i}=50$

and

(ii) We will compute the mean deviation from the mean in the following way:

 Classes Frequency ${f}_{i}$ Midpoints ${x}_{i}$ ${f}_{i}{x}_{i}$ = 95−105 9 100 900 28.58 257.22 105−115 13 110 1430 18.58 241.54 115−125 16 120 1920 8.58 137.28 125−135 26 130 3380 1.42 36.92 135−145 30 140 4200 11.42 342.6 145−155 12 150 1800 21.42 257.04

and

(iii) We will compute the mean deviation from the mean in the following way:

 Classes Frequency ${f}_{i}$ Midpoints ${x}_{i}$ ${f}_{i}{x}_{i}$ = 0−10 6 5 30 22 132 10−20 8 15 120 12 96 20−30 14 25 350 2 28 30−40 16 35 560 8 128 40−50 4 45 180 18 72 50−60 2 55 110 28 56 $\sum _{i=1}^{6}{f}_{i}=50$

and  ​

#### Question 3:

Compute mean deviation from mean of the following distribution:

 Mark 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 No. of students 8 10 15 25 20 18 9 5

Computation of mean deviation from the mean:

 Marks Number of Students ${f}_{i}$ Midpoints ${x}_{i}$ ${f}_{i}{x}_{i}$ = 10−20 8 15 120 34 272 20−30 10 25 250 24 240 30−40 15 35 525 14 210 40−50 25 45 1125 4 100 50−60 20 55 1100 6 120 60−70 18 65 1170 16 288 70−80 9 75 675 26 234 80−90 5 85 425 36 180

and

#### Question 4:

The age distribution of 100 life-insurance policy holders is as follows:

 Age (on nearest birth day) 17-19.5 20-25.5 26-35.5 36-40.5 41-50.5 51-55.5 56-60.5 61-70.5 No. of persons 5 16 12 26 14 12 6 5
Calculate the mean deviation from the median age.

To make this function continuous, we need to subtract 0.25 from the lower limit and add 0.25 to the upper limit of the class.

 Age Number of People ${f}_{i}$ Cumulative Frequency Midpoints ${x}_{i}$ $\left|{d}_{i}\right|=\left|{x}_{i}-38.63\right|$ 16.75−19.75 5 5 18.25 20.38 101.9 19.75−25.75 16 21 22.75 15.88 254.08 25.75−35.75 12 33 30.75 7.88 94.56 35.75−40.75 26 59 38.25 0.38 9.38 40.75−50.75 14 73 45.75 7.12 99.68 50.75−55.75 12 85 53.25 14.62 175.44 55.75−60.75 6 91 58.25 19.62 117.72 60.75−70.75 5 96 65.75 27.12 135.6 $\sum _{i=1}^{8}{f}_{i}\left|{d}_{i}\right|=988.36$

#### Question 5:

Find the mean deviation from the mean and from median of the following distribution:

 Marks 0-10 10-20 20-30 30-40 40-50 No. of students 5 8 15 16 6

Computation of mean distribution from the median:

 Marks Number of Students ${f}_{i}$ Cumulative Frequency Midpoints ${x}_{i}$ $\left|{d}_{i}\right|=\left|{x}_{i}-28\right|$ ${f}_{\mathrm{i}}\left|{d}_{\mathit{i}}\right|$ ${f}_{i}{x}_{i}$ $\left|{x}_{i}-27\right|$ ${f}_{\mathrm{i}}\left|{x}_{\mathit{i}}-27\right|$ 0−10 5 5 5 23 115 25 22 110 10−20 8 13 15 13 104 120 12 96 20−30 15 28 25 3 45 375 2 30 30−40 16 44 35 7 112 560 8 128 40−50 6 50 45 17 102 270 18 108 $N=50$ $\sum _{i=1}^{5}{f}_{i}\left|{d}_{i}\right|=478$ 1350 $\sum _{i=1}^{5}{f}_{i}\left|{x}_{i}-27\right|=472\phantom{\rule{0ex}{0ex}}$

The cumulative frequency just greater than $\frac{N}{2}=25$  is 28 and the corresponding class is 20−30.
Thus, the median class is 20−30.

Using formula:

Mean deviation from the median and the mean are 9.56 and 9.44, respectively.

#### Question 6:

Calculate mean deviation about median age for the age distribution of 100 persons given below:

 Age: 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 Number of persons 5 6 12 14 26 12 16 9

Since the function is not continuous, we subtract 0.5 from the lower limit of the class and add 0.5 to the upper limit of the class so that the class interval remains same, while the function becomes continuous.

Thus, the mean distribution table will be as follows:

 Age Number of Persons ${f}_{i}$ Midpoint ${x}_{i}$ Cumulative Frequency $\left|{d}_{i}\right|=\left|{x}_{i}-38\right|\phantom{\rule{0ex}{0ex}}$ ${f}_{i}{d}_{i}$ 15.5−20.5 5 18 5 20 100 20.5−25.5 6 23 11 15 90 25.5−30.5 12 28 23 10 120 30.5−35.5 14 33 37 5 70 35.5−40.5 26 38 63 0 0 40.5−45.5 12 43 75 5 60 45.5−50.5 16 48 91 10 160 50.5−55.5 9 53 100 15 135 $N=\sum _{i=1}^{8}{f}_{i}=100$ $\sum _{i=1}^{8}{f}_{i}{d}_{i}=735$

Thus, the cumulative frequency slightly greater than 50 is 63 and falls in the median class 35.5−40.5.

Thus, the mean deviation from the median age is 7.35 years.

#### Question 7:

Calculate the mean deviation about the mean for the following frequency distribution:

 Class interval: 0–4 4–8 8–12 12–16 16–20 Frequency 4 6 8 5 2

Let the assumed mean A = 10 and h = 4.

 Class Interval Mid-Value(xi) Frequency(fi) ${d}_{i}=\frac{{x}_{i}-10}{4}$ ${f}_{i}{d}_{i}$ $\left|{x}_{i}-\overline{)X}\right|\phantom{\rule{0ex}{0ex}}=\left|{x}_{i}-9.2\right|$ ${f}_{i}\left|{x}_{i}-\overline{)X}\right|$ 0–4 2 4 −2 −8 7.2 28.8 4–8 6 6 −1 −6 3.2 19.2 8–12 10 8 0 0 0.8 6.4 12–16 14 5 1 5 4.8 24 16–20 18 2 2 4 8.8 17.6 N = 25 $\sum _{}{f}_{i}{d}_{i}$ = −5 $\sum _{}{f}_{i}\left|{x}_{i}-\overline{)X}\right|=$96

Here, N = 25 and $\sum _{}{f}_{i}{d}_{i}$ = −5

Mean,

∴ Mean deviation about mean$=\frac{1}{N}\sum _{}{f}_{i}\left|{x}_{i}-\overline{)X}\right|=\frac{1}{25}×96=3.84$

#### Question 8:

Calculate mean deviation from the median of the following data:                    [NCERT EXEMPLAR]

 Class interval: 0–6 6–12 12–18 18–24 24–30 Frequency 4 5 3 6 2

Calculation of mean deviation about the median.

 Class Interval Mid-Values (xi) Frequency (fi) Cummulative Frequency (c.f.) $\left|{x}_{i}-14\right|$ ${f}_{i}\left|{x}_{i}-14\right|$ 0–6 3 4 4 11 44 6–12 9 5 9 5 25 12–18 15 3 12 1 3 18–24 21 6 18 7 42 24–30 27 2 20 13 26 N = 20 $\sum _{}{f}_{i}\left|{x}_{i}-14\right|=140$

Here, N = 20. So, $\frac{N}{2}=10$

The cummulative frequency just greater than $\frac{N}{2}$ is 12. Thus, 12–18 is the median class.

Now, l = 12, h = 6, f = 3 and F = 9

$\therefore \mathrm{Median}=l+\frac{\frac{N}{2}-F}{f}×h=12+\left(\frac{10-9}{3}\right)×6=14$

Now,

Mean deviation about median = $\frac{1}{N}\sum _{}{f}_{i}\left|{x}_{i}-14\right|=\frac{1}{20}×140=7$

#### Question 1:

Find the mean, variance and standard deviation for the following data:
(i) 2, 4, 5, 6, 8, 17.
(ii) 6, 7, 10, 12, 13, 4, 8, 12.
(iii) 227, 235, 255, 269, 292, 299, 312, 321, 333, 348.
(iv) 15, 22, 27, 11, 9, 21, 14, 9.

(i)    2,4,5,6,8,17

${x}_{i}$ ${\left({x}_{i}-7\right)}^{2}$
2 -5 25
4 -3 9
5 -2 4
6 -1 1
8 1 1
17 10 100

$n=6$

(ii) 6,7,10,12,13,4,8,12

${x}_{i}$ ${\left({x}_{\mathit{i}}\mathit{-}\overline{)\mathit{X}}\right)}^{2}$
6 -3 9
7 -2 4
10 1 1
12 3 9
13 4 16
4 $-$5 25
8 $-$1 1
12 3 9

$n=8$

(iii)      227,235,255,269,292,299,312,321,333,348,

${x}_{i}$ ${\left({x}_{\mathit{i}}-\overline{)X}\right)}^{2}$
227 − 62.1 3856.41
235 − 54.1 2926.81
255 − 34.1 1162.81
269 − 20.1 404.01
292 2.9 8.41
299 9.9 98.01
312 22.9 524.41
321 31.9 1017.61
333 43.9 1927.21
348 58.9 3469.21

(iv)   15,22,27,11,9,21,14,9

${x}_{i}$ $\left({x}_{i}\mathit{-}\overline{)X}\right)\mathit{=}\left({x}_{i}\mathit{-}16\right)$ ${\left({x}_{i}\mathit{-}\overline{X}\right)}^{\mathit{2}}$
15 −1 1
22 6 36
27 11 121
11 5 25
9 −7 49
21 5 25
14 −2 4
9 −7 49

#### Question 2:

The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.

Let   be  the 20 given observations.

Let u1,u2,,u3, ..., u20 be the new observations, such that

Thus, variance of the new observations is 20.

#### Question 3:

The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Let x1,,x2,,x3 , ..., x15 be the given observations.

Variance X  is given as 4.
If  $\overline{X}$ is the mean of the given observations, then we get:

Let u1,u2,u3 ... u15 be the new observations such that

Thus, variance of the new observation is 4.

#### Question 4:

The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

Let be the other two observations.

Mean is 4.4.

$\therefore \frac{1+2+6+x+y}{5}=4.4$

Let Var (X) be the variance of these observations, which is given to be 8.24.

If $\overline{\mathit{X}}$ is the mean, then we have:

Thus, the other two observations are 9 and 4.

#### Question 5:

The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Thus, mean of the new observations is 24.

Thus, standard deviation of the new observations is 12.

#### Question 6:

The mean and variance of 8 observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Let x and y be the remaining two observations.

Thus, the remaining two observations are 8 and 4.

#### Question 7:

For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.

We have:

#### Question 8:

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?

To find the corrected SD:

Corrected mean = 39.9
Corrected standard deviation = 5

#### Question 9:

The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted
(ii) if it is replaced by 12.

(i)     If  observation 8 is omitted, then total 19 observations are left.

Incorrected $\sum _{}{x}_{i}=200$

Thus, if  8 is omitted, then the mean is 10.10 and SD is 1.997.

(ii)  When incorrect observation 8 is replaced by 12:

If 8 is replaced by 12, then the mean  is 10.2 and SD is 1.9899.

#### Question 10:

The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations were omitted.

Thus, after omitting three values, the mean would be 20 and SD would be 3.0357.

#### Question 11:

Show that the two formulae for the standard deviation of ungrouped data

$\sigma =\sqrt{\frac{1}{n}\underset{}{\sum {\left({x}_{i}-\overline{)X}\right)}^{2}}}$ and $\sigma \text{'}=\sqrt{\frac{1}{n}\underset{}{\sum {x}_{i}^{2}-{\overline{)X}}^{2}}}$ are equivalent, where $\overline{)X}=\frac{1}{n}\sum _{}{x}_{i}$

$=\sqrt{\frac{1}{n}\underset{}{\sum {x}_{i}^{2}-2{\overline{)X}}^{2}+}{\overline{)X}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{n}\underset{}{\sum {x}_{i}^{2}-{\overline{)X}}^{2}}}\phantom{\rule{0ex}{0ex}}=\sigma \text{'}$

Hence, the formulae $\sigma =\sqrt{\frac{1}{n}\underset{}{\sum {\left({x}_{i}-\overline{)X}\right)}^{2}}}$ and $\sigma \text{'}=\sqrt{\frac{1}{n}\underset{}{\sum {x}_{i}^{2}-{\overline{)X}}^{2}}}$ are equivalent, where $\overline{)X}=\frac{1}{n}\sum _{}{x}_{i}$.

#### Question 1:

Find the standard deviation for the following distribution:

 x : 4.5 14.5 24.5 34.5 44.5 54.5 64.5 f : 1 5 12 22 17 9 4

 x: 4.5 14.5 24.5 34.5 44.5 54.5 64.5 f: 1 5 12 22 17 9 4

Median value of $x$ is 34.5.

 ${x}_{i}$ ${f}_{i}$ ${d}_{i}={x}_{i}-34.5$ ${u}_{i}=\frac{{x}_{i}-34.5}{10}$ ${f}_{i}{u}_{{i}_{}}$ ${{u}_{i}}^{2}$ ${f}_{i}{{u}_{i}}^{2}$ 4.5 1 $-$30 $-$3 $-$3 9 9 14.5 5 $-$20 $-$2 $-$10 4 20 24.5 12 $-$10 $-$1 $-$12 1 12 34.5 22 0 0 0 0 0 44.5 17 10 1 17 1 17 54.5 9 20 2 18 4 36 64.5 4 30 3 12 9 36 $N=\sum {f}_{i}=70$ $\sum {f}_{i}{u}_{i}=22$ $\sum {f}_{i}{{u}_{i}}^{2}=130$

$\mathrm{Var}\left(X\right)={h}^{2}\left[\frac{1}{N}\sum _{i=1}^{n}{f}_{i}{{u}_{i}}^{2}-{\left(\frac{1}{N}\sum _{i=1}^{n}{f}_{i}{u}_{i}\right)}^{2}\right]$

We have

Plugging all the values in the formula of variance:

Standard deviation, $\mathrm{SD}=\sqrt{\mathrm{Var}\left(X\right)}$

#### Question 2:

Table below shows the frequency f with which 'x' alpha particles were radiated from a diskette

 x : 0 1 2 3 4 5 6 7 8 9 10 11 12 f : 51 203 383 525 532 408 273 139 43 27 10 4 2
Calculate the mean and variance.

Mean, $\overline{X}=\frac{\sum {f}_{i}{x}_{i}}{\sum {f}_{i}}=\frac{10078}{2600}=3.88$

 ${x}_{i}$ ${f}_{i}$ ${f}_{i}{x}_{i}$ ${x}_{i}-\overline{X}$ ${\left({x}_{i}-\overline{X}\right)}^{2}$ ${f}_{i}{\left({x}_{i}-\overline{X}\right)}^{2}$ 0 51 0 −3.88 15.05 767.55 1 203 203 −2.88 8.29 1682.87 2 383 766 −1.88 3.53 1351.99 3 525 1575 −0.88 0.77 404.25 4 532 2128 0.12 0.014 7.448 5 408 2040 1.12 1.25 510 6 273 1638 2.12 4.49 1225.77 7 139 973 3.12 9.73 1352.47 8 43 344 4.12 16.97 729.71 9 27 243 5.12 26.21 707.67 10 10 100 6.12 37.45 374.5 11 4 44 7.12 50.69 202.76 12 2 24 8.12 65.93 131.86 $\sum {f}_{i}=N=2600$ $\sum {f}_{i}{x}_{i}=10078$ $\sum {f}_{i}{\left({x}_{i}-\overline{X}\right)}^{2}=9448.848$

Variance, ${\sigma }^{2}=\frac{\sum {f}_{i}{\left({x}_{i}-\overline{X}\right)}^{2}}{N}=\frac{9448.848}{2600}=3.63$

#### Question 3:

Find the mean, mode, S.D. and coefficient of skewness for the following data:

 Year render: 10 20 30 40 50 60 No. of persons (cumulative): 15 32 51 78 97 109

 Year render No. of persons (cumulative) ${f}_{i}$ ${u}_{i}=\frac{{x}_{i}-35}{10}$ ${f}_{i}{u}_{i}$ ${{u}_{i}}^{2}$ ${f}_{i}{{u}_{i}}^{2}$ 10 15 15 $-$2.5 $-$37.5 6.25 93.75 20 32 17 $-$1.5 $-$25.5 2.25 38.25 30 51 19 $-$0.5 $-$9.5 0.25 4.75 40 78 27 0.5 13.5 0.25 6.75 50 97 19 1.5 28.5 2.25 42.75 60 109 12 2.5 30 6.25 75 $\sum {f}_{i}=N=109$ $\sum {f}_{i}{u}_{i}=-0.5$ $\sum {f}_{i}{{u}_{i}}^{2}=261.25$

Coefficient of skewness = Mean $-$ Mode
= 34.96 $-$ 40
= $-$5.04

#### Question 4:

Find the standard deviation for the following data:

 x : 3 8 13 18 23 f : 7 10 15 10 6

 ${x}_{i}$ ${f}_{i}$ ${f}_{i}{x}_{i}$ $\left({x}_{i}-\overline{X}\right)$ ${\left({x}_{i}-\overline{X}\right)}^{2}$ ${f}_{i}{\left({x}_{i}-\overline{X}\right)}^{2}$ 3 7 21 −9.79 95.84 670.88 8 10 80 −4.79 22.94 229.4 13 15 195 0.21 0.04 0.6 18 10 180 5.21 27.14 271.4 23 6 138 10.21 104.24 625.44 $\sum {f}_{i}=48$ $\sum {f}_{i}{x}_{i}=614$ $\sum {f}_{i}{\left({x}_{i}-\overline{X}\right)}^{2}=1797.32$

Variance, ${\sigma }^{2}=\frac{\sum {f}_{i}{\left({x}_{i}-\overline{X}\right)}^{2}}{\sum {f}_{i}}=\frac{1797.32}{48}=37.44$

#### Question 1:

Calculate the mean and S.D. for the following data:

 Expenditure in Rs: 0-10 10-20 20-30 30-40 40-50 Frequency: 14 13 27 21 15

 Expenditure (Rs) ${f}_{i}$ Midpoint $\left({x}_{i}\right)$ ${f}_{i}{x}_{i}$ ${x}_{i}-\overline{X}$ ${\left({x}_{i}-\overline{X}\right)}^{2}$ ${f}_{i}{\left({x}_{i}-\overline{X}\right)}^{2}$ 0−10 14 5 70 $-$21.1 445.21 6233.94 10−20 13 15 195 $-$11.1 123.21 1601.73 20−30 27 25 675 $-$1.1 1.21 34.67 30−40 21 35 735 8.9 79.21 1663.41 40−50 15 45 675 18.9 357.21 5358.15 $\sum {f}_{i}=90$ $\sum {f}_{i}{x}_{i}=2350$ $\sum {f}_{i}{\left({x}_{i}-\overline{X}\right)}^{2}=14891.9$

Mean,
Variance,

#### Question 2:

Calculate the standard deviation for the following data:

 Class: 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequency: 9 17 43 82 81 44 24

 Class ${f}_{i}$ Midpoint$\left({x}_{i}\right)$ ${y}_{i}=\frac{{x}_{i}-105}{30}$ ${{y}_{i}}^{2}$ ${f}_{i}{y}_{i}$ ${f}_{i}{{y}_{i}}^{2}$ 0−30 9 15 −3 9 −27 81 30−60 17 45 −2 4 −34 68 60−90 43 75 −1 1 −43 43 90−120 82 105 0 0 0 0 120−150 81 135 1 1 81 81 150−180 44 165 2 4 88 176 180−210 24 195 3 9 72 216 $\sum {f}_{i}=N=300$ $\sum {f}_{i}{y}_{i}=137$ $\sum {f}_{i}{{y}_{i}}^{2}=665$

Mean, $\overline{x}=a+h\left(\frac{1}{N}\Sigma {f}_{i}{y}_{i}\right)=105+30\left(\frac{137}{300}\right)=118.7$

Variance:

#### Question 3:

Calculate the A.M. and S.D. for the following distribution:

 Class: 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency: 18 16 15 12 10 5 2 1

 Class ${f}_{i}$ Midpoint$\left({x}_{i}\right)$ ${u}_{i}=\frac{{x}_{i}-35}{10}$ ${f}_{i}{u}_{i}$ ${f}_{i}{{u}_{i}}^{2}$ 0−10 18 5 $-$3 $-$54 162 10−20 16 15 $-$2 $-$32 64 20−30 15 25 $-$1 $-$15 15 30−40 12 35 0 0 0 40−50 10 45 1 10 10 50−60 5 55 2 10 20 60−70 2 65 3 6 18 70−80 1 75 4 4 16 $\sum {f}_{i}=79$ $\sum {f}_{i}{u}_{i}=-71$ $\sum {f}_{i}{{u}_{i}}^{2}=305$

$\overline{X}=a+h\left(\frac{\sum {f}_{i}{u}_{i}}{N}\right)=35+10\left(\frac{-71}{79}\right)=26.01$

AM = 26.01

${\sigma }^{2}={h}^{2}\left[\frac{\sum {f}_{i}{{u}_{i}}^{2}}{N}-{\left(\frac{\sum {f}_{i}{u}_{i}}{N}\right)}^{2}\right]=100\left[\frac{305}{79}-\frac{5041}{6241}\right]=305.20\phantom{\rule{0ex}{0ex}}\sigma =\sqrt{305.20}=17.47$

#### Question 4:

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and S.D.

#### Question 5:

Calculate the mean, median and standard deviation of the following distribution:

 Class-interval: 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70 Frequency: 2 3 8 12 16 5 2 3

 Class Interval ${f}_{i}$ Midpoint ${x}_{i}$ ${u}_{i}=\frac{{x}_{i}-53}{4}$ ui 2 ${f}_{i}{u}_{i}$ ${f}_{i}{{u}_{i}}^{2}$ 31−35 2 33 $-$5 25 $-$10 50 36−40 3 38 $-$3.75 14.06 $-$11.25 42.18 41−45 8 43 $-$2.5 6.25 $-$20 50 46−50 12 48 $-$1.25 1.56 $-$15 18.72 51−55 16 53 0 0 0 0 56−60 5 58 1.25 1.56 6.25 7.8 61−65 2 63 2.5 6.25 5 12.5 66−70 3 68 3.75 14.06 11.25 42.18 N = 51 $\sum _{i=1}^{n}{f}_{i}{u}_{i}=-33.75$ $\sum _{i=1}^{n}{f}_{i}{{u}_{i}}^{2}=223.38$

 ${f}_{i}$ $CF$ (Cumulative frequency) 2 2 3 5 8 13 12 25 16 41 5 46 2 48 3 51

$\sum {f}_{i}=51=N\phantom{\rule{0ex}{0ex}}\frac{N}{2}=25.5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Median class interval is
5155.
$L=51\phantom{\rule{0ex}{0ex}}F=25\phantom{\rule{0ex}{0ex}}f=16\phantom{\rule{0ex}{0ex}}h=4$

$\mathrm{Median}=L+\frac{\frac{N}{2}-F}{f}×h\phantom{\rule{0ex}{0ex}}$

$=51+\frac{25.5-25}{16}×4\phantom{\rule{0ex}{0ex}}=51+\frac{0.5}{4}\phantom{\rule{0ex}{0ex}}=51.125$

#### Question 6:

Find the mean and variance of frequency distribution given below:

 xi: 1 ≤ x < 3 3 ≤ x < 5 5 ≤ x < 7 7 ≤ x < 10 fi: 6 4 5 1

 xi Mid-Values(yi) yi2 fi fi yi fi yi2 1–3 2 4 6 12 24 3–5 4 16 4 16 64 5–7 6 36 5 30 180 7–10 8.5 72.25 1 8.5 72.25 N = $\sum _{}{f}_{i}$= 16 $\sum _{}{f}_{i}{y}_{i}=66.5$ $\sum _{}{f}_{i}{{y}_{i}}^{2}=340.25$

Therefore,

Mean = $\frac{\sum _{}{f}_{i}{y}_{i}}{\sum _{}{f}_{i}}=\frac{66.5}{16}=4.16$

Variance = $\left(\frac{1}{N}\sum _{}{f}_{i}{y}_{i}^{2}\right)-{\left(\frac{1}{N}\sum _{}f{}_{i}y{}_{i}\right)}^{2}=\frac{1}{16}×340.25-{\left(\frac{1}{16}×66.5\right)}^{2}=21.26-17.22=4.04$

#### Question 7:

The weight of coffee in 70 jars is shown in the following table:

 Weight (in grams): 200–201 201–202 202–203 203–204 204–205 205–206 Frequency: 13 27 18 10 1 1

Determine the variance and standard deviation of the above distribution.                                          [NCERT EXEMPLAR]

 Weight (in grams) Mid-Values$\left({x}_{i}\right)$ Frequency$\left({f}_{i}\right)$ ${d}_{i}={x}_{i}-202.5$ ${d}_{i}^{2}$ ${f}_{i}{d}_{i}$ ${f}_{i}{d}_{i}^{2}$ 200–201 200.5 13 −2 4 −26 52 201–202 201.5 27 −1 1 −27 27 202–203 202.5 18 0 0 0 0 203–204 203.5 10 1 1 10 10 204–205 204.5 1 2 4 2 4 205–206 205.5 1 3 9 3 9 N = $\sum _{}{f}_{i}=70$ $\sum _{}{f}_{i}{d}_{i}=-38$ $\sum _{}{f}_{i}{d}_{i}^{2}=102$

Now,

Variance, ${\sigma }^{2}$

Standard deviation, $\sigma$ =

#### Question 8:

Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
[NCERT EXEMPLAR]

Given:

Number of observations, n = 100

Mean, $\overline{)x}$ = 40

Standard deviation, $\sigma$ = 10

We know that

$\overline{)x}=\frac{\sum _{}{x}_{i}}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{\sum _{}{x}_{i}}{100}=40\phantom{\rule{0ex}{0ex}}⇒\sum _{}{x}_{i}=4000$

∴ Correct $\sum _{}{x}_{i}=4000-\left(30+70\right)+\left(3+27\right)=3930$

Correct mean =

Now,

Incorrect variance, ${\sigma }^{2}=\frac{\sum _{}{x}_{i}^{2}}{100}-{\left(40\right)}^{2}$
$⇒\frac{\sum _{}{x}_{i}^{2}}{100}=100+1600=1700\phantom{\rule{0ex}{0ex}}⇒\sum _{}{x}_{i}^{2}=170000$

Correct $\sum _{}{x}_{i}^{2}=170000-{30}^{2}-{70}^{2}+{3}^{2}+{27}^{2}=164939$

∴ Correct standard deviation

$=\sqrt{\frac{164939}{100}-{\left(39.3\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1649.39-1544.49}\phantom{\rule{0ex}{0ex}}=\sqrt{104.9}\phantom{\rule{0ex}{0ex}}=10.24$

#### Question 9:

While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
[NCERT EXEMPLAR]

Given:

Number of observations, n = 10

Mean, $\overline{)x}$ = 45

Variance, ${\sigma }^{2}$ = 16

Now,

Incorrect mean, $\overline{)x}$ = 45

∴ Correct $\sum _{}{x}_{i}=450-52+25=423$

⇒ Correct mean =

Incorrect variance, ${\sigma }^{2}$ = 16

∴ Correct $\sum _{}{x}_{i}^{2}=20410-{\left(52\right)}^{2}+{\left(25\right)}^{2}=20410-2704+625=18331$

Now,

Correct variance = $\frac{18331}{10}-{\left(42.3\right)}^{2}=1833.1-1789.29=43.81$

#### Question 10:

Calculate the mean, variance and standard deviation of the following frequency distribution.

 Class: 1–10 10–20 20–30 30–40 40–50 50–60 Frequency: 11 29 18 4 5 3

Let the assumed mean A = 25.

 Class Mid-Values$\left({x}_{i}\right)$ ${d}_{i}^{2}$ Frequency $\left({f}_{i}\right)$ ${f}_{i}{d}_{i}$ ${f}_{i}{d}_{i}^{2}$ 1–10 5.5 −19.5 380.25 11 −214.5 4182.75 10–20 15 −10 100 29 −290 2900 20–30 25 0 0 18 0 0 30–40 35 10 100 4 40 400 40–50 45 20 400 5 100 2000 50–60 55 30 900 3 90 2700 N = $\sum _{}{f}_{i}$ = 70 $\sum _{}{f}_{i}{d}_{i}$ = −274.5 $\sum _{}{f}_{i}{d}_{i}^{2}=$12182.75

Mean = $A+\frac{\sum _{}{f}_{i}{d}_{i}}{\sum _{}{f}_{i}}=25+\left(\frac{-274.5}{70}\right)=25-3.92=21.08$

Variance = ${\sigma }^{2}=\left(\frac{1}{N}\sum _{}{f}_{i}{d}_{i}^{2}\right)-{\left(\frac{1}{N}\sum _{}{f}_{i}{d}_{i}\right)}^{2}=\frac{12182.75}{70}-{\left(\frac{-274.5}{70}\right)}^{2}=174.02-15.37=158.65$

Standard deviation = $\sigma =\sqrt{158.65}=12.6$

#### Question 1:

Two plants A and B of a factory show following results about the number of workers and the wages paid to them

 Plant A Plant B No. of workers 5000 6000 Average monthly wages Rs 2500 Rs 2500 Variance of distribution of wages 81 100
In which plant A or B is there greater variability in individual wages?

Variance of the distribution of wages in plant $A\left({\sigma }^{2}\right)=81$
Standard deviation of the distribution of wages in plant $A\left(\sigma \right)=9$

Variance of the distribution of wages in plant $B\left({\sigma }^{2}\right)=100$
Standard deviation of the distribution of wages in plant $B\left(\sigma \right)=10$
Average monthly wages in both the plants are Rs 2500.
Thus, the plant with greater value of SD will have more variability in salary.
Plant B has more variability in individual wages than plant A.

#### Question 2:

The means and standard deviations of heights ans weights of 50 students of a class are as follows:

 Weights Heights Mean 63.2 kg 63.2 inch Standard deviation 5.6 kg 11.5 inch
Which shows more variability, heights or weights?

CV in heights is greater than CV in weights.

#### Question 3:

Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?

The coefficient of variation (CV) for the first distribution is 60.
The coefficient of variation (CV) for the second distribution is 70.
$SD\left({\sigma }_{1}\right)=21$
$SD\left({\sigma }_{2}\right)=16$
We know:
$CV=\frac{\sigma }{\overline{X}}×100$

From the above formula, we get:
$\overline{X}=\frac{\sigma }{CV}×100$

$\overline{{X}_{1}}=\frac{21}{60}×100=35\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\overline{{X}_{2}}=\frac{16}{70}×100=22.86$

#### Question 4:

Calculate coefficient of variation from the following data:

 Income (in Rs): 1000-1700 1700-2400 2400-3100 3100-3800 3800-4500 4500-5200 No. of families: 12 18 20 25 35 10

 Income (Rs) ${f}_{i}$ Midpoint$\left({x}_{i}\right)$ ${u}_{i}=\frac{{x}_{i}-3450}{700}$ ${f}_{i}{u}_{i}$ ${f}_{i}{{u}_{i}}^{2}$ 1000−1700 12 1350 −3 −36 108 1700−2400 18 2050 −2 −36 72 2400−3100 20 2750 −1 −20 20 3100−3800 25 3450 0 0 0 3800−4500 35 4150 1 35 35 4500−5200 10 4850 2 20 40 $\sum {f}_{i}=120$ $\sum {f}_{i}{u}_{i}=-37$ $\sum {f}_{i}{{u}_{i}}^{2}=275$

​​$\overline{X}=a+h\left(\frac{\sum {f}_{i}{u}_{i}}{N}\right)=3450+700\left(\frac{-37}{120}\right)=3234.17$

#### Question 5:

An analysis of the weekly wages paid to workers in two firms A and B, belonging to the same industry gives the following results:

 Firm A Firm B No. of wage earners 586 648 Average weekly wages Rs 52.5 Rs. 47.5 Variance of the distribution of wages 100 121
(i) Which firm A or B pays out larger amount as weekly wages?
(ii) Which firm A or B has greater variability in individual wages?

Average weekly wages
Total weekly wages = (Average weekly wages) (Numbers of workers)
Total weekly wages for firm A = Rs 52.5$×$586 = Rs 30765
Total weekly wages for firm B = Rs 47.5$×$648 = Rs 30780
(i) Firm B pays a larger amount as weekly wages.

(ii)  SD (firm A) = 10
SD (firm B) = 11

Since CV of firm B is greater than that of firm A, firm B has greater variability in individual wages.

#### Question 6:

The following are some particulars of the distribution of weights of boys and girls in a class:

 Number Boys Girls 100 50 Mean weight 60 kg 45 kg Variance 9 4
Which of the distributions is more variable?

​We know:
SD (boys) is 3 and SD (girls) is 2.

$\mathrm{CV}=\frac{\sigma }{\overline{X}}×100$

Since CV for boys is greater than that of  girls, distribution of the weights of boys is more variable than that of girls.

#### Question 7:

The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:

 Subject Mathematics Physics Chemistry Mean 42 32 40.9 Standard 12 15 20 Deviation
Which of the three subjects shows the highest variability in marks and which shows the lowest?

We know:

CV of mathematics marks $=\frac{12}{42}×100=\frac{1200}{42}=28.57\phantom{\rule{0ex}{0ex}}$

CV of physics marks ​$=\frac{15}{32}×100=\frac{1500}{32}=46.87$

CV of chemistry marks $=\frac{20}{40.9}×100=\frac{2000}{40.9}=48.89$

Since CV of chemistry is the greatest, the variability of marks in chemistry is the highest and that of mathematics is the lowest.

#### Question 8:

From the data given below state which group is more variable, G1 or G2?

 Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Group G1 9 17 32 33 40 10 9 Group G2 10 20 30 25 43 15 7

 Marks ${f}_{i}$ Midpoint $\left({x}_{i}\right)$ ${u}_{i}=\frac{{x}_{i}-45}{10}$ ${f}_{i}{u}_{i}$ ${f}_{i}{{u}_{i}}^{2}$ 10−20 9 15 −3 −27 81 20−30 17 25 −2 −34 68 30−40 32 35 −1 −32 32 40−50 33 45 0 0 0 50−60 40 55 1 40 40 60−70 10 65 2 20 40 70−80 9 75 3 27 81 N=150 $\sum {f}_{i}{u}_{i}=-6$ $\sum {f}_{i}{{u}_{i}}^{2}=342$

For group 1:
$\overline{X}=a+h\left(\frac{\sum {f}_{i}{u}_{i}}{N}\right)=45+10\left(\frac{-6}{150}\right)=44.6$
${\sigma }^{2}={h}^{2}\left[\frac{\sum {f}_{i}{{u}_{i}}^{2}}{N}-{\left(\frac{\sum {f}_{i}{u}_{i}}{N}\right)}^{2}\right]=100\left[\frac{342}{150}-\frac{36}{22500}\right]=227.84\phantom{\rule{0ex}{0ex}}\sigma =\sqrt{227.84}=15.09$

 Marks ${f}_{i}$ Midpoint$\left({x}_{i}\right)$ ${u}_{i}=\frac{{x}_{i}-45}{10}$ ${f}_{i}{u}_{i}$ ${f}_{i}{{u}_{i}}^{2}$ 10−20 10 15 −3 −30 90 20−30 20 25 −2 −40 80 30−40 30 35 −1 −30 30 40−50 25 45 0 0 0 50−60 43 55 1 43 43 60−70 15 65 2 30 60 70−80 7 75 3 21 63 $\sum {f}_{i}=150$ $\sum {f}_{i}{u}_{i}=-6$ $\sum {f}_{i}{{u}_{i}}^{2}=366$

For group 2:
$\overline{X}=a+h\left(\frac{\sum {f}_{i}{u}_{i}}{N}\right)=45+10\left(\frac{-6}{150}\right)=44.6$
${\sigma }^{2}={h}^{2}\left[\frac{\sum {f}_{i}{{u}_{i}}^{2}}{N}-{\left(\frac{\sum {f}_{i}{u}_{i}}{N}\right)}^{2}\right]=100\left[\frac{366}{150}-\frac{36}{22500}\right]=243.84\phantom{\rule{0ex}{0ex}}\sigma =\sqrt{243.84}=15.62$
Mean of both the groups are same and SD of group 2 is greater than that of group 1.
So, group 2 will be more variable.

#### Question 9:

Find the coefficient of variation for the following data:

 Size (in cms): 10-15 15-20 20-25 25-30 30-35 35-40 No. of items: 2 8 20 35 20 15

 Size (cm) ${f}_{i}$ Midpoint $\left({x}_{i}\right)$ ${u}_{i}=\frac{{x}_{i}-27.5}{5}$ ${f}_{i}{u}_{i}$ ${f}_{i}{{u}_{i}}^{2}$ 10−15 2 12.5 $-$3 $-$6 18 15−20 8 17.5 $-$2 $-$16 32 20−25 20 22.5 $-$1 $-$20 20 25−30 35 27.5 0 0 0 30−35 20 32.5 1 20 20 35−40 15 37.5 2 30 60 $\sum {f}_{i}=N=100$ $\sum {f}_{i}{u}_{i}=8$ $\sum {f}_{i}{{u}_{i}}^{2}=150$

Here,

$\overline{X}=a+h\left(\frac{\sum {f}_{i}{u}_{i}}{N}\right)=27.5+5\left(\frac{8}{100}\right)=27.9$
${\sigma }^{2}={h}^{2}\left[\frac{\sum {f}_{i}{{u}_{i}}^{2}}{N}-{\left(\frac{\sum {f}_{i}{u}_{i}}{N}\right)}^{2}\right]=25\left[\frac{150}{100}-\frac{64}{10000}\right]=37.34$

$\sigma =\sqrt{37.34}=6.11$

We know that
$CV=\frac{\sigma }{\overline{X}}×100\phantom{\rule{0ex}{0ex}}=\frac{6.11}{27.9}×100=21.9$

#### Question 10:

From the prices of shares X and Y given below: find out which is more stable in value:

 X: 35 54 52 53 56 58 52 50 51 49 Y: 108 107 105 105 106 107 104 103 104 101

Let Ax = 51

 ${x}_{i}$ ${d}_{i}={x}_{i}-51$ ${{d}_{i}}^{2}$ 35 $-$16 256 54 3 9 52 1 1 53 2 4 56 5 25 58 7 49 52 1 1 50 $-$1 1 51 0 0 49 $-$2 4 $\sum {d}_{i}=0$ $\sum {{d}_{i}}^{2}=350$

Here, we have

$\sigma =\sqrt{35}=5.91$

Let Ay =105
 ${x}_{i}$ ${d}_{i}={x}_{i}-105$ ${{d}_{i}}^{2}$ 108 3 9 107 2 4 105 0 0 105 0 0 106 1 1 107 2 4 104 $-$1 1 103 $-$2 4 104 $-$1 1 101 $-$4 16 $\sum {d}_{i}=0$ $\sum {{d}_{i}}^{2}=40$

$\sigma =\sqrt{4}=2$

Since CV of prices of share Y is lesser than that of X, prices of shares Y are more stable.

#### Question 11:

Life of bulbs produced by two factories A and B are given below:

 Length of life (in hours): 550–650 650–750 750–850 850–950 950–1050 Factory A: (Number of bulbs) 10 22 52 20 16 Factory B: (Number of bulbs) 8 60 24 16 12

The bulbs of which factory are more consistent from the point of view of length of life?                             [NCERT EXEMPLAR]

For factory A

Let the assumed mean A = 800 and h = 100.

 Length of Life (in hours) Mid-Values$\left({x}_{i}\right)$ ${u}_{i}=\frac{{x}_{i}-800}{100}$ ${u}_{i}^{2}$ Number of bulbs$\left({f}_{i}\right)$ ${f}_{i}{u}_{i}$ ${f}_{i}{u}_{i}^{2}$ 550–650 600 −2 4 10 −20 40 650–750 700 −1 1 22 −22 22 750–850 800 0 0 52 0 0 850–950 900 1 1 20 20 20 950–1050 1000 2 4 16 32 64 $\sum _{}{f}_{i}=120$ $\sum _{}{f}_{i}{u}_{i}=10$ $\sum _{}{f}_{i}{u}_{i}^{2}=146$

Mean, ${\overline{)X}}_{A}=A+h\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)=800+100×\frac{10}{120}=808.33$

Standard deviation, ${\sigma }_{A}=h\sqrt{\left(\frac{\sum _{}{f}_{i}{u}_{i}^{2}}{\sum _{}{f}_{i}}\right)-{\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)}^{2}}=100\sqrt{\left(\frac{146}{120}\right)-{\left(\frac{10}{120}\right)}^{2}}=100×1.0998=109.98$

∴ Coefficient of variation = $\frac{{\sigma }_{A}}{{\overline{)X}}_{A}}×100=\frac{109.98}{808.33}×100=13.61$

For factory B

Let the assumed mean A = 800 and h = 100.

 Length of Life (in hours) Mid-Values$\left({x}_{i}\right)$ ${u}_{i}=\frac{{x}_{i}-800}{100}$ ${u}_{i}^{2}$ Number of bulbs$\left({f}_{i}\right)$ ${f}_{i}{u}_{i}$ ${f}_{i}{u}_{i}^{2}$ 550–650 600 −2 4 8 −16 32 650–750 700 −1 1 60 −60 60 750–850 800 0 0 24 0 0 850–950 900 1 1 16 16 16 950–1050 1000 2 4 12 24 48 $\sum _{}{f}_{i}=120$ $\sum _{}{f}_{i}{u}_{i}=-36$ $\sum _{}{f}_{i}{u}_{i}^{2}=156$

Mean, ${\overline{)X}}_{B}=A+h\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)=800+100×\frac{\left(-36\right)}{120}=800-30=770$

Standard deviation, ${\sigma }_{B}=h\sqrt{\left(\frac{\sum _{}{f}_{i}{u}_{i}^{2}}{\sum _{}{f}_{i}}\right)-{\left(\frac{\sum _{}{f}_{i}{u}_{i}}{\sum _{}{f}_{i}}\right)}^{2}}=100\sqrt{\left(\frac{156}{120}\right)-{\left(\frac{-36}{120}\right)}^{2}}=100×1.1=110$

∴ Coefficient of variation = $\frac{{\sigma }_{B}}{{\overline{)X}}_{B}}×100=\frac{110}{770}×100=14.29$

Since the coefficient of variation of factory B is greater than the coefficient of variation of factory A, therefore, factory B has more variability than factory A. This means bulbs of factory A are more consistent from the point of view of length of life.

#### Question 12:

Following are the marks obtained,out of 100 by two students Ravi and Hashina in 10 tests:

 Ravi: 25 50 45 30 70 42 36 48 35 60 Hashina: 10 70 50 20 95 55 42 60 48 80

Who is more intelligent and who is more consistent?                                                               [NCERT EXEMPLAR]

For Ravi

 Marks $\left({x}_{i}\right)$ ${d}_{i}={x}_{i}-45$ ${d}_{i}^{2}$ 25 −20 400 50 5 25 45 0 0 30 −15 225 70 25 625 42 −3 9 36 −9 81 48 3 9 35 −10 100 60 15 225 $\sum _{}{d}_{i}=-9$ $\sum _{}{d}_{i}^{2}=1699$

Mean, ${\overline{)X}}_{R}=A+\frac{\sum _{}{d}_{i}}{10}=45+\frac{\left(-9\right)}{10}=44.1$

Standard deviation, ${\sigma }_{R}=\sqrt{\frac{\sum _{}{d}_{i}^{2}}{10}-{\left(\frac{\sum _{}{d}_{i}}{10}\right)}^{2}}=\sqrt{\frac{1699}{10}-{\left(\frac{-9}{10}\right)}^{2}}=\sqrt{169.09}=13.003$

Coefficicent of variation = $\frac{{\sigma }_{R}}{{\overline{)X}}_{R}}×100=\frac{13.003}{44.1}×100=29.49$

For Hashina

 Marks $\left({x}_{i}\right)$ ${d}_{i}={x}_{i}-55$ ${d}_{i}^{2}$ 10 −45 2025 70 15 625 50 −5 25 20 −35 1225 95 40 1600 55 0 0 42 −13 169 60 5 25 48 −7 49 80 25 625 $\sum _{}{d}_{i}=-20$ $\sum _{}{d}_{i}^{2}=6368$

Mean, ${\overline{)X}}_{H}=A+\frac{\sum _{}{d}_{i}}{10}=55+\frac{\left(-20\right)}{10}=53$

Standard deviation, ${\sigma }_{H}=\sqrt{\frac{\sum _{}{d}_{i}^{2}}{10}-{\left(\frac{\sum _{}{d}_{i}}{10}\right)}^{2}}=\sqrt{\frac{6368}{10}-{\left(\frac{-20}{10}\right)}^{2}}=\sqrt{632.8}=25.16$

Coefficicent of variation = $\frac{{\sigma }_{H}}{{\overline{)X}}_{H}}×100=\frac{25.16}{53}×100=47.47$

Since the coefficient of variation in mark obtained by Hashima is greater than the coefficient of variation in mark obtained by Ravi, so Hashina is more consistent and intelligent.

#### Question 1:

Write the variance of first n natural numbers.

​Sum of first n natural numbers $=\frac{n\left(n+1\right)}{2}$

Mean,
$=\frac{\frac{n\left(n+1\right)}{2}}{n}=\frac{n+1}{2}$

#### Question 2:

If the sum of the squares of deviations for 10 observations taken from their mean is 2.5, then write the value of standard deviation.

The sum of the squares of deviations for 10 observations, taken from their mean, is 2.5.
Square of each deviation = $\frac{2.5}{10}=0.25$
Standard deviation = $\sqrt{0.25}=0.5$

#### Question 3:

If x1, x2, ..., xn are n values of a variable X and y1, y2, ..., yn are n values of variable Y such that yi = axi + b; i = 1, 2, ..., n, then write Var(Y) in terms of Var(X).

$Var\left(X\right)=\frac{\sum {\left({x}_{i}-\overline{X}\right)}^{2}}{n}\phantom{\rule{0ex}{0ex}}Var\left(Y\right)=\frac{\sum {\left({y}_{i}-\overline{Y}\right)}^{2}}{n}$

We have:
${y}_{i}=a{x}_{i}+b\phantom{\rule{0ex}{0ex}}\overline{y}=\frac{\sum {y}_{i}}{n}=\frac{a\sum {x}_{i}+nb}{n}=a\overline{X}+b\phantom{\rule{0ex}{0ex}}$

#### Question 4:

If X and Y are two variates connected by the relation $Y=\frac{aX+b}{c}$ and Var (X) = σ2, then write the expression for the standard deviation of Y.

​​

We know:

#### Question 5:

In a series of 20 observations, 10 observations are each equal to k and each of the remaining half is equal to − k. If the standard deviation of the observations is 2, then write the value of k.

#### Question 6:

If each observation of a raw data whose standard deviation is σ is multiplied by a, then write the S.D. of the new set of observations.

Standard deviation, $\sigma =\sqrt{\frac{\sum _{i}{\left({x}_{i}-\overline{)x}\right)}^{2}}{n}}$

Here, $\overline{)x}$ represents the arithmetic mean.

Multiplying each ${x}_{i}$ by $a$:

#### Question 7:

If a variable X takes values 0, 1, 2,..., n with frequencies nC0, nC1, nC2 , ... , nCn, then write variance X.

#### Question 1:

For a frequency distribution mean deviation from mean is computed by
(a) M.D. =

(b) M.D. = $\frac{\mathrm{\Sigma }d}{\mathrm{\Sigma }f}$

(c) M.D. = $\frac{\mathrm{\Sigma }fd}{\mathrm{\Sigma }f}$

(d) M.D. =

#### Question 2:

For a frequency distribution standard deviation is computed by applying the formula
(a) $\mathrm{\sigma }=\sqrt{\frac{\mathrm{\Sigma }f{d}^{2}}{\mathrm{\Sigma }f}-{\left(\frac{\mathrm{\Sigma }fd}{\mathrm{\Sigma }f}\right)}^{2}}$

(b) $\mathrm{\sigma }=\sqrt{{\left(\frac{\mathrm{\Sigma }fd}{\mathrm{\Sigma }f}\right)}^{2}-\frac{\mathrm{\Sigma }f{d}^{2}}{\mathrm{\Sigma }f}}$

(c) $\mathrm{\sigma }=\sqrt{\frac{\mathrm{\Sigma }f{d}^{2}}{\mathrm{\Sigma }f}-\frac{\mathrm{\Sigma }fd}{\mathrm{\Sigma }f}}$

(d) $\sqrt{{\left(\frac{\mathrm{\Sigma }fd}{\mathrm{\Sigma }f}\right)}^{2}-\frac{\mathrm{\Sigma }f{d}^{2}}{\mathrm{\Sigma }f}}$

#### Question 3:

If v is the variance and σ is the standard deviation, then
(a) $v=\frac{1}{{\mathrm{\sigma }}^{2}}$

(b) $v=\frac{1}{\mathrm{\sigma }}$

(c) v = σ2

(d) v2 = σ

The variance is the square of the standard deviation.

#### Question 4:

The mean deviation from the median is
(a) equal to that measured from another value
(b) maximum if all observations are positive
(c) greater than that measured from any other value.
(d) less than that measured from any other value.

(d) less than that measured from any other value.

In a frequency distribution, the sum of absolute values of deviations from the mean and mode is always more than the sum of the deviations from the median.

#### Question 5:

If n = 10, $\overline{)X}=12$  and $\Sigma {x}_{i}^{2}\mathit{=}1530$, then the coefficient of variation is
(a) 36%
(b) 41%
(c) 25%
(d) none of these

(c) 25%

Standard deviation is expressed in the following manner:

#### Question 6:

The standard deviation of the data:

 x: 1 a a2 .... an f: nC0 nC1 nC2 .... nCn
is
(a) ${\left(\frac{1+{a}^{2}}{2}\right)}^{n}-{\left(\frac{1+a}{2}\right)}^{n}$

(b) ${\left(\frac{1+{a}^{2}}{2}\right)}^{2n}-{\left(\frac{1+a}{2}\right)}^{n}$

(c) ${\left(\frac{1+a}{2}\right)}^{2n}-{\left(\frac{1+{a}^{2}}{2}\right)}^{n}$

(d) none of these

(d)   None of these

xi fi fixi ${{x}_{i}}^{2}$ ${f}_{i}{{x}_{i}}^{2}$
1 ${}^{n}C_{0}$ ${}^{n}C_{0}$ 1 1
a ${}^{n}C_{1}$ ${}^{n}C_{1}$ a2 a2  ${}^{n}C_{1}$
a2 ${}^{n}C_{2}$ a2 ${}^{n}C_{2}$ a4   a4  ${}^{n}C_{2}$
a3 ${}^{n}C_{3}$  a3 ${}^{n}C_{3}$ a6  a6   ${}^{n}C_{3}$
:
:
:
:
:
:
:
:
:
:
:
:
:

:
:
:
:
an ${}^{n}C_{n}$ an  ${}^{n}C_{n}$ a2n a2n ${}^{n}C_{n}$

$\sum _{i=1}^{n}{f}_{i}={2}^{n}$

$\sum _{i=1}^{n}{f}_{i}{x}_{i}={\left(1+a\right)}^{n}$

#### Question 7:

The mean deviation of the series a, a + d, a + 2d, ..., a + 2n from its mean is
(a)

(b) $\frac{nd}{2n+1}$

(c)

(d)

${x}_{i}$ $\left|{x}_{i}-\overline{)X}\right|=\left|{x}_{i}-\left(a+nd\right)\right|$
a nd
a + d (n-1)d
a + 2d (n-2)d
a + 3d (n-3)d
: :
: :
a + (n - 1)d d
a + nd 0
a + (n+1)d d
: :
: :
a + 2nd nd
$\sum _{}{x}_{i}=\left(2n+1\right)\left(a+nd\right)$

#### Question 8:

A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is
(a) 8.6
(b) 6.4
(c) 10.6
(d) 7.6

Disclaimer: No option is matching the answer.

 xi di =$\left|{x}_{i}-49.4\right|$ 34 15.4 38 11.4 42 7.4 44 5.4 46 3.4 48 1.4 54 4.6 55 5.6 63 13.6 70 20.6 $\sum _{i=}^{n}{d}_{i}=88.8$

#### Question 9:

The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
(a) 25
(b) 5
(c) 1.2
(d) 0

(c) 1.2

Taking the absolute value of deviation of each term from the mean, we get:

#### Question 10:

The sum of the squares deviations for 10 observations taken from their mean 50 is 250. The coefficient of variation is
(a) 10 %
(b) 40 %
(c) 50 %
(d) none of these

(a) 10%

Using  $\mathrm{CV}=\frac{\sigma }{\overline{)X}}×100$

#### Question 11:

Let x1, x2, ..., xn be values taken by a variable X and y1, y2, ..., yn be the values taken by a variable Y such that yi = axi + b, i = 1, 2,..., n. Then,
(a) Var (Y) = a2 Var (X)
(b) Var (X) = a2 Var (Y)
(c) Var (X) = Var (X) + b
(d) none of these

#### Question 12:

If the standard deviation of a variable X is σ, then the standard deviation of variable is
(a) a σ

(b) $\frac{a}{c}\mathrm{\sigma }$

(c)

(d) $\frac{a\mathrm{\sigma }+b}{c}$

(c) $\left|\frac{a}{c}\right|\sigma$

#### Question 13:

If the S.D. of a set of observations is 8 and if each observation is divided by −2, the S.D. of the new set of observations will be
(a) −4
(b) −8
(c) 8
(d) 4

(d) 4

If a set of observations, with SD $\sigma$, are multiplied with a non-zero real number a, then SD of the new observations will be
Dividing the set of observations by − 2 is same as multiplying the observations by $\frac{1}{-2}$.

#### Question 14:

If two variates X and Y are connected by the relation , where a, b, c are constants such that ac < 0, then
(a) ${\mathrm{\sigma }}_{Y}=\frac{a}{c}{\mathrm{\sigma }}_{X}$

(b) ${\mathrm{\sigma }}_{Y}=-\frac{a}{c}{\mathrm{\sigma }}_{X}$

(c) ${\mathrm{\sigma }}_{Y}=\frac{a}{c}{\mathrm{\sigma }}_{X}+b$

(d) none of these

#### Question 15:

If for a sample of size 60, we have the following information $\sum _{}{x}_{i}^{2}=18000$ and $\sum _{}{x}_{i}=960$, then the variance is

(a) 6.63                                  (b) 16                                   (c) 22                                  (d) 44

Given: $\sum _{}{x}_{i}^{2}=18000$, $\sum _{}{x}_{i}=960$ and n = 60

∴ Variance

$=\frac{\sum _{}{x}_{i}^{2}}{n}-{\left(\frac{\sum _{}{x}_{i}}{n}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{18000}{60}-{\left(\frac{960}{60}\right)}^{2}\phantom{\rule{0ex}{0ex}}=300-256\phantom{\rule{0ex}{0ex}}=44$

Hence, the correct answer is option (d).

#### Question 16:

Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is

(a) s                                  (b) ks                                  (c) s + k                                  (d) $\frac{s}{k}$

The given observations are a, b, c, d, e.

Mean = m = $\frac{a+b+c+d+e}{5}$

$⇒\sum _{}{x}_{i}=a+b+c+d+e=5m$            .....(1)

Standard deviation, s = $\sqrt{\frac{\sum _{}{x}_{i}^{2}}{5}-{m}^{2}}$

Now, consider the observations a + k, b + k, c + k, d + k, e + k.

New mean $=\frac{\left(a+k\right)+\left(b+k\right)+\left(c+k\right)+\left(d+k\right)+\left(e+k\right)}{5}$

$=\frac{a+b+c+d+e+5k}{5}\phantom{\rule{0ex}{0ex}}=\frac{5m+5k}{5}\phantom{\rule{0ex}{0ex}}=m+k$

∴ New standard deviation

$=\sqrt{\frac{\sum _{}{\left({x}_{i}+k\right)}^{2}}{5}-{\left(m+k\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{\sum _{}\left({x}_{i}^{2}+{k}^{2}+2{x}_{i}k\right)}{5}-\left({m}^{2}+{k}^{2}+2mk\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{\sum _{}{x}_{i}^{2}}{5}+\frac{\sum _{}{k}^{2}}{5}+\frac{\sum _{}2{x}_{i}k}{5}-\left({m}^{2}+{k}^{2}+2mk\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{\sum _{}{x}_{i}^{2}}{5}-{m}^{2}+\frac{5{k}^{2}}{5}-{k}^{2}+\frac{2k\sum _{}{x}_{i}}{5}-2mk}$

Hence, the correct answer is option (a).

#### Question 17:

The standard deviation of first 10 natural numbers is

(a) 5.5                                  (b) 3.87                                   (c) 2.97                                  (d) 2.87

We know that the standard deviation of first n natural number is $\sqrt{\frac{{n}^{2}-1}{12}}$.

∴ Standard deviation of first 10 natural numbers

$=\sqrt{\frac{{10}^{2}-1}{12}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{99}{12}}\phantom{\rule{0ex}{0ex}}=\sqrt{8.25}\phantom{\rule{0ex}{0ex}}=2.87$

Hence, the correct answer is option (d).

#### Question 18:

Consider the first 10 positive integers. If we multiply each number by −1 and then add 1 to each number, the variance of the numbers so obtained is

(a) 8.25                                  (b) 6.5                                   (c) 3.87                                  (d) 2.87

The first 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Multiplying each number by −1, we get

−1, −2, −3, −4, −5, −6, −7, −8, −9, −10

Adding 1 to each of these numbers, we get

0, −1, −2, −3, −4, −5, −6, −7, −8, −9

Now,

$\sum _{}{x}_{i}=0+\left(-1\right)+\left(-2\right)+\left(-3\right)+\left(-4\right)+\left(-5\right)+\left(-6\right)+\left(-7\right)+\left(-8\right)+\left(-9\right)=-45$

$\sum _{}{x}_{i}^{2}=0+1+4+9+16+25+36+49+64+81=285$

∴ Variance of the obtained numbers

$=\frac{\sum _{}{x}_{i}^{2}}{10}-{\left(\frac{\sum _{}{x}_{i}}{10}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{285}{10}-{\left(\frac{-45}{10}\right)}^{2}\phantom{\rule{0ex}{0ex}}=28.5-20.25\phantom{\rule{0ex}{0ex}}=8.25$

Hence, the correct answer is option (a).

#### Question 19:

Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is

(a) 6.5                                  (b) 2.87                                   (c) 3.87                                 (d) 8.25

The given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

If 1 is added to each number, then the new numbers obtained are

2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Now,

$\sum _{}{x}_{i}=2+3+4+5+6+7+8+9+10+11=65$

$\sum _{}{x}_{i}^{2}=4+9+16+25+36+49+64+81+100+121=505$

∴ Variance of the numbers so obtained

$=\frac{\sum _{}{x}_{i}^{2}}{10}-{\left(\frac{\sum _{}{x}_{i}}{10}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{505}{10}-{\left(\frac{65}{10}\right)}^{2}\phantom{\rule{0ex}{0ex}}=50.5-42.25\phantom{\rule{0ex}{0ex}}=8.25$

Hence, the correct answer is option (d).

#### Question 20:

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is

(a) 50,000                                  (b) 250,000                                   (c) 252500                                  (d) 255000

Let $\overline{)x}$ and $\sigma$ be the mean and standard deviation of 100 observations, respectively.

and n = 100

Mean, $\overline{)x}=50$

Now,

Standard deviation, $\sigma =5$

Thus, the sum of all squares of all the observations is 252500.

Hence, the correct answer is option (c).

#### Question 21:

Let x1, x2, ..., xn be n observations. Let ${y}_{i}=a{x}_{i}+b$ for i = 1, 2, 3, ..., n, where a and b are constants. If the mean of ${x}_{i}\text{'}\mathrm{s}$ is 48 and their standard deviation is 12, the mean of ${y}_{i}\text{'}\mathrm{s}$ is 55 and standard deviation of ${y}_{i}\text{'}\mathrm{s}$ is 15, the values of a and b are

(a) a = 1.25, b = −5                 (b) a = −1.25, b = 5                 (c) a = 2.5, b = −5                 (d) a = 2.5, b = 5

It is given that ${y}_{i}=a{x}_{i}+b$ for i = 1, 2, 3, ..., n, where a and b are constants.

$\overline{){x}_{i}}$ = 48 and ${\sigma }_{{x}_{i}}=12$

$\overline{){y}_{i}}=55$ and ${\sigma }_{{y}_{i}}=15$

Now,

Standard deviation of yi = Standard deviation of $a{x}_{i}+b$

$⇒{\sigma }_{{y}_{i}}=a×{\sigma }_{{x}_{i}}\phantom{\rule{0ex}{0ex}}⇒15=12a\phantom{\rule{0ex}{0ex}}⇒a=\frac{15}{12}=1.25$

Putting a = 1.25 in (1), we get

$b=55-48×1.25=55-60=-5$

Thus, the values of a and b are 1.25 and −5, respectively.

Hence, the correct answer is option (a).

#### Question 22:

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is

(a) 2                                  (b) 2.57                                   (c) 3                                 (d) 3.57

The given observations are 3, 10, 10, 4, 7, 10, 5.

∴ Mean, $\overline{)x}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$

Now,

Mean deviation from mean, MD

$=\frac{\sum _{}\left|{x}_{i}-7\right|}{7}$

$=\frac{\left|3-7\right|+\left|10-7\right|+\left|10-7\right|+\left|4-7\right|+\left|7-7\right|+\left|10-7\right|+\left|5-7\right|}{7}\phantom{\rule{0ex}{0ex}}=\frac{4+3+3+3+0+3+2}{7}\phantom{\rule{0ex}{0ex}}=\frac{18}{7}\phantom{\rule{0ex}{0ex}}=2.57$

Hence, the correct answer is (b).

#### Question 23:

The mean deviation for n observations from their mean $\overline{)X}$ is given by

(a) $\sum _{i=1}^{n}\left({x}_{i}-\overline{)X}\right)$                               (b) $\frac{1}{n}\sum _{i=1}^{n}\left({x}_{i}-\overline{)X}\right)$                                   (c) $\sum _{i=1}^{n}{\left({x}_{i}-\overline{)X}\right)}^{2}$                                  (d) $\frac{1}{n}\sum _{i=1}^{n}{\left({x}_{i}-\overline{)X}\right)}^{2}$

The mean deviation for n observations from their mean $\overline{)X}$ is $\frac{1}{n}\sum _{i=1}^{n}\left|{x}_{i}-\overline{)X}\right|$.

Disclaimer: There is some printing error in option (b) given in the question. The answer would be option (b) if it given as $\frac{1}{n}\sum _{i=1}^{n}\left|{x}_{i}-\overline{)X}\right|$.

#### Question 24:

Let be n observations and $\overline{)X}$ be their arithmetic mean. The standard deviation is given by

(a) $\sum _{i=1}^{n}{\left({x}_{i}-\overline{)X}\right)}^{2}$                 (b) $\frac{1}{n}\sum _{i=1}^{n}{\left({x}_{i}-\overline{)X}\right)}^{2}$                 (c) $\sqrt{\frac{1}{n}\sum _{i=1}^{n}{\left({x}_{i}-\overline{)X}\right)}^{2}}$                 (d) $\sqrt{\frac{1}{n}\sum _{i=1}^{n}{x}_{i}^{2}-{\overline{)X}}^{2}}$

It is given that are n observations and $\overline{)X}$ is their arithmetic mean.

The standard deviation of given observations is $\sqrt{\frac{1}{n}\sum _{i=1}^{n}{\left({x}_{i}-\overline{)X}\right)}^{2}}$.

Also,

$\sqrt{\frac{1}{n}\sum _{i=1}^{n}{\left({x}_{i}-\overline{)X}\right)}^{2}}$ = $\sqrt{\frac{1}{n}\sum _{i=1}^{n}{x}_{i}^{2}-{\overline{)X}}^{2}}$

Hence, the correct answers are options (c) and (d).

Disclaimer: For option (c) to be the only correct answer, option (d) should be different from the given value.

#### Question 25:

The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is

(a) 6                             (b) $\sqrt{6}$                             (c) $\frac{52}{7}$                            (d) $\sqrt{\frac{52}{7}}$

The given observations are 6, 5, 9, 13, 12, 8, 10.

Now,

$\sum _{}{x}_{i}=6+5+9+13+12+8+10=63$

$\sum _{}{x}_{i}^{2}=36+25+81+169+144+64+100=619$

∴ Standard deviation of the observations, $\sigma$

$=\sqrt{\frac{1}{N}\sum _{}x{}_{i}^{2}-{\left(\frac{1}{N}\sum _{}{x}_{i}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{7}×619-{\left(\frac{1}{7}×63\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{619}{7}-81}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{619-567}{7}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{52}{7}}$

Hence, the correct answer is option (d).

#### Question 1:

Calculate the mean deviation about the median of the following observations:
(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000
(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

Formula used for mean deviation:

M = Median

i) Arranging the data in ascending order:
2354, 2780, 3011, 3020, 3541, 4150, 5000

Here, median $\left(M\right)=3020$ and n = 7.

${x}_{i}$ $\left|{d}_{i}\right|$ =
3011 9
2780 240
3020 0
2354 666
3541 521
4150 1130
5000 1980
Total 4546

$MD=\frac{1}{n}\sum _{i=1}^{n}\left|{d}_{i}\right|$

ii) Arranging the data in ascending order:
34, 38, 42, 44, 46, 48, 54, 55, 63, 70

Here, n is equal to 10.
Median is the arithmetic mean of the fifth and the sixth observation.

 xi 38 9 70 23 48 1 34 13 42 5 55 8 63 16 46 1 54 7 44 3 Total 86

$MD=\frac{1}{10}×86=8.6$

iii) Arranging the data in ascending order:
30, 34, 38, 40, 42, 44, 50, 51, 60, 66

Here,
$n=10$
Also, median is the AM of the fifth and the sixth observation.

 xi 34 9 66 23 30 13 38 5 44 1 50 7 40 3 60 17 42 1 51 8 Total 87

$MD=\frac{1}{10}×87=8.7$

iv) Arranging the data in ascending order.
22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Also, median is the AM of the fifth and the sixth observation.

 xi 22 6.5 24 4.5 30 1.5 27 1.5 29 0.5 31 2.5 25 3.5 28 0.5 41 12.5 41 13.5 Total 47

$MD=\frac{1}{10}×47=4.7$

v) Arranging the data in ascending order:
34, 38, 42, 44, 47, 48, 53, 55, 63, 70

Here, $n=10$.
Also, median is the AM of the fifth and the sixth observation.

 xi 38 9.5 70 22.5 48 0.5 34 13.5 63 15.5 42 5.5 55 7.5 44 3.5 53 5.5 47 0.5 Total 84

$MD=\frac{1}{10}×84=8.4$

#### Question 2:

Calculate the mean deviation from the mean for the following data:
(i) 4, 7, 8, 9, 10, 12, 13, 17
(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Formula used for finding the mean deviation about the mean is given below:

i)
Let $\overline{)x}$ be the mean of the given data.

 ${x}_{i}$ 4 6 7 3 8 2 9 1 10 0 12 2 13 3 17 7 Total 24

$MD=\frac{1}{8}×24=3$

ii)
Let $\overline{)x}$ be the mean of the given data.

$\overline{)x}=\frac{13+17+16+14+11+13+10+16+11+18+12+17}{12}=14$

 ${x}_{i}$ 13 1 17 3 16 2 14 0 11 3 13 1 10 4 16 2 11 3 18 4 12 2 17 3 Total 28

iii)
Let $\overline{)x}$ be the mean of the given data.

$\overline{)x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}=50$

 ${x}_{i}$ 38 12 70 20 48 2 40 10 42 8 55 5 63 13 46 4 54 4 44 6 Total 84

iv)
Let $\overline{)x}$ be the mean of the given data.

$\overline{)x}=\frac{36+746+42+60+45+53+46+51+59}{10}=50$

 ${x}_{i}$ 36 14 72 22 46 4 42 8 60 10 45 5 53 3 46 4 51 1 49 1 Total 72

#### Question 3:

Calculate the mean deviation of the following income groups of five and seven members from their medians:

 I Income in Rs. II Income in Rs. 4000 4200 4400 4600 4800 300 4000 4200 4400 4600 4800 5800

Calculate the mean deviation for the first data set.
The data is already arranged in ascending order.
For this data set, n is equal to 5.
Also, median, $M=4400$

 ${x}_{i}$ 4000 400 4200 200 4400 0 4600 200 4800 400 Total 1200

$MD=\frac{1}{5}×1200=240$
Therefore, for the income of families in the first group, the mean deviation from the median is Rs 240.

Now, consider the second data set. This is also arranged in ascending order.
Here, $n=7$.
Also, median, $M=4400$

 xi 300 4100 4000 400 4200 200 4400 0 4600 200 4800 400 5800 1400 Total 6700

$MD=\frac{1}{7}×6700=957.14$
Therefore, for the income of families in the second group, the mean deviation from the median is Rs 957.14.

#### Question 4:

The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
(i) Find mean deviation from median
(ii) Find mean deviation from the mean also.

i) Formula for the mean deviation from the median is as follows:

Arranging the data in ascending order for finding the median:
15.2, 27.9, 30.2, 32.5, 40, 52.3, 52.8, 55.2, 72.9, 79

Here, $n=10$.
Therefore, median is the average of the fifth and the sixth observations.

$M=\frac{40+52.3}{2}=46.15$

 ${x}_{i}$ $\left|{d}_{i}\right|=\left|{x}_{i}-46.15\right|$ 40 6.15 52.3 6.15 55.2 9.05 72.9 26.75 52.8 6.65 79 32.85 32.5 13.65 15.2 30.95 27.9 18.25 32 14.15 Total 164.6

$MD=\frac{1}{10}×164.6=16.46$

Mean deviation from median in 16.4 cm.

ii)
Let $\overline{x}$ be the mean of the given data set.

$\overline{)x}=\frac{40+52.3+55.2+72.9+52.8+79+32.5+15.2+27.9+30.2}{10}=45.98$

 ${x}_{i}$ $\left|{d}_{i}\right|=\left|{x}_{i}-45.98\right|$ 40 5.98 52.3 6.32 55.2 9.22 72.9 26.92 52.8 6.82 79 33.02 32.5 13.48 15.2 30.78 27.9 18.08 32 13.98 Total 164.6

Mean deviation from the mean is 16.4 cm.

#### Question 5:

In question 1 (iii), (iv), (v) find the number of observations lying between $\overline{)X}$ − M.D. and $\overline{)X}$ + M.D, where M.D. is the mean deviation from the mean.

​iii)
Let $\overline{x}$ be the mean of the data set.

$\overline{)x}=\frac{34+66+30+38+44+50+40+60+42+51}{10}=45.5$

 ${x}_{i}$ 34 11.5 66 20.5 30 15.5 38 7.5 44 1.5 50 4.5 40 5.5 60 14.5 42 3.5 51 5.5 Total 90

Hence, there are 6 observations between 36.5 and 54.5.

iv)
Let $\overline{x}$ be the mean of the data set.

$\overline{)x}=\frac{22+24+30+27+29+31+25+28+41+42}{10}=29.9$

 ${x}_{i}$ 22 7.9 24 5.9 30 0.1 27 2.9 29 0.9 31 1.1 25 4.9 28 1.9 41 11.9 42 12.1 Total 48.8

There are 5 observations between 25.02 and 34.78.

v)
Let $\overline{x}$ be the mean of the data set.

$\overline{)x}=\frac{38+70+48+34+63+42+55+44+53+47}{10}=49.4$

 ${x}_{i}$ 38 11.4 70 20.6 48 1.4 34 15.4 63 13.6 42 7.4 55 5.6 44 5.4 53 3.6 47 2.4 Total 86.8

There are 6 observations between 40.72 and 58.08.

View NCERT Solutions for all chapters of Class 13