NCERT Solutions for Class 11 Science Math Chapter 15 Statistics are provided here with simple step-by-step explanations. These solutions for Statistics are extremely popular among class 11 Science students for Math Statistics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 11 Science Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class 11 Science Math are prepared by experts and are 100% accurate.

#### Question 1:

Find the mean deviation about the mean for the data

4, 7, 8, 9, 10, 12, 13, 17

The given data is

4, 7, 8, 9, 10, 12, 13, 17

Mean of the data,

The deviations of the respective observations from the mean are

–6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e., are

6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is

#### Question 2:

Find the mean deviation about the mean for the data

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data,

The deviations of the respective observations from the mean are

–12, 20, –2, –10, –8, 5, 13, –4, 4, –6

The absolute values of the deviations, i.e. , are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

#### Question 3:

Find the mean deviation about the median for the data.

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

The deviations of the respective observations from the median, i.e.are

–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations,, are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is

#### Question 4:

Find the mean deviation about the median for the data

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

The given data is

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

The deviations of the respective observations from the median, i.e.are

–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations,, are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Thus, the required mean deviation about the median is

#### Question 5:

Find the mean deviation about the mean for the data.

 xi 5 10 15 20 25 fi 7 4 6 3 5

 xi fi fi xi 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 25 350 158

#### Question 6:

Find the mean deviation about the mean for the data

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

 xi fi fi xi 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 80 4000 1280

#### Question 7:

Find the mean deviation about the median for the data.

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

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

 xi fi c.f. 5 8 8 7 6 14 9 2 16 10 2 18 12 2 20 15 6 26

Here, N = 26, which is even.

Median is the mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

The absolute values of the deviations from median, i.e.are

 |xi – M| 2 0 2 3 5 8 fi 8 6 2 2 2 6 fi |xi – M| 16 0 4 6 10 48

and

#### Question 8:

Find the mean deviation about the median for the data

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

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

 xi fi c.f. 15 3 3 21 5 8 27 6 14 30 7 21 35 8 29

Here, N = 29, which is odd.

observation = 15th observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.

∴ Median = 30

The absolute values of the deviations from median, i.e.are

 |xi – M| 15 9 3 0 5 fi 3 5 6 7 8 fi |xi – M| 45 45 18 0 40

#### Question 9:

Find the mean deviation about the mean for the data.

 Income per day Number of persons 0-100 4 100-200 8 200-300 9 300-400 10 400-500 7 500-600 5 600-700 4 700-800 3

The following table is formed.

 Income per day Number of persons fi Mid-point xi fi xi 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 50 17900 7896

Here,

#### Question 10:

Find the mean deviation about the mean for the data

 Height in cms Number of boys 95-105 9 105-115 13 115-125 26 125-135 30 135-145 12 145-155 10

The following table is formed.

 Height in cms Number of boys fi Mid-point xi fi xi 95-105 9 100 900 25.3 227.7 105-115 13 110 1430 15.3 198.9 115-125 26 120 3120 5.3 137.8 125-135 30 130 3900 4.7 141 135-145 12 140 1680 14.7 176.4 145-155 10 150 1500 24.7 247

Here,

#### Question 11:

Find the mean deviation about median for the following data:

 Marks Number of girls 0-10 6 10-20 8 20-30 14 30-40 16 40-50 4 50-60 2

The following table is formed.

 Marks Number of girls fi Cumulative frequency (c.f.) Mid-point xi |xi – Med.| fi |xi – Med.| 0-10 6 6 5 22.85 137.1 10-20 8 14 15 12.85 102.8 20-30 14 28 25 2.85 39.9 30-40 16 44 35 7.15 114.4 40-50 4 48 45 17.15 68.6 50-60 2 50 55 27.15 54.3 50 517.1

The class interval containing theor 25th item is 20 – 30.

Therefore, 20 – 30 is the median class.

It is known that,

Here, l = 20, C = 14, f = 14, h = 10, and N = 50

∴ Median =

Thus, mean deviation about the median is given by,

#### Question 12:

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

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

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

The table is formed as follows.

 Age Number fi Cumulative frequency (c.f.) Mid-point xi |xi – Med.| fi |xi – Med.| 15.5-20.5 5 5 18 20 100 20.5-25.5 6 11 23 15 90 25.5-30.5 12 23 28 10 120 30.5-35.5 14 37 33 5 70 35.5-40.5 26 63 38 0 0 40.5-45.5 12 75 43 5 60 45.5-50.5 16 91 48 10 160 50.5-55.5 9 100 53 15 135 100 735

The class interval containing theor 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100

Thus, mean deviation about the median is given by,

#### Question 1:

Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12

6, 7, 10, 12, 13, 4, 8, 12

Mean,

The following table is obtained.

 xi 6 –3 9 7 –2 4 10 –1 1 12 3 9 13 4 16 4 –5 25 8 –1 1 12 3 9 74

#### Question 2:

Find the mean and variance for the first n natural numbers

The mean of first n natural numbers is calculated as follows.

#### Question 3:

Find the mean and variance for the first 10 multiples of 3

The first 10 multiples of 3 are

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Here, number of observations, n = 10

The following table is obtained.

 xi 3 –13.5 182.25 6 –10.5 110.25 9 –7.5 56.25 12 –4.5 20.25 15 –1.5 2.25 18 1.5 2.25 21 4.5 20.25 24 7.5 56.25 27 10.5 110.25 30 13.5 182.25 742.5

#### Question 4:

Find the mean and variance for the data

 xi 6 10 14 18 24 28 30 f i 2 4 7 12 8 4 3

The data is obtained in tabular form as follows.

 xi f i fixi 6 2 12 –13 169 338 10 4 40 –9 81 324 14 7 98 –5 25 175 18 12 216 –1 1 12 24 8 192 5 25 200 28 4 112 9 81 324 30 3 90 11 121 363 40 760 1736

Here, N = 40,

#### Question 5:

Find the mean and variance for the data

 xi 92 93 97 98 102 104 109 f i 3 2 3 2 6 3 3

The data is obtained in tabular form as follows.

 xi f i fixi 92 3 276 –8 64 192 93 2 186 –7 49 98 97 3 291 –3 9 27 98 2 196 –2 4 8 102 6 612 2 4 24 104 3 312 4 16 48 109 3 327 9 81 243 22 2200 640

Here, N = 22,

#### Question 6:

Find the mean and standard deviation using short-cut method.

 xi 60 61 62 63 64 65 66 67 68 fi 2 1 12 29 25 12 10 4 5

The data is obtained in tabular form as follows.

 xi fi yi2 fiyi fiyi2 60 2 –4 16 –8 32 61 1 –3 9 –3 9 62 12 –2 4 –24 48 63 29 –1 1 –29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80 100 220 0 286

Mean,

#### Question 7:

Find the mean and variance for the following frequency distribution.

 Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequencies 2 3 5 10 3 5 2

 Class Frequency fi Mid-point xi yi2 fiyi fiyi2 0-30 2 15 –3 9 –6 18 30-60 3 45 –2 4 –6 12 60-90 5 75 –1 1 –5 5 90-120 10 105 0 0 0 0 120-150 3 135 1 1 3 3 150-180 5 165 2 4 10 20 180-210 2 195 3 9 6 18 30 2 76

Mean,

#### Question 8:

Find the mean and variance for the following frequency distribution.

 Classes 0-10 10-20 20-30 30-40 40-50 Frequencies 5 8 15 16 6

 Class Frequency fi Mid-point xi yi2 fiyi fiyi2 0-10 5 5 –2 4 –10 20 10-20 8 15 –1 1 –8 8 20-30 15 25 0 0 0 0 30-40 16 35 1 1 16 16 40-50 6 45 2 4 12 24 50 10 68

Mean,

#### Question 9:

Find the mean, variance and standard deviation using short-cut method

 Height in cms No. of children 70-75 3 75-80 4 80-85 7 85-90 7 90-95 15 95-100 9 100-105 6 105-110 6 110-115 3

 Class Interval Frequency fi Mid-point xi yi2 fiyi fiyi2 70-75 3 72.5 –4 16 –12 48 75-80 4 77.5 –3 9 –12 36 80-85 7 82.5 –2 4 –14 28 85-90 7 87.5 –1 1 –7 7 90-95 15 92.5 0 0 0 0 95-100 9 97.5 1 1 9 9 100-105 6 102.5 2 4 12 24 105-110 6 107.5 3 9 18 54 110-115 3 112.5 4 16 12 48 60 6 254

Mean,

#### Question 10:

The diameters of circles (in mm) drawn in a design are given below:

 Diameters No. of children 33-36 15 37-40 17 41-44 21 45-48 22 49-52 25

 Class Interval Frequency fi Mid-point xi fi2 fiyi fiyi2 32.5-36.5 15 34.5 –2 4 –30 60 36.5-40.5 17 38.5 –1 1 –17 17 40.5-44.5 21 42.5 0 0 0 0 44.5-48.5 22 46.5 1 1 22 22 48.5-52.5 25 50.5 2 4 50 100 100 25 199

Here, N = 100, h = 4

Let the assumed mean, A, be 42.5.

Mean,

#### Question 1:

From the data given below state which group is more variable, A or B?

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

Firstly, the standard deviation of group A is calculated as follows.

 Marks Group A fi Mid-point  xi ${\mathbit{y}}_{\mathbit{i}}\mathbf{=}\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{45}}{\mathbf{10}}$ yi2 fiyi fiyi2 10-20 9 15 –3 9 –27 81 20-30 17 25 –2 4 –34 68 30-40 32 35 –1 1 –32 32 40-50 33 45 0 0 0 0 50-60 40 55 1 1 40 40 60-70 10 65 2 4 20 40 70-80 9 75 3 9 27 81 Σ 150 –6 342

Here, h = 10, N = 150, A = 45

The standard deviation of group B is calculated as follows.

 Marks Group B fi Mid-point  xi ${\mathbit{y}}_{\mathbit{i}}\mathbf{=}\frac{{\mathbit{x}}_{\mathbit{i}}\mathbf{-}\mathbf{45}}{\mathbf{10}}$ yi2 fiyi fiyi2 10-20 10 15 –3 9 –30 90 20-30 20 25 –2 4 –40 80 30-40 30 35 –1 1 –30 30 40-50 25 45 0 0 0 0 50-60 43 55 1 1 43 43 60-70 15 65 2 4 30 60 70-80 7 75 3 9 21 63 Σ 150 –6 366

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group B has more variability in the marks.

#### Question 2:

From the prices of shares X and Y 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

The prices of the shares X are

35, 54, 52, 53, 56, 58, 52, 50, 51, 49

Here, the number of observations, N = 10

The following table is obtained corresponding to shares X.

 xi 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 4 350

The prices of share Y are

108, 107, 105, 105, 106, 107, 104, 103, 104, 101

The following table is obtained corresponding to shares Y.

 yi 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 40

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.

Thus, the prices of shares Y are more stable than the prices of shares X.

#### Question 3:

An analysis of monthly 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 Mean of monthly wages Rs 5253 Rs 5253 Variance of the distribution of wages 100 121

(i) Which firm A or B pays larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

(i) Monthly wages of firm A = Rs 5253

Number of wage earners in firm A = 586

∴Total amount paid = Rs 5253 × 586

Monthly wages of firm B = Rs 5253

Number of wage earners in firm B = 648

∴Total amount paid = Rs 5253 × 648

Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A = 100

∴ Standard deviation of the distribution of wages in firm

A ((σ1) =

Variance of the distribution of wages in firm = 121

∴ Standard deviation of the distribution of wages in firm

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.

Thus, firm B has greater variability in the individual wages.

#### Question 4:

The following is the record of goals scored by team A in a football session:

 No. of goals scored 0 1 2 3 4 No. of matches 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard

deviation 1.25 goals. Find which team may be considered more consistent?

The mean and the standard deviation of goals scored by team A are calculated as follows.

 No. of goals scored No. of matches fixi xi2 fixi2 0 1 0 0 0 1 9 9 1 9 2 7 14 4 28 3 5 15 9 45 4 3 12 16 48 25 50 130

Thus, the mean of both the teams is same.

The standard deviation of team B is 1.25 goals.

The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.

Thus, team A is more consistent than team B.

#### Question 5:

The sum and sum of squares corresponding to length x (in cm) and weight y

(in gm) of 50 plant products are given below:

Which is more varying, the length or weight?

Here, N = 50

∴ Mean,

Mean,

Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.

#### Question 1:

The mean and variance of eight 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 the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

From (1), we obtain

x2 + y2 + 2xy = 144 …(3)

From (2) and (3), we obtain

2xy = 64 … (4)

Subtracting (4) from (2), we obtain

x2 + y2 – 2xy = 80 – 64 = 16

xy = ± 4 … (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 4, when xy = 4

x = 4 and y = 8, when xy = –4

Thus, the remaining observations are 4 and 8.

#### Question 2:

The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.

Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

From (1), we obtain

x2 + y2 + 2xy = 196 … (3)

From (2) and (3), we obtain

2xy = 196 – 100

⇒ 2xy = 96 … (4)

Subtracting (4) from (2), we obtain

x2 + y2 – 2xy = 100 – 96

⇒ (xy)2 = 4

xy = ± 2 … (5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 6 when xy = 2

x = 6 and y = 8 when xy = – 2

Thus, the remaining observations are 6 and 8.

#### Question 3:

The mean and standard deviation of six 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.

Let the observations be x1, x2, x3, x4, x5, and x6.

It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are yi, then

From (1) and (2), it can be observed that,

Substituting the values of xi and in (2), we obtain

Therefore, variance of new observations =

Hence, the standard deviation of new observations is

#### Question 4:

Given that is the mean and σ2 is the variance of n observations x1, x2xn. Prove that the mean and variance of the observations ax1, ax2, ax3axn are and a2 σ2, respectively (a ≠ 0).

The given n observations are x1, x2xn.

Mean =

Variance = σ2

If each observation is multiplied by a and the new observations are yi, then

Therefore, mean of the observations, ax1, ax2axn, is .

Substituting the values of xiand in (1), we obtain

Thus, the variance of the observations, ax1, ax2axn, is a2 σ2.

#### Question 5:

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) Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

∴ Correct mean

(ii) When 8 is replaced by 12,

Incorrect sum of observations = 200

∴ Correct sum of observations = 200 – 8 + 12 = 204

#### Question 6:

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 deviation 12 15 20

Which of the three subjects shows the highest variability in marks and which shows the lowest?

Standard deviation of Mathematics = 12

Standard deviation of Physics = 15

Standard deviation of Chemistry = 20

The coefficient of variation (C.V.) is given by .

The subject with greater C.V. is more variable than others.

Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

#### Question 7:

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 are omitted.

Number of observations (n) = 100

Incorrect mean

Incorrect standard deviation

∴ Incorrect sum of observations = 2000

⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940

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