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#### Question 1:

Find average for following individual data.

 2 3 5 6 8 10 11 13 17 20

Calculating the average using direct method:

Hence, the average of the above individual data is 9.5

#### Question 2:

Daily income of 10 families is given as follows:

 S. No. 1 2 3 4 5 6 7 8 9 10 Daily Income (in ₹) 100 120 80 85 95 130 200 250 225 275
Calculate average daily income

 Serial No. Daily Income (in Rs) (X) 1 2 3 4 5 6 7 8 9 10 100 120 80 85 95 130 200 250 225 275 N=10 $\mathbf{\sum }_{}\mathbit{X}$= 1560

Calculating the average daily income using direct method:

Hence, the average daily income is Rs 156

#### Question 3:

The following table gives the marks obtained by 10 students of a class:

 S. No. 1 2 3 4 5 6 7 8 9 10 Marks 43 60 37 48 65 48 57 78 31 59
Find out mean marks, using Direct Method as well as Short-Cut Method

 S.N. Marks X deviation from assumed mean (X − A) 1 2 3 4 5 6 7 8 9 10 43 60 37 48 $\overline{)65=A}$ 48 57 78 31 59 −22 −5 −28 −17 0 −17 −8 13 −34 −6 N=10 $\sum _{}\mathbit{X}$=526 Σd=−124

(i) Calculating mean marks using direct method:

(ii) Calculating mean marks using short cut method:

$\overline{)X}=A+\frac{\Sigma d}{N}$
Here,
A represents assumed mean
d represents the deviation of the values from the assumed mean i.e. X − A

Here, we take 65 as the assumed mean. So, we take deviations of each item in the series from 65.

Thus, mean marks is 52.6

#### Question 4:

The following figures are the heights in cms of 7 children chose at random:
64, 59, 67, 69, 65, 70, 68
Calculate the simple arithmetic mean of the heights by (i) Direct method, (ii) Short-c ut Method, and (iii) Step Deviation Method.

 S.N. Height (X) Deviation from assumed mean d=(X − A) $\mathbit{d}\mathbf{\text{'}}\mathbf{=}\frac{{\mathbit{d}}_{}}{\mathbit{i}}\mathbf{=}\frac{{\mathbit{d}}_{}}{\mathbf{2}}$ 1 2 3 4 5 6 7 64 59 67 $\overline{)69=A}$ 65 70 68 −5 −10 −2    0 −4    1 −1 −2.5 −5 −1   0 −2   0.5 −0.5 N=7 $\sum _{}\mathbit{X}$=462 Σd=−21 Σd'= −10.5

(i) Calculating mean height using direct method:

(ii) Calculating mean height using short cut method:

$\overline{)X}=A+\frac{\sum _{}^{}d}{N}$

Here,

A represents assumed mean.
d represents the deviation of the values from the assumed mean.

Here, we take 69 as the assumed mean. So, we take deviations of each item in the series from 69.

(iii) Calculating mean height using step-deviation method:

$\overline{)X}=A+\frac{\mathrm{\Sigma }d\text{'}}{N}×i$

Hence, the arithmetic mean of the heights is 66 cms.

#### Question 5:

Find average for following discrete series.

 X 3 5 6 7 8 f 2 4 3 8 10

 X f fX 3 5 6 7 8 2 4 3 8 10 6 20 18 56 80 N=Σf= 27 ΣfX= 180

Calculating average for the above discrete series:

$\overline{)X}=\frac{\Sigma fX}{\Sigma f}$

$\overline{)X}=\frac{180}{27}\phantom{\rule{0ex}{0ex}}⇒\overline{)X}=6.66$

Thus, the average of the above series is 6.66

#### Question 6:

Compute the arithmetic mean from the following frequency table:

 Hight (in cms.) 58 60 62 64 66 68 Number of Plants 12 14 20 13 8 5

 Height (X) Plants (f) fX 58 60 62 64 66 68 12 14 20 13 8 5 696 840 1240 832 528 340 N=Σf= 72 ΣfX=4476

Calculating the arithmetic mean of the above series by direct method:

Hence, the mean height of the plant is 62.16 cms

#### Question 7:

Compute mean marks from the data given below by: (i) Direct method, (ii) Short-cut Method, and (iii) Step Deviation Method.

 Marks 5 15 25 35 45 55 65 Students 4 6 10 20 10 6 4

 Marks (X) Students (f) fX d=X−A fd $\mathbit{d}\mathbf{\text{'}}\mathbf{=}\frac{\mathbit{d}}{\mathbit{i}}\mathbf{=}\frac{\mathbit{d}}{\mathbf{10}}$ fd' 5 15 25 $\overline{)35=A}$ 45 55 65 4 6 10 20 10 6 4 20 90 250 700 450 330 260 $-$30 −20 −10 0 10 20 30 −120 −120 −100 0 100 120 120 −3 −2 −1 0 1 2 3 −12 −12 −10 0 10 12 12 N=Σf= 60 ΣfX= 2100 $\mathbf{\Sigma }\mathbit{f}\mathbit{d}$=0 $\mathbf{\Sigma }\mathbit{f}\mathbit{d}\mathbf{\text{'}}$=0

i) Calculating mean marks using direct method:

(ii) Calculating mean marks using short cut method:

$\overline{)X}=A+\frac{\Sigma fd}{\Sigma f}$

Here,

A represents assumed mean.
d represents the deviation of the values from the assumed mean.

Here, we take 35 as the assumed mean. So, we take deviations of each item in the series from 35.

(iii) Calculating mean marks using step-deviation method:

$\overline{)X}=A+\frac{\Sigma fd\text{'}}{\Sigma f}×i$

Hence, the mean marks of the above data is 35

#### Question 8:

The distribution fo age at marriage of 50 males is given below:

 Age in Years 20 21 22 23 24 25 No. Males 1 2 4 5 15 23
Calculate arithmetic mean using direct method.

 Age (X) No. of Males (f) fX 20 21 22 23 24 25 1 2 4 5 15 23 20 42 88 115 360 575 Σf= 50 ΣfX= 1200

Calculating mean age at marriage:

Hence, mean age at marriage is 24 years

#### Question 9:

Compute the mean marks obtained by the students from the following data:

 Marks 0−10 10−20 20−30 30−40 40−50 No. of Students 4 6 10 20 10

 Class Interval Mid Values $\mathbit{m}\mathbf{=}\frac{{\mathbit{l}}_{\mathbf{1}}\mathbf{+}{\mathbit{l}}_{\mathbf{2}}}{\mathbf{2}}$ Students (f) fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 5 15 25 35 45 4 6 10 20 10 20 90 250 700 450 Σf= 50 Σfm= 1510

Calculating mean marks using direct method:

Hence, mean marks obtained by the students are 30.2

#### Question 10:

Compute the mean marks obtained by the students from the following data:

 Marks 0−4 4−8 8−12 12−16 16−20 20−24 No. of Students 7 9 16 8 6 4

 Class Interval (Marks) Mid-Values (m) Students (f) fm 0 − 4 4 − 8 8 − 12 12 − 16 16 − 20 20 − 24 2 6 10 14 18 22 7 9 16 8 6 4 14 54 160 112 108 88 $\mathbf{\Sigma }\mathbit{f}$=50 $\mathbf{\Sigma }\mathbit{f}\mathbit{m}$=586

Calculating the mean marks using direct method:

Hence, mean marks obtained by the students are 10.72

#### Question 11:

Calculate the arithmetic average from the following data:

 Daily wages (in ₹) 2−4 4−6 6−8 8−10 10−12 12−14 14−16 16−18 No. of Workers 11 14 20 32 25 7 5 2

 Class Interval (Wages) Mid Values (m) Workers (f) fm 2 − 4 4 − 6 6 − 8 8 − 10 10 − 12 12 − 14 14 − 16 16 − 18 3 5 7 9 11 13 15 17 11 14 20 32 25 7 5 2 33 70 140 288 275 91 75 34 $\mathbf{\Sigma }\mathbit{f}$=116 $\mathbf{\Sigma }\mathbit{f}\mathbit{m}$=1006

Calculating the average wage using direct method:

Hence, average daily wage is Rs 8.67

#### Question 12:

Calculate mean from following data:

 Marks 5−15 15−25 25−35 35−45 45−55 55−65 No. of Students 8 12 6 14 7 3

 Class Interval (Marks) Mid-Values (m) Students (f) fm 5 − 15 15 − 25 25 − 35 35 − 45 45 − 55 55 − 65 10 20 30 40 50 60 8 12 6 14 7 3 80 240 180 560 350 180 Σf= 50 Σfm= 1590

Calculating mean marks by using direct method:

Hence, mean of the above series is 31.8 marks

#### Question 13:

Find mean for the following data by using: (i) Direct Method: (ii) Short-cut Method; (iii) Step Deviation Method.

 X 100−200 200−300 300−400 400−500 500−600 f 10 18 12 20 40

 Class Interval Mid values (m) (f) fm d=m−A fd $\mathbit{d}\mathbf{\text{'}}\mathbf{=}\frac{\mathbit{d}}{\mathbit{i}}\mathbf{=}\frac{\mathbit{d}}{\mathbf{100}}$ fd' 100−200 200 −300 300 −400 400 −500 500 −600 150 250 $\overline{)350=A}$ 450 550 10 18 12 20 40 1500 4500 4200 9000 22000 −200 −100 0 100 200 −2000 −1800 0 2000 8000 −2 −1 0 1 2 −20 −18 0 20 80 $\mathbit{\Sigma }\mathbit{f}$=100 $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=41200 $\mathbit{\Sigma }\mathbit{f}\mathbit{d}$=6200 $\mathbit{\Sigma }\mathbit{f}\mathbit{d}\mathbf{\text{'}}$=62

(i) Calculating mean using direct method:

(ii) Calculating mean using short cut method:

$\overline{)X}=A+\frac{\Sigma fd}{\Sigma f}$

Here,

A represents assumed mean.
d represents the deviation of the values from the assumed mean.

Here, we take 350 as the assumed mean. So, we take deviations of each item in the series from 350.

(iii) Calculating mean height step-deviation method:

$\overline{)X}=A+\frac{\Sigma fd\text{'}}{\Sigma f}×i$

Hence, the mean of the above series is 412.

#### Question 14:

The following table shows the marks obtained by 90 students in a certain examination. Calculate the average marks per students by step deviation method.

 Marks 10−20 20−30 30−40 40−50 50−60 60−70 70−80 80−90 No. of students 7 13 20 25 10 8 6 1

 Class Interval (Marks) Mid Values (m) Students (f) d=m−A $\mathbit{d}\mathbf{\text{'}}\mathbf{=}\frac{\mathbit{d}}{\mathbit{i}}\mathbf{=}\frac{\mathbit{d}}{\mathbf{10}}$ fd' 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 15 25 35 $\overline{)45=A}$ 55 65 75 85 7 13 20 25 10 8 6 1 −30 −20 −10 0 10 20 30 40 −3 −2 −1 0 1 2 3 4 −21 −26 −20 0 10 16 18 4 $\mathbit{\Sigma }\mathbit{f}$=90 $\mathbit{\Sigma }\mathbit{f}\mathbit{d}\mathbf{\text{'}}$=−19

Calculating mean marks using step-deviation method:

Hence, the mean of the above series is 42.89 marks.

#### Question 15:

Find the mean form the following data:

 Find the mean form the following data No. of Students Less than 10 5 Less than 20 20 Less than 30 45 Less than 40 70 Less than 50 80 Less than 60 88 Less than 70 98 Less than 80 100

Converting less than cumulative frequency distribution into a simple frequency distribution

 Marks No. of students (c.f) Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 5 20 45 70 80 88 98 100

 Class Interval (Marks) Mid-Values (m) Students (f) fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 5 15 25 35 45 55 65 75 0-5=5 20-5=15 45-20=25 70-45=25 80-70=10 88-80=8 98-88=10 100-98=2 25 225 625 875 450 440 650 150 Σf=100 Σfm= 3440

Calculating mean from the above data by using direct method:

Hence, the mean of the above data is 34.4 marks.

#### Question 16:

Calculate arithemtic eman from the foolowing data:

 Marks No. Students More than 0 150 More than 10 140 More than 20 100 More than 30 80 More than 40 80 More than 50 70 More than 60 30 More than 70 14

Converting more than cumulative frequency distribution into a simple frequency distribution

 Marks No. of students (c.f) More than 0 More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 150 140 100 80 80 70 30 14

 Class Interval Mid-Values (m) No. of students (f) fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 5 15 25 35 45 55 65 75 150-140=10 140-100=40 100-80=20 80-80=0 80-70=10 70-30=40 30-14=16 14-0=14 50 600 500 0 450 2200 1040 1050 $\mathbit{\Sigma }\mathbit{f}$=150 $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=5890

Calculating the mean marks by using direct method:

Hence, the mean marks of the above series is 39.27

#### Question 17:

Find the mean from the following data:

 Marks No. of Students Above 0 80 Above 10 77 Above 20 72 Above 30 65 Above 40 55 Above 50 43 Above 60 28 Above 70 16 Above 80 10 Above 90 8 Above 100 0

 Marks Students Above 0 Above 10 Above 20 Above 30 Above 40 Above 50 Above 60 Above 70 Above 80 Above 90 Above 100 80 77 72 65 55 43 28 16 10 8 0

 Class Interval (Marks) Mid-Values (m) Students (f) fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 90 − 100 100 − 110 5 15 25 35 45 55 65 75 85 95 105 80−77=3 77−72=5 72−-65=7  65−55=10 55−43=12 43−28=15 28−16=12 16−10=6 10−8=2 8−0=8 0−0=0 15 75 175 350 540 825 780 450 170 760 0 Σf= 80 Σfm= 4140

Calculating mean marks by using direct method:

#### Question 18:

The following table shows the age of workers in a factory. Find out the average age of workers.

 Age (in Years) 20−29 30−39 40−49 50−59 60−69 Workers 10 8 6 4 2

Calculation of mean in an inclusive series:

 Class Interval (Age) Mid-Values (m) Workers (f) fm 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 24.5 34.5 44.5 54.5 64.5 10 8 6 4 2 245 276 267 218 129 $\mathbit{\Sigma }\mathbit{f}$=30 $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=1135

Computing the average age of workers:

Hence, average age of workers is 37.83 years

#### Question 19:

Find the average age from the following data:

 Age (in Years) Less than 10 10−20 20−30 30−40 More than 40 No. of Persons 5 8 12 6 4

It must be noted that the given distribution is an open ended distribution, that is, the first and the last class interval are not explicitly defined. As mean is based on all the items in the series, so for such distributions mean cannot be calculated accurately. ​

However, going by the symmetry of the distribution we assume the first and the last class interval to be (0- 10) and (40 - 50), respectively

 Age Class Interval Mid-Values (m) Persons (f) fm Less than 10 10 − 20 20 − 30 30 − 40 more than 40 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 5 15 25 35 45 5 8 12 6 4 25 120 300 210 180 $\mathbit{\Sigma }\mathbit{f}$=35 $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=835

Computing average age:

Hence, average age is 23.85 years.

#### Question 20:

Calculate the mean from the following data:

 Mid-value 10 20 30 40 50 Frequency 8 12 15 9 6

 Mid-Values (m) (f) fm 10 20 30 40 50 8 12 15 9 6 80 240 450 360 300 $\mathbit{\Sigma }\mathbit{f}$=50 $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=1430

Calculating mean using direct method:

Hence, mean of the above series is 28.6

#### Question 21:

Calculate mean form the following data:

 X 0−10 10−20 20−30 30−60 60−90 Frequency 5 9 20 12 4

 Class Interval Mid-Values (m) (f) fm 0 − 10 10 − 20 20 − 30 30 − 60 60 − 90 5 15 25 45 75 5 9 20 12 4 25 135 500 540 300 $\mathbit{\Sigma }\mathbit{f}$=50 $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=1500

Calculating the mean using direct method:

Hence, mean of the above series is 30

#### Question 22:

The man height of 25 male workers in a factory is 61 inches and the mean height of 35 female workers in the same factory is 58 inches. Find the combined mean height of 60 workers in the factory

Combined Mean =
Here,

Hence, the combined mean height of the factory is 59.25 inches.

#### Question 23:

The mean age of a combined group of men and women is 30.5 years. If the mean age of the group of men is 35 and that of the group of women is 25, find out the percentage of men and women in the group.

Let the percentage of male in the combined mean be x
∵ percentage of women = (100 − x)

Note: Total of the percentage= 100

Given:

${\overline{)X}}_{m,w}=30.5\phantom{\rule{0ex}{0ex}}{\overline{)X}}_{m}=35\phantom{\rule{0ex}{0ex}}{\overline{)X}}_{w}=25$

We know,

∴ Males = 55%
∴ Women= (100 − x) = (100 − 55) = 45%

Therefore, there are 55% men and 45% women in the group.

#### Question 24:

The mean weight of 150 students in a class is 60 kg. The mean of boys in the class is 70 kg and that of girls is 55 kg. Find the number of boys and girls in the class.

Given:

Nb + Ng = 150     (Equation-1)

Ng
= 150 − Nb      (From eqn-1)

We know,

Therefore,

No. of boys = Nb = 50
No. of Girls = Ng = 100

#### Question 25:

The average weight of a group of 20 boys was calculated to be 89.4. It was later discovered that one weight was misread as 78 kg instead of the correct on of 87 kg. Calculate the correct average weight.

Given:

Mean, $\overline{)\mathbf{X}}$=89.4
Number of observations, N=20
Incorrect observation= 78
Correct observation = 87

Thus, wrong summation of X, i.e., ${\left(\Sigma X\right)}_{wrong}$ is 1788.
Now, Correct mean is calculated using the following formula.

Hence, the correct average weight is 89.95 kgs

#### Question 26:

The mean of 100 observation was found to be 40. Later on, it was discovered that tow items were wrongly taken as 30 and 27 instead of 3 and 72. Find correct mean.

Given:

Mean, $\overline{)\mathbf{X}}$=40
Number of observations, N=100
Incorrect observations= 30 and 27
Correct observations = 3 and 72
Wrong X=Wrong ΣXNor, 89.4=Wrong ΣX20Wrong ΣX=1

Thus, wrong summation of X, i.e., $\mathbf{\left(}\mathbit{\Sigma }\mathbit{X}{\mathbf{\right)}}_{\mathbf{w}\mathbf{r}\mathbf{o}\mathbf{n}\mathbf{g}}$=4000
Now, correct mean is calculated using the following formula.

Hence, the correct mean is 40.18

#### Question 27:

The average height of a group of 40 students was calculated as 155 cm. It was later discovered that the height of one student was read as 157 instead of 137 cm. Calculate the correct average height.

Given:

Mean, $\overline{)\mathbf{X}}$= 155
Number of observations =40
Incorrect observation= 157
Correct Observation = 137

Thus, wrong summation of X, i.e., $\mathbit{\Sigma }{\mathbit{X}}_{\mathbf{\left(}\mathbf{w}\mathbf{r}\mathbf{o}\mathbf{n}\mathbf{g}\mathbf{\right)}}$ is 6200
Now, correct mean is calculated using the following method.

Hence, the correct average height is 154.5 cm

#### Question 28:

From the following data, calculate the weighted mean.

 Marks 62 77 65 62 57 Weights 2 1 2 3 4

 Marks (X) Weight (W) WX 62 77 65 62 57 2 1 2 3 4 124 77 130 186 228 $\mathbit{\Sigma }\mathbit{W}$=12 $\mathbit{\Sigma }\mathbit{W}\mathbit{X}$=745

${\overline{)X}}_{W}=\frac{\Sigma WX}{\Sigma W}$

${\overline{)\mathrm{X}}}_{w}=\frac{745}{12}\phantom{\rule{0ex}{0ex}}⇒{\overline{)\mathrm{X}}}_{w}=62.08$

Hence, the weighted mean is 62.08 marks

#### Question 29:

Find out weighted mean by weighting each price by the quantity consumed:

 Items Quantity consumed (in kg) Price in Rupees (per kg) Apple 12 30 Mango 10 50 Tomato 10 40 Orange 15 35 Grapes 12 45

 Items Quantity (W) Price (X) WX Apple Mango Tomato Orange Grapes 12 10 10 15 12 30 50 40 35 45 360 500 400 525 540 $\mathbit{\Sigma }\mathbit{W}$=59 ΣWX= 2325

Now, weighted mean can be calculating using the following formula.

Hence, the weighted mean of the above series is 39.40

#### Question 30:

An examination was held to decide the most eligible student for scholarship. The weights of various subjects along with the marks obtained by the three candidate (out of 100) are given below:

 Subjects Weighs Marks of Ram Marks of Shyam Marks of Manish Maths 3 65 64 70 Statistics 4 63 60 65 Economics 1 70 80 52 English 2 58 56 63

 Subjects Weights (W) Ram (X1) Shyam (X2) Manish (X3) WX1 WX2 WX3 Math Stats Eco English 3 4 1 2 65 63 70 58 64 60 80 56 70 65 52 63 195 252 70 116 192 240 80 112 210 260 52 126 10 633 624 648

Manish should get the scholarship as he has the maximum weighted mean marks.

#### Question 31:

Arithmetic mean of the following series is 41:

 Marks 20 30 40 ? 60 70 No. of Students 8 12 20 10 6 4
Find the missing item

 Marks (X) No. of Students (f) fX 20 30 40 x 60 70 8 12 20 10 6 4 160 360 800 10x 360 280 $\mathbit{\Sigma }\mathbit{f}\mathbf{=}$60 $\mathbit{\Sigma }\mathbit{f}\mathbit{X}$=1960 + 10x

Given:

Mean, $\overline{)\mathbf{X}}$= 41

Hence, the missing item is 50 marks

#### Question 32:

Calculate the number of students against the class 30−40 of the following data, where $\overline{)X}$ = 28.

 Marks 0−10 10−20 20−30 30−40 40−50 50−60 Frequency 12 18 27 ? 17 6

 Class Interval (Marks) Mid-Values (m) Frequency (f) fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 5 15 25 35 45 55 12 18 27 f 17 6 60 270 675 35f 765 330 $\mathbit{\Sigma }\mathbit{f}$=80 + f $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=2100 + 35f

Let the missing frequency be f

Given:
Mean, $\overline{)\mathbit{X}}$=28

Hence, the missing frequency is 20.

#### Question 33:

Find the missing frequency form the following data, if athematic mean is 25.4.

 Class-interval 10−20 20−30 30−40 40−50 50−60 Frequency 20 15 10 ? 2

 Class Interval Mid-Values (m) Frequency (f) fm 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 15 25 35 45 55 20 15 10 f 2 300 375 350 45f 110 $\mathbit{\Sigma }\mathbit{f}$=47+f $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=1135 + 45 f

Let missing frequency be f

Given:

Mean, $\overline{)\mathbit{X}}$ =25.4

Hence, missing frequency is 3

#### Question 34:

Find the mi9ssing frequency from the followinjg table:

 Marks No. of Students Below 10 25 Below 20 40 Below 30 60 Below 40 75 Below 50 95 Below 60 125 Below 70 190 Below 80 240

 Marks Students Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80 25 40 60 75 95 125 190 240

 Class Interval Mid-Values (m) No. of Students (f) fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 5 15 25 35 45 55 65 75 25-0=25 40-25=15 60-40=20 75-60=15 95-75=20 125-95=30 190-125=65 240-190=50 125 225 500 525 900 1650 4225 3750 $\mathbit{\Sigma }\mathbit{f}$=240 $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=11,900

Calculating average marks using direct method:

Hence, the average marks are 49.58

#### Question 35:

Calculate the mean from the following frequency distribution:

 X 10−19 20−29 30−39 40−49 50−59 60−69 70−79 80−89 No. Persons 32 42 40 56 20 6 2 2

 Class interval (X) Mid-Values (m) No. of persons (f) fm 10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 70 − 79 80 − 89 14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 32 42 40 56 20 6 2 2 464 1029 1380 2492 1090 387 149 169 $\mathbit{\Sigma }\mathbit{f}$=200 $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=7160

Calculating the mean from the above inclusive series by using direct method:

Hence, the mean of the above series is 35.8

#### Question 36:

The Average age of 20 students in a class is 16 yars.One student whose age is 18 years has left the class. Find out average age of rest of the students.

Given:
Mean, $\overline{)\mathbf{X}}$ = 16 years
Number of observations, N= 20

We know,
$\overline{)X}=\frac{\Sigma X}{N}$

ΣX of 20 students = ${\overline{)X}}_{20}×N$= 20 × 16 = 320
ΣX of 19 students = =320 − 18 = 302

Thus, the average age of rest of the students is:

Hence, the correct average age is 15.89 years

#### Question 37:

Locate the missing frequency, if arithmetic mean of the following series is 44.8.

 X 20 30 40 50 60 70 f 5 ? 15 10 8 5

 X f fX 20 30 40 50 60 70 5 f 15 10 8 5 100 30f 600 500 480 350 $\Sigma f$=43 + f $\Sigma fX$=2030 + 30f

Let missing frequency be f

Given:
Mean, $\overline{)\mathrm{X}}=44.8$

We know,

Hence, the missing frequency is 7

#### Question 38:

The average marks of 30 students in a class were 52. The top six students had an average of 31. What were the average marks of the other students?

Given:

N= 30

Average of top 6 students = 31

ΣX of top 6 students, $\Sigma {X}_{6}$= 31 × 6 = 186

ΣX of remaining 24 (30-6) students, $\Sigma {X}_{24}$ = $\Sigma {X}_{30}-\Sigma {X}_{6}$= 1560 − 186 = 1374

Now, we will calculate the average marks of the other students by using the direct method:

#### Question 39:

Calculate the mean of the given data:

 Size of item 2 4 6 8 10 12 14 16 Frequency 5 7 10 13 17 9 6 3

 X f fX 2 4 6 8 10 12 14 16 5 7 10 13 17 9 6 3 10 28 60 104 170 108 84 48 $\mathbf{\Sigma }\mathbit{f}$=70 $\mathbf{\Sigma }\mathbit{f}\mathbit{X}$=612

Calculating the mean of the above data by using the direct method:

Hence, the mean of the above series is 8.74

#### Question 40:

A Candidate obtained the following percentage following of marks in an examination: English 60; Business Studies 75; Maths 63; Accounts 59; Economics 55. Find the candidate's weighted arithmetic mean, if weights 1, 2, 1, 3, 3 respectively are allotted to the subjects

The information given in the question can be presented as follows:

 Subject Marks (X) Weights (W) WX English Business studies Maths Accounts Economics 60 75 63 59 55 1 2 1 3 3 60 150 63 177 165 $\mathbit{\Sigma }\mathbit{W}$=10 $\mathbit{\Sigma }\mathbit{W}\mathbit{X}$=615

Hence, the weighted arithmetic mean is 61.5 marks

#### Question 41:

The mean height of 50 student of a college is 68 inches. The height of 30 students is given below. Find the mean height of the remaining 20 students.

 Height (in inches) 64 66 68 70 72 Frequency 4 12 4 8 2

 Height (X) Frequency (f) fX 64 66 68 70 72 4 12 4 8 2 256 792 272 560 144 $\mathbit{\Sigma }\mathbit{f}$=30 $\mathbit{\Sigma }\mathbit{f}{\mathbit{X}}_{\mathbf{30}}$=2024

Given:
Σf = 50

${\overline{)X}}_{50}=68$

$\Sigma f{X}_{50}={\overline{)X}}_{50}×\Sigma f=68×50=3400\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
ΣfX of remaining 20 students

$\Sigma f{X}_{20}$= $\Sigma f{X}_{50}-\Sigma f{X}_{30}$=3400 − 2024 = 1376

Hence, mean height of 20 students is 68.8 inches

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