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# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 2

• Last Updated : 03 Mar, 2021

### Question 14. Find the mean of the following frequency distribution:

Solution:

Letâ€™s consider the assumed mean (A) = 42

From the table itâ€™s seen that,

A = 42 and h = 5

Mean = A + h x (Î£fi ui/N)

= 42 + 5 x (-79/70)

= 42 â€“ 79/14

= 42 â€“ 5.643

= 36.357

### Question 15. For the following distribution, calculate mean using all suitable methods:

Solution:

By direct method

Mean = (sum/N) + A

= 848/64

= 13.25

By assuming mean method

Let the assumed mean (A) = 65

Mean = A + sum/N

= 6.5 + 6.75

= 13.25

### Question 16. The weekly observation on cost of living index in a certain city for the year 2004 – 2005 are given below. Compute the weekly cost if living index.

Solution:

Let the assumed mean (A) = 1650

We have

A = 16, h = 100

Mean = A + h (sum/N)

= 1650 + (175/13)

= 21625/13

= 1663.46

### Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

Solution:

(i) Direct method:

Mean = sum/N

= 3620/140

= 25.857

(ii) Assumed mean method:

Let the assumed mean = 25

Mean = A + (sum/N)

Mean = A + (sum/N)

= 25 + (120/140)

= 25 + 0.857

= 25.857

(iii) Step deviation method:

Let the assumed mean (A) = 25

Mean = A + h(sum/N)

= 25 + 10(12/140)

= 25 + 0.857

= 25.857

### Question 18. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the miss frequency f1 and f2.

Solution:

Given,

Sum of frequency = 50

5 + f1 + 10 + f2 + 7 + 8 = 50

f1 + f2 = 20

3f1 + 3f2 = 60        —(1) [Multiply both side by 3]

and mean = 62.8

Sum/N = 62.8

(30f1 + 70f2 + 2060)/50 = 62.8

30f1 + 70f2 = 3140 – 2060

30f1 + 70f2 = 1080

3f1 + 7f2 = 108      —(2) [divide it by 10]

Subtract equation (1) from equation (2)

3f1 + 7f2 – 3f1 – 3f2 = 108 – 60

4f2 = 48

f2 = 12

Put value of f2 in equation (1)

3f1 + 3(12) = 60

f1 = 24/3 = 8

f1 = 8, f2 = 12

### Question 19. The following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is Rs 18.00. Find out the missing frequency.

Solution:

Given mean = 18,

Let the missing frequency be v

Mean = sum/N

18 = 752 + 20×44 + x

792 + 18x = 752 + 20x

2x = 40

x = 20

### Question 20. If the mean of the following distribution is 27. Find the value of p.

Solution:

Given mean = 27

Mean = sum/N

1245 + 15p43 + p = 27

1245 + 15p = 1161 + 27p

12p = 84

P =7

### Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?

Solution:

We may observe that class internals are not continuous

There is a gap between two class intervals. So we have to add Â½ from lower class limit to each interval and class mark (xi) may be obtained by using the relation

xi = upperlimit + lowerclasslimit2

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculated di, ui, fiui as follows

Now we have N

Sum = 25

Mean = A +h (sum/N)

= 57 + 3 (45/400)

= 57 + 3/16

= 57 + 0.1875

= 57.19

Clearly mean number of mangoes kept in packing box is 57.19

### Find the mean daily expenditure on food by a suitable method.

Solution:

We may calculate class mark (xi) for each interval by using the relation

xi = upperlimit + lowerclasslimit2

Class size = 50

Now, Taking 225 as assumed mean (xi) we may calculate di, ui, fiui as follows

Now we may observe that

N = 25

Sum = -7

225 + 50 (-7/25)

225 – 14 = 211

So, mean daily expenditure on food is Rs 211

### Find the mean concentration of SO2 in the air

Solution:

We may find class marks for each interval by using the relation

x = upperlimit + lowerclasslimit2x =

Class size of this data = 0.04

Now taking 0.04 assumed mean (xi) we may calculate di, ui, fiui as follows

From the table we may observe that

N = 30

Sum = -31

= 0.14 + (0.04)(-31/30)

= 0.099 ppm

So mean concentration of SO2 in the air is 0.099 ppm.

### Question 24. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

We may find class mark of each interval by using the relation

x =  upperlimit + lowerclasslimit2x =

Now, taking 16 as assumed mean (a) we may

Calculate di and fidi as follows

Now we may observe that

N = 40

Sum = -145

= 16 + (-145/40)

= 16 – 3.625

= 12.38

So mean number of days is 12.38 days, for which student was absent.

### Question 25. The following table gives the literacy rate (in percentage) of 35 cities. find the mean literacy rate.

Solution:

We may find class marks by using the relation

x = upperlimit + lowerclasslimit2x =

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) wrong

Calculate di, ui, fiui as follows

Now we may observe that

N = 35

Sum = -2

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43%

### Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?

Solution:

We may observe that class internals are not continuous

There is a gap between two class intervals. So we have to add Â½ from lower class limit to each interval and class mark (xi) may be obtained by using the relation

xi = upperlimit + lowerclasslimit2

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculated di, ui, fiui as follows

Now we have N

Sum = 25

Mean = A +h (sum/N)

= 57 + 3 (45/400)

= 57 + 3/16

= 57 + 0.1875

= 57.19

Clearly mean number of mangoes kept in packing box is 57.19

### Find the mean daily expenditure on food by a suitable method.

Solution:

We may calculate class mark (xi) for each interval by using the relation

xi = upperlimit + lowerclasslimit2

Class size = 50

Now, Taking 225 as assumed mean (xi) we may calculate di, ui, fiui as follows

Now we may observe that

N = 25

Sum = -7

225 + 50 (-7/25)

225 – 14 = 211

So, mean daily expenditure on food is Rs 211

### Find the mean concentration of SO2 in the air

Solution:

We may find class marks for each interval by using the relation

x = upperlimit + lowerclasslimit2x =

Class size of this data = 0.04

Now taking 0.04 assumed mean (xi) we may calculate di, ui, fiui as follows

From the table we may observe that

N = 30

Sum = -31

= 0.14 + (0.04)(-31/30)

= 0.099 ppm

So mean concentration of SO2 in the air is 0.099 ppm.

### Question 24. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

We may find class mark of each interval by using the relation

x =  upperlimit + lowerclasslimit2x =

Now, taking 16 as assumed mean (a) we may

Calculate di and fidi as follows

Now we may observe that

N = 40

Sum = -145

= 16 + (-145/40)

= 16 – 3.625

= 12.38

So mean number of days is 12.38 days, for which student was absent.

### Question 25. The following table gives the literacy rate (in percentage) of 35 cities. find the mean literacy rate.

Solution:

We may find class marks by using the relation

x = upperlimit + lowerclasslimit2x =

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) wrong

Calculate di, ui, fiui as follows

Now we may observe that

N = 35

Sum = -2

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43%

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