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RD Chapter 7 Statistics Ex 7.3 Solutions

Question - 21 : -
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
 
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose ?

Answer - 21 : -

We shall apply the assumed mean deviation method
Let assumed mean (A) = 57
We shall choose the method of assumed mean deviation :
 
= 57 + 3 x (25/100)
= 57 + 3/16
= 57 + 0.1875 = 57.1875 = 57.19

Question - 22 : - The table below shows the daily expenditure on food of 25 households in a locality

Answer - 22 : -

Let assumed mean (A) = 225
 
∴ Mean expenditure on food = Rs. 211

Question - 23 : -
To find out the concentration of S02 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below :
 
Find the mean concentration of S02 in the air.

Answer - 23 : -

Let assumed mean (A) = 0.10
  
= 0.10 – 0.00133 = 0.09867 = 0.099 (approx)

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 student was absent.

Answer - 24 : -


∴ Mean number of days a students was absent = 12.475

Question - 25 : - The following table gives the literacy rate (in percentage) of 3§ cities. Find the mean

Answer - 25 : -


Question - 26 : - The following is the cummulative frequency distribution (of less than type) of 1000 persons each of age 20 years and above. Determine the mean age.[NCERT Exemplar]

Answer - 26 : -

First, we make the frequency distribution of the given data and then proceed to calculate mean by computing class marks (xi), ui’s and fiui‘s as follows:
 
= 45 + 6.3 = 51.3
Thus, the mean age is 51.3 years.

Question - 27 : - If the mean of the following frequency distribution is 18, find the missing frequency.

Answer - 27 : -


Question - 28 : - Find the missing frequencies in the following distribution, if the sum of the frequencies is 120 and the mean is 50.

Answer - 28 : -


⇒ 4+ f2 -f1 =0
⇒ -f2+f1 = 4 ……..(ii)
On adding Eqs. (i) and (ii), we get
⇒ 2f1 =56
⇒ f1=28
Put the value of f1 in Eq. (i), we get
f2 = 52-28
⇒ f2 =24
Hence, f1 = 28 and f2 = 24

Question - 29 : -
The daily income of a sample of 50 employees are tabulated as follows:
 
Find the mean daily income of employees. [NCERT Exemplar]

Answer - 29 : -

Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Now we first, find the class mark xt of each class and then proceed as follows:
 
∴ Assumed mean, a = 300.5
Class width, h = 200
and total observation, N = 50
By step deviation method,
 
= 300.5 + 200 x (1/50) x 14
= 300.5 + 56 = 356.5

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