Math, asked by suyashsingh1234, 2 months ago

The expenditure of 1000 families is given below:
Expenditure(InRs.) 40-59 60-79 80-99 100-119 120-139 No. Of Families ? 150 ? 250 50
The mean for the distribution is Rs 87.50.Calculate the missing frequencies.

Answers

Answered by anjalikapurwanstu102
3

Here is the answer of the given question

Attachments:
Answered by Syamkumarr
1

Answer:

Missing frequencies are 50 and 500.

Step-by-step explanation:

Given data

Expenditure(In Rs.)     40-59    60-79    80-99    100-119     120-139

No. Of Families                ?           150          ?           250            50  

Mean of the given data  = 87.50

Total number of families = 1000

here we need to find missing frequencies  

Let missing frequencies  is a and b  

And here we will use step deviation method to calculate mean

from given data mean lies in  80-99 class,

so 89.5 taken as assumed mean A    

now calculate following values in given table

   C.I            frequency(f_{i})        mid point(x_{i})        u_{i}=\frac{x_{i} - A}{c}         f_{i} u_{i}    

 40-59              a                          49.5                   -2             -2a    

 60-79             150                       69.5                   -1             -150

 80-99              b                            89.5(A)               0                  0  

 100-119            250                      109.5                   1                250

 120-139           50                        129.5                   2                 100

               ∑f_{i} = a+b+450                              ∑f_{i}u_{i} = -2a+200                                              

from given data total number of families = 1000

                              ⇒   a+150+b+250+50 = 1000

                              ⇒  a+b+ = 1000-450 = 550

                              ⇒  a + b = 550 _(1)

As we know mean of a data in step deviation method

                                 = A+ [ ∑f_{i}u_{i}/∑f_{i}]×C    

                                 = 89.5 + [ \frac{-2a+200}{1000} ]20  

                                 = 89.5 + [ \frac{-2a+200}{50} ]

from given data      89.5 + [ \frac{-2a+200}{50} ] = 87.50

                                [ \frac{-2a+200}{50} ] = 87.50 - 89.5    

                                  [ \frac{-2a+200}{50} ] = 2

                                  -2a+200 = 100

                                           -2a = -100

                                               a = 50

                    (1)  ⇒ 50 + b = 550  

                         ⇒  b = 500    

#SPJ3

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