# Case Study Question 08

## Chapter 14: Statistics

### Class 10

A survey was conducted by an NGO to know the monthly expenditure of families living in slums in Delhi. A total of 200 families were interviewed and it was found that their minimum monthly expenditure was Rs.1000. The result is tabulated as given below:

Question.1.
Find the number of families whose monthly expenditure is more than or equal to Rs. 8000.

We can see from the given frequency distribution that the number of families having monthly expenditure less than Rs. 8000 is 163 out of a total of 200 families. Therefore, number of families whose monthly expenditure is more than or equal to Rs. 8000 = 200 – 163 = 37

Question.2.
Find the number of families whose monthly expenditure is in the range Rs. (6000 – 7000).

Let us prepare a frequency distribution table from the given cumulative frequency table as below:

Question.3.
Find the lower limit of median class.

Median class is the class whose cumulative frequency is just greater than half of sum of all frequencies. Here \frac{N}{2}=100. As the cululative frequency of the class 5000 – 6000 is 115, which is just greater than 100, therefore the median class is 5000 – 6000 and thus the lower limit of the median class is 5000.

Question.4.
Find the median monthly expenditure of the families as per the frequency distribution table.

The median class is 5000 – 6000. The formula for calculating the median is:
Median = l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h
Where, l= lower limit of the median class,
h = size of clas interval,
n = number of observations
cf = cumulative frequency of class preceding the median class,
f = frequency of the median class

Using the table,

error: Content is protected !!
Scroll to Top