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Q3:
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
Solution :
Given: The frequency distribution of ages of 100 policy holders, provided as a cumulative frequency table (less than type):
| Age (years) | Number of policy holders (Cumulative Frequency) |
|---|---|
| Below 20 | 2 |
| Below 25 | 6 |
| Below 30 | 24 |
| Below 35 | 45 |
| Below 40 | 78 |
| Below 45 | 89 |
| Below 50 | 92 |
| Below 55 | 98 |
| Below 60 | 100 |
To Find: The median age of the policy holders.
Step 1: Convert the cumulative frequency distribution into a standard frequency distribution table.
To calculate the median, we need the class intervals and their corresponding frequencies ($f_i$).
| Class Interval | Frequency ($f_i$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 15-20 | 2 | 2 |
| 20-25 | 6 - 2 = 4 | 6 |
| 25-30 | 24 - 6 = 18 | 24 |
| 30-35 | 45 - 24 = 21 | 45 |
| 35-40 | 78 - 45 = 33 | 78 |
| 40-45 | 89 - 78 = 11 | 89 |
| 45-50 | 92 - 89 = 3 | 92 |
| 50-55 | 98 - 92 = 6 | 98 |
| 55-60 | 100 - 98 = 2 | 100 |
| Total | $N = 100$ | - |
Step 2: Identify the median class.
The total number of observations is $N = 100$.
We calculate $\frac{N}{2} = \frac{100}{2} = 50$.
Looking at the cumulative frequency column, the first cumulative frequency greater than 50 is 78, which corresponds to the class interval 35-40.
Therefore, the median class is 35-40.
Step 3: Apply the Median Formula.
The formula for the median of a grouped frequency distribution is:
$\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
Where:
- $l$ (lower limit of median class) = $35$
- $N$ (total frequency) = $100$
- $cf$ (cumulative frequency of the class preceding the median class) = $45$
- $f$ (frequency of the median class) = $33$
- $h$ (class size) = $40 - 35 = 5$
Step 4: Perform the calculation.
Substitute the values into the formula:
$\text{Median} = 35 + \left( \frac{50 - 45}{33} \right) \times 5$
$\text{Median} = 35 + \left( \frac{5}{33} \right) \times 5$
$\text{Median} = 35 + \frac{25}{33}$
$\text{Median} = 35 + 0.7575...$
$\text{Median} \approx 35.76$
Final Answer: The median age of the policy holders is 35.76 years.
More Questions from Class 10 Mathematics Statistics EXERCISE 13.3
- Q1: The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
- Q2: If the median of the distribution given below is 28.5, find the values of $x$ and $y$.
- Q4: The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table : Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)
- Q5: The following table gives the distribution of the life time of 400 neon lamps : Find the median life time of a lamp.
- Q6: 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
- Q7: The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
CBSE Solutions for Class 10 Mathematics Statistics
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