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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.
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.
Solution :
Given: A frequency distribution of the number of letters in 100 surnames.
| Number of letters | Number of surnames ($f_i$) |
|---|---|
| 1 - 4 | 6 |
| 4 - 7 | 30 |
| 7 - 10 | 40 |
| 10 - 13 | 16 |
| 13 - 16 | 4 |
| 16 - 19 | 4 |
To Find: The Median, Mean, and Mode of the given data.
Step 1: Calculation of Mean ($\bar{x}$)
We use the Assumed Mean Method. Let the assumed mean $a = 11.5$. The class size $h = 3$.
| Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $d_i = x_i - a$ | $u_i = \frac{x_i - a}{h}$ | $f_i u_i$ |
|---|---|---|---|---|---|
| 1-4 | 6 | 2.5 | -9 | -3 | -18 |
| 4-7 | 30 | 5.5 | -6 | -2 | -60 |
| 7-10 | 40 | 8.5 | -3 | -1 | -40 |
| 10-13 | 16 | 11.5 | 0 | 0 | 0 |
| 13-16 | 4 | 14.5 | 3 | 1 | 4 |
| 16-19 | 4 | 17.5 | 6 | 2 | 8 |
| Total | $\sum f_i = 100$ | - | - | - | $\sum f_i u_i = -106$ |
Formula for Mean: $\bar{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h$
$\bar{x} = 11.5 + \left( \frac{-106}{100} \right) \times 3 = 11.5 - 3.18 = 8.32$
Step 2: Calculation of Median
Total frequency $N = 100$. So, $N/2 = 50$.
| Class Interval | Frequency ($f_i$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 1-4 | 6 | 6 |
| 4-7 | 30 | 36 |
| 7-10 | 40 | 76 |
The cumulative frequency just greater than 50 is 76, so the median class is $7-10$.
Lower limit ($l$) = 7, $cf$ of preceding class = 36, $f$ of median class = 40, $h = 3$.
Median = $l + \left( \frac{N/2 - cf}{f} \right) \times h = 7 + \left( \frac{50 - 36}{40} \right) \times 3 = 7 + \left( \frac{14}{40} \right) \times 3 = 7 + 1.05 = 8.05$
Step 3: Calculation of Mode
The modal class is the class with the highest frequency, which is $7-10$ ($f_1 = 40$).
$l = 7, f_1 = 40, f_0 = 30, f_2 = 16, h = 3$.
Mode = $l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h = 7 + \left( \frac{40 - 30}{2(40) - 30 - 16} \right) \times 3$
Mode = $7 + \left( \frac{10}{80 - 46} \right) \times 3 = 7 + \left( \frac{10}{34} \right) \times 3 = 7 + 0.88 = 7.88$
Final Answer: The Mean number of letters is 8.32, the Median is 8.05, and the Mode is 7.88.
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 $y$.
- 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.
- 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.
- 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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