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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.)
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.)
Solution :
Given: The frequency distribution of the lengths of 40 leaves measured to the nearest millimetre.
| Length (mm) | Number of leaves ($f$) |
|---|---|
| 118 - 126 | 3 |
| 127 - 135 | 5 |
| 136 - 144 | 9 |
| 145 - 153 | 12 |
| 154 - 162 | 5 |
| 163 - 171 | 4 |
| 172 - 180 | 2 |
To Find: The median length of the leaves.
Step 1: Converting to Continuous Classes
The given classes are discontinuous (e.g., 126 and 127). To make them continuous, we subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class interval.
| Class Interval | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 117.5 - 126.5 | 3 | 3 |
| 126.5 - 135.5 | 5 | 8 |
| 135.5 - 144.5 | 9 | 17 |
| 144.5 - 153.5 | 12 | 29 |
| 153.5 - 162.5 | 5 | 34 |
| 162.5 - 171.5 | 4 | 38 |
| 171.5 - 180.5 | 2 | 40 |
| Total | $n = 40$ | - |
Step 2: Identifying the Median Class
The total number of observations is $n = 40$.
We calculate $\frac{n}{2} = \frac{40}{2} = 20$.
Looking at the cumulative frequency column, the first cumulative frequency greater than 20 is 29, which corresponds to the class interval $144.5 - 153.5$.
Thus, the median class is $144.5 - 153.5$.
Step 3: Applying the Median Formula
The formula for the median is:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
Where:
$l$ (lower limit of median class) = $144.5$
$n$ (total number of observations) = $40$
$cf$ (cumulative frequency of the class preceding the median class) = $17$
$f$ (frequency of the median class) = $12$
$h$ (class size) = $153.5 - 144.5 = 9$
Step 4: Calculation
$\text{Median} = 144.5 + \left( \frac{20 - 17}{12} \right) \times 9$
$\text{Median} = 144.5 + \left( \frac{3}{12} \right) \times 9$
$\text{Median} = 144.5 + \left( \frac{1}{4} \right) \times 9$
$\text{Median} = 144.5 + 2.25$
$\text{Median} = 146.75$
Final Answer: The median length of the leaves is 146.75 mm.
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.
- 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
Chapters in CBSE - Class 10 Mathematics
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