4 Marks Questions:
1. A frequency distribution of the life times of 400 T.V. picture tubes tested in a tube company is given below. Find the average life of tube.
Solution:
2. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency ‘f’.
Solution:
⇒ 18(44 + f) = 752 + 20 f
⇒ 792 + 18 f = 752 + 20 f
⇒ 792 – 752 = 20 f – 18 f
⇒ 2f = 40
⇒ f = 20
3. To find out the concentration of SO2 in the air (in parts per million i.e. ppm), the data was collected for 30 localities in a certain city and is presented below:
Solution:
∴ The mean concentration of SO2 in the air = 0.099 ppm
4. Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarized as shown. Find the mean heart beats per minute for these women, choosing a suitable method.
Solution:
5. The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?
Thus, the average number of wickets taken by these 45 bowlers in one-day cricket is 152.89
Note: Even if the class sizes are unequal, and xi are large numerically, we can still apply the step-deviation method by taking ‘h’ to be a suitable divisor of all the di s.
6. A student noted the number of cars passing through a spot on a road for 100 periods; each of 3 minutes, and summarized this in the table given below.
Find the mode of the data.
Solution:
Here is the maximum class frequency is 20, and the class corresponding to this frequency is 40-50. So, the modal class is 40-50.
7. The median of the following data is 525. Find the values of x and y, if the total frequency is 100.
Solution:

8. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Find the median length of the leaves.
Solution:
2 Marks Questions:
1. Following table shows the weight of 12 students is given. Find the mean weight of the students.
Solution:
2. If the mean of the following distribution is 6, find the value of ‘p’
Solution:
3. Find the median for the following frequency distribution
Solution:
Here N = 120 ⇒ = 60
We find that the cumulative frequency just greater than i.e. 60 is 65 and the value of x corresponding to 65 is 5. Therefore Median = 5.
4. Prepare tables to draw less than cumulative frequency curve and greater than cumulative frequency curve for the following frequency distribution (No need to draw graphs)
Solution:
For Less than cumulative frequency curve
For Greater than cumulative frequency curve
1 Mark Questions:
1. Write the principle to find mean of the grouped data by direct method and explain the terms in the principle.
Solution:
Here fi = frequencies and xi = observations
2. Write the principle to find mean of the grouped data by deviation method and explain the terms in the principle.
Solution:
Here A = Assumed Arithmetic Mean, fi = frequencies and di = xi – A (deviations)
3. Write the principle to find mean of the grouped data by step deviation method and explain the terms in the principle.
Solution:
Here A = Assumed Arithmetic Mean, fi = frequencies and (deviations), h = height of the class.
4. Write the principle to find mode of the grouped data and explain the terms in the principle.
Solution:
Here l = lower boundary of the modal class, h = size of the modal class
f1 = frequency of the modal class, f0 = frequency of the class preceding the modal class
f2 = frequency of the class succeeding the modal class.
5. Write the principle to find median of the grouped data and explain the terms in the principle.
Solution:
Here l = lower boundary of median class, n = number of observations
cf = cumulative frequency of class preceding the median class
f = frequency of median class, h = class size (size of the median class)
Dr. T.S.V.S. Suryanarayana Murthy