Statistics

EXERCISE : 14.3

QUESTION 1. 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.

Sol. Median

Let us prepare a cumulative frequency table:

Now, we have n = 68

This observation lies in the class 125–145.

125–145 is the median class.

l = 125, cf = 22

f = 20 and h = 20

Using the formula,

Median =

= 125 + 12 = 137 units.

Mean

Assumed mean (a) = 135

Class interval (h) = 20

Now, we have the following table:

= 137.05 units.

Mode

Class 125–145 has the highest frequency.

This is the modal class.

We have:

h = 20

l = 125

f₁ = 20

f₀ = 13

f₂ = 14

Mode =

We observe that the three measures are approximately equal in this case.

QUESTION 2. If the median of the distribution given below is 28.5, find the values of x and y.

Sol. Here, we have n = 60

Now, cumulative frequency table is:

Since, median = 28.5

Median class is 20 – 30

We have: l = 20

h = 10

f = 20

cf = 5 + x

Median =

Also 45 + x + y = 60

Thus, x = 8

y = 7

QUESTION 3. 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 years.

Sol. The given table is cumulative frequency distribution. We write the frequency distribution as given below:

We have

The cumulative frequency just greater than

i.e., just greater than 50 is 78.

The median class is 78.

Now,

Now, Median =

Median = 35 +

Thus, the median age = 35.76 years.

Statistics EXERCISE : 14.3

QUESTION 4. 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 class then change to 117.5–126.5, 126.5–135.5, …, 171.5–180.5.]

Sol. After changing the given table as continuous classes we prepare the cumulative frequency table:

Now,

The cumulative frequency just above

i.e., 20 is 29 and it corresponds to the class 144.5–153.5.

So, 144.5–153.5 is the median class.

We have:

and h = 9

Median =

= 146.75 mm.

Statistics EXERCISE : 14.3

QUESTION 5. The following table gives the distribution of the life time of 400 neon lamps:

Find the median life time of a lamp.

Sol. To compute the median, let us write the cumulative frequency distribution as given below:

We have

Since, the cumulative frequency just greater than

i.e., greater than 200 is 216.

The median class is 3000–3500

l = 3000, cf = 130, f = 86, h = 500

and

Now, Median =

= 3406.98

Thus, median life = 3406.98 hours.

Statistics EXERCISE : 14.3

QUESTION. 6. 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.

Sol. Median