NCERT Solution for Statistics

Which method did you use for finding the mean, and why?

Thus, mean number of plants per house is 8.1.

Since, values of x_{i} and f_{i} are small

We have used the direct method.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Class interval h = 20

We have the following table:

Now

= 145.2

Class interval h = 2

Now, we have the following table:

Since a = 16 and h = 2

Thus, the missing frequency is 20.

Class interval h = 3

Now, we have the following table:

Thus, the mean heart beat per minute is 75.9.

Find the mean number of mangoes kept in a packing box. Which of finding the mean did you choose?

Here, class interval h = 3

Now, we have the following table:

= 57.19 (approx).

Thus, the average number of mangoes per box = 57.19.

Find the mean daily expenditure on food by a suitable method.

And class interval h = 50

We have the following table:

= 211

Thus, the mean daily expenditure of food is Rs. 211.

Find the mean concentration of SO_{2} in the air.

Here, class interval h = 0.04

We have the following table:

Thus, mean concentration of SO_{2} in air is 0.099 ppm.

Thus, mean number of days a student remaind absent = 12.47. (Approx.)

Class interval h = 10

Now, we have the following table:

= 69.43% (approx.)

Thus, the mean literacy rate is 69.43%.

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Here, the highest frequency is 23.

The frequency 23 corresponds to the class interval 35 – 45.

The modeal class is 35 – 45

Now, Class size (h) = 10

Lower limit (l) = 35

Frequency of the modal class (f_{1}) = 23

Frequency of the class preceding the modal class

f_{0} = 21

Frequency of the class succeeding the modal class

f_{2} = 14

Mode =

= 36.8 years (approx.)

Let assumed mean a = 40

h = 10

Required mean = 35.37 years.

Determine the modal lifetimes of the components.

The frequency 61 corresponds to class 60 – 80

The modal class = 60 – 80

We have: l = 60

h = 20

f_{1} = 61

f_{0} = 52

f_{2} = 38

Mode =

= 65.625 hours.

Thus, the required modal life times of the components is 65.625 hours.

The maximum number of families 40 have their total monthly expenditure is in interval 1500 – 2000.

Modal class is 1500 – 2000.
**4.** The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
**Sol.** Mode:
**Mean:**
**5. ** The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
**Sol.** The class 4000–5000 has the highest frequency i.e., 18
**6.** A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
**Sol.** The class 40 – 50 has the maximum frequency i.e., 20
**EXERCISE : 14.3**
**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
**Mean**
**Mode**
**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
**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:
**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:
**Sol.** After changing the given table as continuous classes we prepare the cumulative frequency table:
**5.** The following table gives the distribution of the life time of 400 neon lamps:
**Sol.** To compute the median, let us write the cumulative frequency distribution as given below:
**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:
**Sol.** **Median**
**Mode**
**7.** The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
**Sol.** We have
**EXERCISE : 14.4**
**1.** The following distribution gives the daily income of 50 workers of a factory.
**Sol.** We have the cumulative frequency distribution as:
**2.** During the medical check-up of 35 students of a class, their wrights were recorded as follows:
**Sol.** Here, the values 38, 40, 42, 44, 46, 48, 50 and 52 are the upper limits of the respective class-intervals.
**Verfication**
**3.** The following table gives production yield per hectare of wheat of 100 farms of a village.
**Sol.** For `more than type' distribution, we have:

I = 1500, h = 500

f_{1} = 40, f_{0} = 24

f_{2} = 33

Mode =

Thus, the required modal monthly expenditure of the families is Rs. 1847.83.

Mean: Let assumed mean (a) = 3250

h = 500

We have the following table:

= 2662.5

Thus, the mean monthly expenditure = Rs. 2662.50.

Since the class 30 – 35 has the greatest frequency and h = 5

l = 30

f_{1} = 10

f_{0} = 9

f_{2} = 3

Mode =

= 30 + 0.625 = 30.6 (approx).

Let the assumed mean (a) = 37.5

Since, h = 5

We have the following table:

Find the mode of the data.

h = 1000

l = 4000

f_{1} = 18

f_{0} = 4

f_{2} = 9

Now, Mode =

= 4000 + 608.695

= 4608.7 (approx.)

Thus, the required mode is 4608.7.

f_{1} = 20, f_{0} = 12, f_{2} = 11 and h = 10

Also l = 40

Mode =

= 40 + 4.7 = 44.7

Thus, the required mode is 44.7.

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.

Assumed mean (a) = 135

Class interval (h) = 20

Now, we have the following table:

= 137.05 units.

Class 125–145 has the highest frequency.

This is the modal class.

We have:

h = 20

l = 125

f_{1} = 20

f_{0} = 13

f_{2} = 14

Mode =

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

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

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.

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.]

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.

Find the median life time of a lamp.

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.

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.

Since, the cumulative frequency just greater than i.e., greater than 50 is 76.

The class 7 – 10 is the median class,

We have

l = 7

cf = 36

f = 40

and h = 3

Median =

= 7 + 1.05 = 8.05

Mean:

Since the class 7 – 10 has the maximum frequency.

The modal class is 7 – 10

So, we have

l = 7, h = 3

f_{1} = 40

f_{0} = 30

f_{2} = 16

Mode =

= 7.88

Thus, the required

Median = 8.05, Mean = 8.32 and Mode = 7.88.

n = 30

The cumulative frequency just more than i.e., more than 15 is 19, which corresponds to the class 55–60.

l = 15

f = 6

cf = 13 and h = 5

Median =

Thus, the required median weight = 56.67 kg.

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Now, we plot the points corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on a graph paper and join them by a free hand smooth curve as shown below:

The curve so obtained is called the less than ogive.

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

We plot the points (ordered pairs) (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them by a free hand smooth curve.

The curve so obtained is the less than type ogive.

From the point (i.e., from 17.5) we draw a line parallel to the x-axis which cuts the curve at P. From this point (i.e., from P), draw a perpendicular to the x-axis, meeting the x-axis at Q. The point Q represents the median of the data which is 47.5.

To verify the result using the formula, let us make the following table in order to find median using the formula:

Here,

Since, the observation lies in the class 46 – 48.

The median class is 46 – 48 such that

l = 46, h = 2, f = 14, cf = 14

Median =

= 46.5

Thus, the median = 46.5 kg is approximately verified.

Change the distribution to a more than type distribution, and draw its ogive.

Now, we plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) and join the point with a free hand curve.

The curve so obtained is the `more than type ogive'.