Statistics Class 10th Mathematics CBSE Solution

Class 10^th Mathematics CBSE Solution

Exercise 14.1

1. A survey was conducted by a group of students as a part of their environment…
2. Consider the following distribution of daily wages of 50 workers of a factory…
3. The following distribution shows the daily pocket allowance of children of a…
4. Thirty women were examined in a hospital by a doctor and the number of heart…
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes.…
6. The table below shows the daily expenditure on food of 25 households in a…
7. To find out the concentration of SO2 in the air (in parts per million, i.e.,…
8. A class teacher has the following absentee record of 40 students of a class for…
9. The following table gives the literacy rate (in percentage) of 35 cities. Find…

Exercise 14.2

1. The following table shows the ages of the patients admitted in a hospital…
2. The following data gives the information on the observed lifetimes (in hours)…
3. The following data gives the distribution of total monthly household…
4. The following distribution gives the state-wise teacher-student ratio in higher…
5. The given distribution shows the number of runs scored by some top batsmen of…
6. A student noted the number of cars passing through a spot on a road for…

Exercise 14.3

1. The following frequency distribution gives the monthly consumption of…
2. If the median of the distribution given below is 28.5, find the values of x and…
3. A life insurance agent found the following data for distribution of ages of 100…
4. The lengths of 40 leaves of a plant are measured correct to the nearest milli…
5. The following table gives the distribution of the life time of 400 neon lamps:…
6. 100 surnames were randomly picked up from a local telephone directory and the…
7. The distribution below gives the weights of 30 students of a class. Find the…

Exercise 14.4

1. The following distribution gives the daily income of 50 workers of a factory…
2. During the medical check-up of 35 students of a class, their weights were…
3. The following table gives production yield per hectare of wheat of 100 farms of…

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Exercise 14.1

Question 1.

A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house
Which method did you use for finding the mean, and why?

Answer:

The above given data can be represented in the form of table as below:

Mean can be calculated as follows:

where f_i = frequency of ith class and
x_i = mid value of ith class

= 8.1

We will use the direct method in this as the values of xi and fi are small. 
You can also use assumed mean method, but it's not necessary as the values are very small and assumed mean method is better for large values.

Question 2.

Consider the following distribution of daily wages of 50 workers of a factory
Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer:

The above given data can be represented in the form of table as below:
let a = 150 [assumed mean]

Now, mean of the deviation can be calculated as follows:

= -4.8

Mean can be calculated as follows:

x = d + a

x = -4.8 + 150

x = 145.20

Method 2:
Now we can also calculate mean by the formula:

Therefore, from the table,

∑f_ix_i = 7260

∑f_i = 50

Therefore,

Mean = 7260/50

Mean = 145.20

Question 3.

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

Answer:

The above given data can be represented in the form of table as below:

We can find the value of f as follows:

18

18( 44 + f ) = 752 + 20f

792 + 18f = 752 + 20f

2f = 40

f = 20

Question 4.

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method

Answer:

The above given data can be represented in the form of table as below:

Let the assumed mean for the given data be, a = 75.5
Now, mean of the deviation can be calculated as follows:

where, f_i = frequency of the ith class
d_i = deviation from assumed mean of the ith class = x_i - a

deviation 

= 0.4

Mean can be calculated as follows:

 = 

= 0.4 + 75.5

= 75.9

Mean heartbeat for women = 75.9 
Question 5.

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes

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

Answer:

The above-given data can be represented in the form of a table as below:
Let a = 57

Now, the mean of the deviation can be calculated as follows:

where ,
f_i = frequency of the ith class
d_i = x_i - a
a = assumed mean of the data

= 0.1875

Mean can be calculated as follows:

 = 

where, a = assumed mean of the data

= 0.1875 + 57

Mean = 57.1875
≈ 57.19
Hence, the mean of the given data is 57.19.

Question 6.

The table below shows the daily expenditure on food of 25 households in a locality
Find the mean daily expenditure on food by a suitable method

Answer:

Solving by short-cut method:
The above given data can be represented in the form of table as below:

Formula of mean is given by

where, a = assumed mean
f_i = frequency of the ith class
h = class width 

= Rs. 211
So, the mean of the data is Rs. 211

Question 7.

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:
Find the mean concentration of SO₂ in the air

Answer:

The above given data can be represented in the form of table as below:

Mean can be calculated as follows:

where f_i = frequency of ith class
and x_i = middle point of ith class

= 0.099 ppm

Therefore, Mean concentration of SO₂ in the air is 0.099 ppm
Assumed Mean method is not feasible for this question, because the values are too small and taking the steps will make the calculation tougher. Use Assumed mean or step deviation method when the values are larger and can be reduced with the help of steps. 
Question 8.

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent

Answer:

The above given data can be represented in the form of table as below:

Mean can be calculated as follows:

where, f_i = frequency of the ith class
and, x_i = midpoint of the ith class

= 12.4 (approx.)

Question 9.

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate

Answer:

The above-given data can be represented in the form of a table as below:

The formula for mean is given by



where, a is the assumed mean, f_i is the frequency ,


u_i = ( x_i - a) / h

= 69.43

Therefore, the mean literacy rate is 69.43 %.

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Exercise 14.2

Question 1.

The following table shows the ages of the patients admitted in a hospital during a year:
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer:

As per the question:

Modal class = 35 -45

l = 35

h = 10

f1 = 23

f0 = 21

f2 = 14

= 

= 36.8

The above given data can be represented in the form of table as below:

= 35.37

The mode of the data shows that maximum number of patients in the age group of 36.8, whereas the average age of all the patients is 35.37.

Question 2.

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Determine the modal lifetimes of the components

Answer:

As per the question:

Modal class = 60-80

l = 60

h = 20

f₁ = 61

f₀ = 52

f₂ = 38

Or Mode = 60 + 5.625

or Mode = 65.62

Thus, the modal lifetime of 225 electrical components is 65.62 hours

Question 3.

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Answer:

As per the question:

Modal class = 1500-2000

l = 1500

h = 500

f1 = 40

f0 = 24

f2 = 33

Formula for calculating mode is

where,
l = lower limit of modal class
f₁ = frequency of the modal class
f₀ = frequency of the class before modal class
f₂ = frequency of the class after modal class
h = width of modal class

Therefore,
Mode 

Mode 

= 1500 + 347.82

= 1847.82

Mode of the data is Rs.1847.82

The above given data can be represented in the form of table as below:

Hence, the mean can be calculated as below:

where a = assumed mean
f_i = frequency of the ith class
h is the class width

Mean 

Mean = 2750 -87.5

Mean = Rs. 2662.50

Question 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

Answer:

As per the question:

Modal class = 30-35

l = 30

h = 5

f1 = 10

f0 = 9

f2 = 3

= 

= 30.625

Hence, the mean can be calculated as below:

= 32.5 -

= 29.22

Question 5.

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches


Find the mode of the data

Answer:

As per the question:

Modal class = 4000-5000

l = 4000

h = 1000

f1 = 18

f0 = 4

f2 = 9

= 

= 4608.70s

Question 6.

A student noted the number of cars passing through a spot on a road for 100periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

Answer:

For finding the mode, first we find the modal class i.e. class with maximum frequency
In the given data, Modal class is 40 - 50

and then we use the following formula for finding the mode


Where

l, lower limit of modal class = 40

h, width of modal class = 10

f1, frequency of modal class = 20

f0, frequency of class preceding modal class = 12

f2, frequency of class exceeding modal class = 11

Putting the values, we get

= 

= 44.70

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Exercise 14.3

Question 1.

The following frequency distribution gives the monthly consumption of electricity of68 consumers of a locality. Find the median, mean and mode of the data and compare them

Answer:

The cumulative frequency of the table can be represented as:

N= 68

where,
l = lower limit of the median group
n = total frequency
c.f = cumulative frequency of the group before median group
f = frequency of median group
W = Group Width

Hence,

Median class = 125 - 145

Cumulative frequency = 42

Lower limit, l = 125

cf = 22

f = 20

h = 20

Hence,

Median can be calculated as:

= 125 + 12

= 137

Now, mode can be calculated as:

where,
l = lower limit of the modal class
f₁ = absolute frequency of the modal class
f₀ = absolute frequency of the class before modal class
f₂ = absolute frequency of the class after modal class
h = class width

Modal class = 125-145

l = 125

h = 20

f₁ = 20

f₀ = 13

f₂ = 14

= 

= 125 + 10.76

= 135.76

Now, mean of the following data can be calculated as:

where, a = assumed mean
f_i = frequency of ith term
u_i = a - x_i / h
h = class width

= 137.05

Hence,

Mean, Median and Mode are more or less equal in this distribution.

Question 2.

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

Answer:

Let's make a cumulative frequency table for the above problem

Total frequency, N= 60

Now,
Given median = 28.5, lies in 20 - 30

Median class = 20-30
frequency corresponding to median class, f = 20
cumulative frequency of the class preceding the median class, cf = 5 + x
Lower limit, l = 20
class height, h = 10

Now,

Median can be calculated as:

28.5 

28.5 - 20 = 
8.5 = 

25 –x =8.5 × 2
⇒ 25 - x = 17

⇒ x = 25-17

⇒ x = 8

Now,

From the cumulative frequency we can find the value of x + y as:

45 +x +y = 60
⇒ x + y = 60 - 45

⇒ x + y = 15

⇒ y = 15 – x
as, x = 8

⇒ y = 15 – 8

⇒y = 7

Hence,

Value of x = 8 and y = 7

Question 3.

A life insurance agent found the following data for distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years on wards but less than 60 year

Answer:

In this case, we are given less than (or below) cumulative frequency distribution,we need to convert it into normal frequency distribution.
So, we need to find class intervals and corresponding frequency.

Since, the difference between ages in each class is 5, we can take the first class interval as 15 - 20 and its frequency will be same as frequency of below 20 class.
Also, for other class, class interval will can be found as following and corresponding frequency can be find by subtracting the previous frequency from the cumulative frequency.

As per the question,

N=100

Hence,

Median class = 35-45

Cumulative frequency = 100

Lower limit, l = 35

cf = 45

f = 33

h = 5

Now,

Median can be calculated as:

where,
l = lower limit of median class
n = total frequency of the data
cf = cumulative frequency of the class before median class
f = frequency of the median class

Median = 35 + 

Median = 35.75 years

Question 4.

The lengths of 40 leaves of a plant are measured correct to the nearest milli meter, 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 to117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5)

Answer:

The cumulative frequency of the data can be calculated as:

As per the question,

N= 40

Hence,

Median class =144.5-153.5

Lower limit, l = 144.5

cf = 17

f = 12

h = 9

Now,

Median can be calculated as:

= 144.5 + 

= 146.75

Question 5.

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


Find the median life time of a lamp

Answer:

The cumulative frequency of the given data can be calculated as:

As per the question,

N= 400

Hence,

Median class = 3000-3500

Now,

Median class = 3000-3500
frequency corresponding to median class, f = 86
cumulative frequency of the class preceding the median class, cf = 130
Lower limit, l = 3000
class height, h = 500

Now,

Median can be calculated as:

= 3000 + 406.97

= 3406.97

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

Answer:

The cumulative frequency of the given data can be calculated as:

As per the question,

N= 100

Hence,

Median class = 7-10

Lower limit, l = 7

cf = 36

f = 40

h = 3

Now,

Median can be calculated as:

h

= 8.05

Now, mode can be calculated as: class corresponding to maximum frequency.

Modal class = 7-10

l = 7

h = 3

f1 = 40

f0 = 30

f2 = 16

= 

= 7.88

Now, mean of the following data can be calculated as:

= 8.25

Question 7.

The distribution below gives the weights of 30 students of a class. Find the median weight of the students

Answer:

The cumulative frequency of the given data can be calculated as:

As per the question,

N= 30

Hence,

Median class = 55-60

Lower limit, l = 55

cf = 13

f = 6

h = 5

Now,

Median can be calculated as:

Median = 56.67
Median weight is 56.57 kg

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Exercise 14.4

Question 1.

The following distribution gives the daily income of 50 workers of a factory


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

Answer:

The less than type cumulative frequency distribution of given data can be found as follows, Here previous cumulative frequencies are added to current frequency to find the cumulative frequency of any class.

Now,

Taking upper class interval on x-axis and their respective frequencies on y-axis, ogive will be:

Question 2.

During the medical check-up of 35 students of a class, their weights were recorded as follows:


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

Answer:

The frequency distribution table of less than type graph is as follows:

Now,

Taking upper class interval on x-axis and their respective frequencies on y-axis, ogive will be:

Here, N = 35

Mark the point A whose ordinate is 17.5 and is x-ordinate is 46.5.

Hence,

Median of the data is 46.5

Now,

It can be observed that the difference between two consecutive upper class limits is 2

The class marks with respective frequencies are obtained below:

We can see that the cumulative frequency is greater than n/2 and is 28 which belongs to the interval 46-48

Hence,

Median class = 46-48

Lower limit, l = 46

cf = 14

f = 14

h = 2

Now,

Median can be calculated as:

= 46.5

Question 3.

The following table gives production yield per hectare of wheat of 100 farms of a village

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

Answer:

The frequency distribution table of more than type graph is as follows:

Now,

Taking lower limit on x-axis,

Cumulative Frequencies on y- axis,

Its ogive can be drawn as: