NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Q1 ) 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 :

The class 35-45 has the maximum frequency, therefore, this is the modal class.

Here l = 35 , h = 10 , f₁ = 23 , f₀ = 21 , f₂ =14

Now, let us substitute these values in the formula

\(Mode \ = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 35 \ + \ \frac{23 \ - \ 21}{2 \ × \ 23 \ - \ 21 \ - \ 14} \ × \ 10 \)

\(= \ 35 \ + \ \frac{2}{46 \ - \ 21 \ - \ 14} \ × \ 10 \)

\(= \ 35 \ + \ \frac{2}{11} \ × \ 10 \)

\(= \ 35 \ + \ 1.8 \ = \ 36.8 \)

Now, to calculate the Mean,

The formula for mean is :

\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= \ \frac{2830}{80} \ = \ 35.37 \) years

\(\therefore \) The mean of the given data = 35.37 years

Q2 ) The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetimes of the components.

Answer :

The class 60-80 has the maximum frequency, therefore, this is the modal class.

Here l = 60 , h = 20 , f₁ = 61 , f₀ = 52 , and f₂ = 38

Now, let us substitute these values in the formula

\(Mode \ = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\( = \ 60 \ + \ ( \frac{61 \ - \ 52}{2 \ × \ 61 \ - \ 52 \ - \ 38}) \ × \ 20 \)

\(= \ 60 \ + \ ( \frac{61 \ - \ 52}{122 \ - \ 52 \ - \ 38}) \ × \ 20 \)

\(= \ 60 \ + \ \frac{9}{32} \ × \ 20 \)

\( = \ 60 \ + \ 5.625 \ = \ 65.625 \)

\(\therefore \) The modal lifetimes of the components is 65.625 hours.

Q3 ) 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 :

The class 1500 – 2000 has the maximum frequency. Therefore, it is the modal class.

Here l = 1500,
h = 500,
f₁ =40 ,
f₀ =24 and
f₂ =33

Now, let us substitute these values in the formula

Mode \(= \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 1500 \ + \ ( \frac{40 \ - \ 24}{2 \ × \ 40 \ - \ 24 \ - \ 33}) \ × \ 500\)

\(= \ 1500 \ + \ \frac{40 \ -\ 24}{80 \ - \ 24 \ - \ 33} \ × \ 500 \)

\(= \ 1500 \ + \ \frac{16}{23} \ × \ 500 \)

\(= \ 1500 \ + \ 347.83 \ = \ 1847.83 \)

\(\therefore \) Modal monthly expenditure = Rs. 1847.83

Let the assumed mean be A = 2750 and h = 500.

The formula to calculate the mean,

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= \ 2750 \ + \ ( \frac{-35}{200}) \ × \ 500 \)

\( = \ 2750 \ - \ 87.50 \ = \ 2662.50 \)

So, the mean monthly expenditure of the families is Rs 2662.50

Q4 ) 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 :

The class 30-35 has the maximum frequency, therefore, this is the modal class.

Here l = 30,
h = 5, f₁ = 10,
f₀ = 9 and
f₂ = 3

Now, let us substitute these values in the formula

Mode \(= \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\( = \ 30 \ + \ ( \frac{10 \ - \ 9}{2 \ × \ 10 \ - \ 9 \ - \ 3}) \ × \ 5 \)

\(= \ 30 \ + \ \frac{10 \ - \ 9}{20 \ - \ 9 \ - \ 3} \ × \ 5 \)

\(= \ 30 \ + \ \frac{1}{8} \ × \ 5 \ = \ 30 \ + \ 0.625 \)

\( = \ 30.625 \ = \ 30.6 \)

Calculation of mean :

\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= \ \frac{1022.5}{35} \ = \ 29.2 \)

\(\therefore \) mean = 29.2

Q5 ) 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 :

The class 4000-5000 has the maximum frequency ,
therefore, this is the modal class.

Here l = 4000,
h = 1000,
f₁ = 18,
f₀ = 4 and
f₂ = 9

Now, let us substitute these values in the formula

Mode \(= l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 4000 \ + \ ( \frac{18 \ - \ 4}{2 \ × \ 18 \ - \ 4 \ - \ 9}) \ × \ 1000 \)

\(= \ 4000 \ + \ \frac{18 \ - \ 4}{36 \ - \ 4 \ - \ 9} \ × \ 1000 \)

\(= \ 4000 \ + \ \frac{14}{23} \ × \ 1000 \)

\(= \ 4000 \ + \ 608.695 \ = 4608.7 \)

\(\therefore \) The mode of the given data is 4608.7 runs.

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:

Answer :

The class 40-50 has the maximum frequency, therefore, this is the modal class.

Here l = 40, h =10, f₁ = 20, f₀ = 12 and f₂ = 11

Now, let us substitute these values in the formula
Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 40 \ + \ ( \frac{20 \ - \ 12}{40 \ - \ 12 \ - \ 11}) \ × \ 10 \)

\( = \ 40 \ + \ \frac{8}{17} \ × \ 10 \)

\( = \ 40+4.7 \ = \ 44.7 \)

∴ Mode = 44.7 cars

Q1 ) The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.

Answer :

We have n = 68
\( \Rightarrow \ \frac{n}{2} \ = \ \frac{68}{2} \ = \ 34 \)

The cumulative frequency just greater than \( \frac{n}{2} \) , is 42 and the corresponding class is 125-145.

Thus, 125-145 is the median class such that \( \frac{n}{2} \ = \ 34 , \)
l = 125 ,
c.f. = 22 ,
f = 20 , h = 20.

Median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f}) \ × \ h \),

Median \( = \ 125 \ + \ ( \frac{34 \ - \ 22}{20}) \ × \ 20 \ \)
\( = \ 125 \ + \ 12 \ = \ 137 \) units

To calculate the mode :

Modal class = \(125 - 145 , \)
\( f_1 = 20 , \)
\( f_0 = 13 ,\)
\( f_2 = 14, \) and
\( \ h = 20 \)

Mode formula :

Mode = \( l + \frac{(f_1- f_0)}{(2f_1- f_0 - f_2)} × h \)

Mode = \( 125 + \frac{(20 -13)}{(40 -13 -14)} × 20 \)

\( = 125 + \frac{140}{13} \)

\( = 125 + 10.77 = 135.77 \)

Therefore, mode = 135.77

Calculation of mean :

Let the assumed mean, A = 135, class interval, h = 20

so \(u_i \ = \ \frac{x_i \ - \ A}{h} \ = \ \frac{x_i \ - \ 135}{20} \)

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\( = \ 135 \ + \ 20( \frac{7}{68}) \)

Mean = 137.05

In this case, mean, median and mode are more/less equal in this distribution.

Q2 ) If the median of a distribution given below is 28.5 then, find the value of an x &y.

Answer :

Given that median is 28.5 and \(n \ = \ \sum f_i \ = \ 60 \)

Here n = 60
\( \Rightarrow \ \frac{n}{2} \ = \ 30 \)

Since the median is given to be 28.5, thus the median class is 20-30.

\(\therefore \) l = 20, h = 10, f = 20 and cf = 5 + x

\(\therefore \) median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(\Rightarrow \ 28.5 \ = \ 20 \ + \ \frac{30 \ - \ 5 \ - \ x}{20} \ × \ 10 \)

\( \Rightarrow \ 57 \ = \ 40 \ + \ 25 \ - \ x \)

\( \Rightarrow \ x \ = \ 65 \ - \ 57 \ \)

\(\Rightarrow \ x \ = \ 8 \)

Also, 45 + x + y = 60
[ As n = 60 ]

\(\Rightarrow \ 45 \ + \ 8 \ + \ y \ = \ 60 \)

\(\Rightarrow \ y \ = \ 60 \ - \ 53 \ \)

\( \Rightarrow \ y \ = \ 7 \)

\(\therefore \) x = 8, y = 7

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.

Answer :

We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute median .

Here \( n \ = \ \sum f_i \ = \ 100 \ => \ \frac{n}{2} \ = \ 50 \)

We see that the cumulative frequency just greater than \( \frac{n}{2} \), i.e., 50 is 78 and the corresponding class is 35 - 40.

So, 35 - 40 is the median class.

\(\therefore \frac{n}{2} \ = \ 50 , l = 35, c.f = 45 , f = 33, h = 5 \)

Now, let us substitute these values in the formula

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(= \ 35 \ + \ \frac{50 \ - \ 45}{33} \ × \ 5 \)

\( = \ 35 \ + \ \frac{5}{33} \ × \ 5 \)

\( = \ 35 \ + \ 0.76 \ = \ 35.76 \)

\(\therefore \) The median age = 35.76 years

Q4 ) 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 leaves.

Answer :

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

So, n = 40 and \( \frac{n}{2} \ = \ 20 \)

Median class = 144.5 -153.5

then, \( l = 144.5, cf = 17, f = 12, h = 9 \)

Now we know that, Median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(\Rightarrow \) Median \( = \ 144.5 \ + \ ( \frac{20 \ - \ 17}{12}) \ × \ 9 \)

\( = \ 144.5 \ + \ \frac{3}{12} \ × \ 9 \)

\( = \ 144.5 \ + \ 2.25 \ = \ 146.75 \)

\(\therefore \) The median length of the leaves = 146.75 mm.

Q5 ) The following table gives the distribution of the life time of 400 neon lamps :

Find the median lifetime of a lamp.

Answer :

n = 400
\(\Rightarrow \frac{n}{2} \ = \ 200 \)

Median class = 3000 - 3500

\(\therefore \) \( l = 3000, cf = 130, f = 86, h = 500 \)

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

Median \(= \ 3000 \ + \ ( \frac{200 \ - \ 130}{86}) \ × \ 500 \)

\(= \ 3000 \ + \ \frac{70}{86} \ × \ 500 \)

\(= \ 3000 \ + \ 406.98 \)

\(= \ 3406.98 \)

\(\therefore \) median lifetime of lamps = 3406.98 hours

Q6 ) In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Answer :

We have n = 100
\(\Rightarrow \frac{n}{2} \ = \ 50 \)

The cumulative frequency just greater than \( \frac{n}{2} \) is 76 and the corresponding class is 7-10 .

Thus 7-10 is the median class such that

\( \frac{n}{2} \ = \ 50 , l = 7, cf = 36 , f = 40 and h = 3 \)

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\( = \ 7 \ + \ ( \frac{50 \ - \ 36}{40}) \ × \ 3 \)

\( = \ 7 \ + \ \frac{14}{40} \ × \ 3 \)

\(= \ 7 \ + \ 1.05 \ = \ 8.05 \)

Now, to calculate the mean,

\(\therefore \) Mean = \( \overline{x} \ = \ \frac{ \sum f_i x_i}{ \sum f_i} \)

\( = \ \frac{832}{100} \)

Therefore, Mean = \( 8.32 \)

Calculation of mode :

The class 7-10 has the maximum frequency therefore, this is the modal class.

Here l = 7 , h = 3, f₁ = 40 , f₀ = 30 and f₂ = 16

Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 7 \ + \ \frac{40 \ - \ 30}{80 \ - \ 30 \ - \ 16} \ × \ 3 \)

\(= \ 7 \ + \ \frac{10}{34} \ × \ 3 \)

\( = \ 7 \ + \ 0.88 \ = \ 7.88 \)

\(\therefore \) Median = 8.05 , Mean = 8.32 and Mode = 7.88

Q7 ) The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer :

We have n = 30 \( \Rightarrow \ \frac{n}{2} \ = \ 15 \)

The cumulative frequency just greater than \( \frac{n}{2} \) is 19 and the corresponding class is 55 - 60.

Thus, 55 - 60 is the median class such that :

\( \frac{n}{2} = 15 , l = 55 , f = 6 , cf = 13 \ and \ h = 5 \)

Median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(= \ 55 \ + \ ( \frac{15 \ - \ 13}{6}) \ × \ 5 \)

\(= \ 55 \ + \ \frac{2}{6} \ × \ 5 \ = \)

\( 55 \ + \ 1.67 \ = \ 56.67 \)

\(\therefore \) The median weight = 56.67 kg .

Q1 ) 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 :

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve

Q2 ) 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 :

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale.

Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.



Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the mode by making a table.

The class 46-48 has the maximum frequency, therefore, this is modal class

Here, l = 46, h = 2, f₁ = 14, f₀ = 5 and f₂ = 4

Now, Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 46 \ + \ \frac{14 \ - \ 5}{28 \ - \ 5 \ - \ 4} \ × \ 2 \ \)

\( = \ 46 \ + \ \frac{9}{19} \ × \ 2 \)

\(= \ 46 \ + \ 0.95 \ = \ 46.95 \)

Thus, mode is verified.

Q3 ) The following tables 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 :

Converting the given distribution to a more than type distribution, we get

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.