NCERT solution for class 10 maths statistics ( Chapter 14)

Answer :

In order to find the mean value, we will use direct method because the numerical value of \(f_i \) and \( x_i \) are small.

The midpoint of the given interval is found by formula: \(Midpoint \ (x_i) \ = \ \frac{upper \ limit \ + \ lower \ limit}{2} \)

The formula to find the mean is:    \( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \ = \ \frac{162}{20} \)

\( = \ 8.1 \)

∴ The mean number of plants per house is 8.1

Answer :

The midpoint of the given interval is found by formula: \(Midpoint \ (x_i) \ = \ \frac{upper \ limit \ + \ lower \ limit}{2} \)

In this case, the value of mid-point ( \(x_i\)) is very large,so let us assume the mean value, A = 150 and class interval is h = 20.

So, \(u_i \ = \ \frac{x_i \ - \ A}{h} \ => \ u_i \ = \ \frac{x_i \ - \ 150}{20} \)

So, the formula to find out the mean is: \( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)
\( = \ 150 \ + \ 20 \ × \ \frac{-12}{50} \)
\(= \ 150 \ - \ 4.8 \)
\(= \ 145.20 \)
∴ Mean daily wage of the workers = Rs. 145.20

Answer :

Given, \( \overline{x} \ = \ 18 \)

The mean formula is\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \ => \ 18 \ = \ \frac{752 \ + \ 20f}{44 \ + \ f} \)

\(=> \ 18(44 \ + \ f) \ = \ (752 \ + \ 20f) \ => \ 792 \ + \ 18f \ = \ 752 \ + \ 20f \)

\(=> \ 792 \ - \ 752 \ = \ 20f \ - \ 18f \ => \ 40 \ = \ 2f \)

\( => \ f \ = \ 20 \)

So, the missing frequency, f = 20.

Answer :

From the given data, let us assume the mean as A = 75.5
Class size (h) = 3

So, \(u_i \ = \ \frac{x_i \ - \ A}{h} \ => \ u_i \ = \ \frac{x_i \ - \ 75.5}{h} \)

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)
\(= 75.5 \ + \ 3 \ × \ \frac{4}{30} \ = \ 75.5 \ + \ \frac{4}{10} \ = \ 75.9 \)
∴ The mean heart beats per minute for these women is 75.9

Answer :

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals are 1

Here, assumed mean (A) = 57, Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

The formula to find out the Mean is:

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i d_i}{ \sum f_i} \ = \ 57 \ + \ 3 \frac{75}{400} \)

\(= \ 57 \ + \ 0.1875 \ = \ 57.19 \)

∴ The mean number of mangoes kept in a packing box is 57.19

Answer :

Let is assume the mean (A) = 225, Class size (h) = 50

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i u_i}{ \sum f_i} \ = \ 225 \ + \ 50 \frac{-7}{25} \)

\(= \ 225 \ - \ 14 \ = \ 211 \)

∴ The mean daily expenditure on food is 211

Answer :

Calculate mean by direct method:

The formula to find out the mean is
\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \ = \ \frac{2.96}{30} \)

\( = \ 0.099 \) ppm

∴ The mean concentration of SO₂ in air is 0.099 ppm.

Answer :

The mean formula is,
\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \ = \ \frac{499}{40} \)

\(= \ 12.48 \) days

∴ The mean number of days a student was absent = 12.48.

Answer :

In this case, the value of mid-point (x_i) is very large,

so let us assume the mean value, A = 70 and class interval is h = 10.

So, \( u_i \ = \ \frac{x_i \ - \ A}{h} \)

\(=> \ u_i \ = \ \frac{x_i \ - \ 70}{10} \)

So, \( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \ = \ 70 \ + \ \frac{-2}{35} \ ? \ 10 \)

\(= \ 69.42 \)

∴ The mean literacy part is 69.42

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 calculate Mean

The mean formula is

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

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

∴ The mean of the given data = 35.37 years

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

\(Mode \ = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \ = \ 60 \ + \ ( \frac{61 \ - \ 52}{2 \ × \ 61 \ - \ 52 \ - \ 38}) \ × \ 20 \)

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

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

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

∴ The modal lifetimes of the components is 65.625 hours.

Answer :

The class 1500 – 2000 has the maximum frequency,therefore, this 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 \(= \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

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

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

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

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

∴ 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

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 \(= \ l \ + \ ( \frac{f_1 \ - \ f0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

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

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

\(= \ 30 \ + \ \frac{1}{8} \ × \ 5 \ = \ 30 \ + \ 0.625 \)
\( = \ 30.625 \ = \ 30.6 \)

Calculation of mean :

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

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

∴ mean = 29.2

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

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

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

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

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

∴ The mode of the given data is 4608.7 runs.

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

Answer :

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

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

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.

Answer :

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

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

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

∴ l = 20, h = 10, f = 20 and cf = 5 + x

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

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

\( => \ 57 \ = \ 40 \ + \ 25 \ - \ x \ => \ x \ = \ 65 \ - \ 57 \ => \ x \ = \ 8 \)

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

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

\(=> \ y \ = \ 60 \ - \ 53 \ => \ y \ = \ 7 \)

∴ x = 8, y = 7

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.
∴ \( \frac{n}{2} \ = \ 50 \) , l = 35, c.f = 45 , f = 33, h = 5
Now, let us substitute these values in the formula
Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)
\(= \ 35 \ + \ \frac{50 \ - \ 45}{33} \ × \ 5 \ = \ 35 \ + \ \frac{5}{33} \ × \ 5 \ = \ 35 \ + \ 0.76 \ = \ 35.76 \)
∴ The median age = 35.76 years

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

=> Median \( = \ 144.5 \ + \ ( \frac{20 \ - \ 17}{12}) \ × \ 9 \ = \ 144.5 \ + \ \frac{3}{12} \ × \ 9 \)

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

∴ The median length of the leaves = 146.75 mm.

Answer :

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

Median class = 3000-3500

∴ l = 3000, cf = 130, f = 86, h = 500

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

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

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

\(= \ 3000 \ + \ 406.98 \)

\(= \ 3406.98 \)

∴ median lifetime of lamps = 3406.98 hours

Answer :

We have n = 100 => \( \frac{n}{2} \ = \ 50 \)
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 \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\( = \ 7 \ + \ ( \frac{50 \ - \ 36}{40}) \ × \ 3 \ = \ 7 \ + \ \frac{14}{40} \ × \ 3 \)

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

∴ Mean = \( \overline{x} \ = \ \frac{ \sum f_i x_i}{ \sum f_i} \ = \ \frac{832}{100} \ = \ 8.32 \)

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

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

\(= \ 7 \ + \ \frac{10}{34} \ × \ 3 \ = \ 7 \ + \ 0.88 \ = \ 7.88 \)

∴ median = 8.05, mean = 8.32 and mode = 7.88

Answer :

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

The cumulative frequency just greater than n2 is 19 and the corresponding class is 55-60.

Thus, 55-60 is the median class such that n2=15, l = 55, f = 6, cf = 13 and h = 5

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

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

\(= \ 55 \ + \ \frac{2}{6} \ × \ 5 \ = \ 55 \ + \ 1.67 \ = \ 56.67 \)

∴ The median weight = 56.67 kg .

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

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.

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.