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CBSE 10th Standard Maths Subject Statistics Ncert Exemplar 4 Marks Questions 2021

Published on: 26/05/2021

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Questions:

Calculate the mean of the following data.

Class	4-7	8-11	12-15	16-19
Frequency	5	4	9	10

Calculate the mean of the scores of 20 students in a Mathematics test.

Marks	10-20	20-30	30-40	40-50	50-60
Number of students	2	4	7	6	1

Determine the mean of the following distribution.

Marks	Number of students
Below 10	5
Below 20	9
Below 30	17
Below 40	29
Below 50	45
Below 60	60
Below 70	70
Below 80	78
Below 90	83
Below 100	85

The mileage (in km/L) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below:

Mileage (in km/L)	10-12	12-14	14-16	16-18
Number of cars	7	12	18	13

Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 km/L. Do you agree with this claim?

The mean of the following frequency distribution is 50, but the frequencies f₁ and f₂ in classes 20-40 and 60-80 respectively are missing. Find the missing frequencies.

Class interval	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	f₁	32	f₂	19	120

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Answers Key:

Here, class interval are not continuous. But it does not affect mid-values. So, we will solve it without making it continuous.

Class	Class marks	Frequency	f_ix_i
4-7	5.5	5	27.5
8-11	9.5	4	38
12-15	13.5	9	121.5
16-19	17.5	10	175
		\(\sum { f_{ i } } =28\)	\(\sum { f_{ i }x_{ i } } =362\)

Mean \(\left( \overline { x } \right) =\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } =\frac { 362 }{ 28 } =12.93\)

Here, we observe that 5 students have scored marks below 10, i.e it lies between class interval 0-10 and 9 students have scored marks below 20.
So, (9-5)=4 students lie in the class interval 10-20. Continuing in the same manner, we get the following frequency distribution table for given data.

Marks	Number of students (f_i)	Class marks (x_i)	\(=\frac { x_{ i }^{ u_{ i } }-45 }{ h } \)	f_iu_i
0-10	5	5	-4	-20
10-20	9-5=4	15	-3	-12
20-30	17-9=8	25	-2	-16
30-40	29-17=12	35	-1	-12
40-50	45-29=16	45	0	0
50-60	60-45=15	55	1	15
60-70	70-60=10	65	2	20
70-80	78-70=8	75	3	24
80-90	83-78=5	85	4	20
90-100	85-83=2	95	5	10
Total	\(n=\sum { f_{ i } } =85\)			\(\sum { f_{ i }u_{ i } } =29\)

Here, assumed mean(a)=45 and class width (h)=10
By step deviation method,
Mean \(\left( \overline { x } \right) =a+\left\{ \frac { \sum { f_{ i }u_{ i } } }{ \sum { f_{ i } } } \right\} \times h\)
\(=45+\left\{ \frac { 29 }{ 85 } \right\} \times 10=45+\frac { 58 }{ 17 } \\ =45+3.41=48.41\)

14.48km/L No, the manufacturer is claiming mileage 1.52km/L more than the average mileage.

f₁=28, f₂=24