Measures of Central Tendency (Mean) - ICSE STD 10 Mathematics Questions

The daily incomes of a group of 10 labourers are tabulated as below. Find the mean daily income of the labourers.

Income (in ₹)	100 - 200	200 - 300	300 - 400	400 - 500
No. of labourers	2	4	3	1

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Q 2[3 marks sum]3 Marks

If the mean of the following distribution is 22, then find the value of f.

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequency	12	16	6	f	9

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Q 3[3 marks sum]3 Marks

Find the mean of the following data:

Height (in cm)	More than10	More than 20	More than 30	More than 40
No. of plants	50	36	18	5

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Q 4[3 marks sum]3 Marks

Find the mean of the following distribution:

Marks	Less than 20	Less than 40	Less than 60	Less than 80
No. of students	4	16	24	30

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Q 5[3 marks sum]3 Marks

Find the mean of the following data using step-deviation method:

Class	1 - 3	3 - 5	5 - 7	7 - 9	9 - 11
Frequency	7	8	2	2	1

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Q 6[4 marks sum]4 Marks

By using step-deviation method, find the mean marks obtained by a student from the following data:

Marks	0 and above	10 and above	20 and above	30 and above	40 and above	50 and above	60 and above	70 and above	80 and above	90 and above	100 and above
Number of students	80	77	72	65	55	43	28	16	10	8	0

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Q 7[4 marks sum]4 Marks

By using step-deviation method, compute the arithmetic mean for the following data:

Marks obtained	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50	Less than 60
Number of students	14	22	37	58	67	75

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Q 8[4 marks sum]4 Marks

Calculate the mean of the following frequency distribution, using the step-deviation method :

Class	0 - 50	50 - 100	100 - 150	150 - 200	200 - 250	250 - 300
Frequency	17	35	43	40	21	24

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Q 9[4 marks sum]4 Marks

The following table gives the wages (in ₹) of workers (per day) in a factory:

Wages (₹/day)	130 -134	134 - 138	138 - 142	142 - 146	146 - 150
No. of workers	5	8	14	12	11

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Q 10[4 marks sum]4 Marks

Using the assumed-mean method, find the mean of the following data:

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequency	7	8	12	13	10

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Q 11MCQ1 Mark

The mean of the following distribution is:

Class	1 - 3	4 - 6	7 - 9	10 - 12
(f)	5	2	1	4

A
10
B
8
C
6
D
5

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Q 12MCQ1 Mark

The mean of the following frequency distribution to the nearest integer is :

Class	0 - 2	2 - 4	4 - 6	6 - 8
(f)	2	1	4	3

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Q 13MCQ1 Mark

If $x_i^{\prime}$ s are the mid-points of the class-intervals of a grouped data, $f_i^{\prime}$ s are corresponding frequencies and $\overline{\mathrm{X}}$ is the mean, then $\Sigma\left(f_i x_i-\overline{\mathrm{X}}\right)$ is equal to:

A
0
B
1
C
-1
D
2

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Q 14MCQ1 Mark

While computing mean of a grouped data, we assume that the frequencies are:

A
evenly distributed over the classes
B
centered at the class-marks of the classes
C
centered at the upper limits of the classes
D
centered at the lower limits of the classes

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Q 15MCQ1 Mark

For a grouped frequency distribution we use $\overline{\mathrm{X}}=\mathrm{A}+\frac{\Sigma f t}{\Sigma f}$ to find the mean using step deviation method. Here $t$ is given by :

A
$x-\mathrm{A}$
B
$\frac{x+\mathrm{A}}{h}$
C
$\frac{x-\mathrm{A}}{h}$
D
$\frac{x \mathrm{~A}}{h}$

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Q 16Assertion (A) & Reason (B) MCQ1 Mark

Assertion (A) : The mean of $n$ observations is $\bar{x}$. If the first observation, is increased by 1 , second by 2 , and so on, then the new mean is $\bar{x}+\frac{n+1}{2}$.
Reason (R): Mean of $n$ observations $x_1, x_2, x_3, \ldots x_n$, with corresponding frequencies $f_1, f_2, f_3, \ldots f_n$ is given by
$
\bar{x}=\frac{x_1 f_1+x_2 f_2+\ldots x_n f_n}{f_1+f_2+\ldots f_n}
$

A
A is true, R is false.
B
A is false, R is true.
C
Both A and R are true, and R is the correct reason for A .
✓
Both A and R are true, and R is incorrect reason for A .

Answer: D.

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Q 17Assertion (A) & Reason (B) MCQ1 Mark

Assertion (A) : The mean of 10 numbers is 64 . If 5 is added to each number, the mean of new set of numbers becomes 69 .
Reason (R) : On adding or subtracting a number to each of the given observation, the mean of numbers does not change.

✓
A is true, R is false.
B
A is false, R is true.
C
Both A and R are true, and R is the correct reason for A .
D
Both A and R are true, and R is incorrect reason for A .

Answer: A.

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Q 18Assertion (A) & Reason (B) MCQ1 Mark

Assertion (A) : A student calculated the mean of a given data using short-cut method as 20.4 . He then calculated the mean of the same data using step deviation method as 20 .
Reason (R) : Mean $(\bar{x})=\mathrm{A}+\frac{\Sigma f d}{\Sigma d}$ and Mean $(\bar{x})=\mathrm{A}+\frac{\Sigma f t}{\Sigma f} \times h$

A
A is true, R is false.
✓
A is false, R is true.
C
Both A and R are true, and R is the correct reason for A .
D
Both A and R are true, and R is incorrect reason for A .

Answer: B.

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