Check the below NCERT MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Statistics Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

## Statistics Class 11 MCQs Questions with Answers

**Statistics Class 11 MCQ Question 1.**

The sum of 10 items is 12 and the sum of their squares is 18. The standard deviation is

(a) 1/5

(b) 2/5

(c) 3/5

(d) 4/5

## Answer

Answer: (c) 3/5

Hint:

Given, ∑x = 12 and ∑x² = 18

Now, varience = ∑x²/n – (∑x/n)²

⇒ varience = 18/10 – (12/10)²

⇒ varience = 9/5 – (6/5)²

⇒ varience = 9/5 – 36/25

⇒ varience = (9 × 5 – 36)/25

⇒ varience = (45 – 36)/25

⇒ varience = 9/25

⇒ Standard deviation = √(9/25)

⇒ Standard deviation = 3/5

**MCQ On Statistics Class 11 Question 2.**

The algebraic sum of the deviation of 20 observations measured from 30 is 2. So, the mean of observations is

(a) 30.0

(b) 30.1

(c) 30.2

(d) 30.3

## Answer

Answer: (b) 30.1

Hint:

Given, algebraic sum of of the deviation of 20 observations measured from 30 is 2

⇒ ∑(x_{i} – 30) = 2 {1 ≤ i ≤ 20}

⇒ ∑x_{i} – 30 × 20 = 2

⇒ (∑x_{i})/20 – (30 × 20)/20 = 2/20

⇒ (∑x_{i})/20 – 30 = 0.1

⇒ Mean – 30 = 0.1

⇒ Mean = 30 + 0.1

⇒ Mean = 30.1

**Statistics MCQ Class 11 Question 3.**

The coefficient of variation is computed by

(a) S.D/.Mean×100

(b) S.D./Mean

(c) Mean./S.D×100

(d) Mean/S.D.

## Answer

Answer: (b) S.D./Mean

Hint:

The coefficient of variation = S.D./Mean

**Class 11 Statistics MCQ Question 4.**

When tested the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623. The mean of the lives of 5 bulbs is

(a) 1445

(b) 1446

(c) 1447

(d) 1448

## Answer

Answer: (b) 1446

Hint:

Given, lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623

Now, mean = (1357 + 1090 + 1666 + 1494 + 1623)/5

= 7230/5

= 1446

**MCQ Of Statistics Class 11 Question 5.**

If mode of a series exceeds its mean by 12, then mode exceeds the median by

(a) 4

(b) 8

(c) 6

(d) 12

## Answer

Answer: (b) 8

Hint:

Given, Mode = Mean + 12

⇒ Mode – 12 = Mean

Now, Mode = 3×Median – 2×Mean

⇒ Mode = 3×Median – 2(Mode – 12)

⇒ Mode = 3×Median – 2×Mode + 24

⇒ Mode + 2×Mode = 3×Median + 24

⇒ 3×Mode = 3×Median = 24

⇒ Mode = Median + 8

So, mode exceeds the median by 8

**Class 11 Maths Chapter 15 MCQ Question 6.**

The median and SD of a distributed are 20 and 4 respectively. If each item is increased by 2, the new median and SD are

(a) 20, 4

(b) 22, 6

(c) 22, 4

(d) 20, 6

## Answer

Answer: (c) 22, 4

Hint:

Since each value is increased by 2, therefore the median value is also increased by

2. So, new median = 22

Again, the variance is independent of the change of origin. So it remains the same.

**Class 11 Maths Statistics MCQ Question 7.**

Range of the data 4, 7, 8, 9, 10, 12, 13 and 17 is

(a) 4

(b) 17

(c) 13

(d) 21

## Answer

Answer: (c) 13

Hint:

Give, data are: 4, 7, 8, 9, 10, 12, 13 and 17

Range = Maximum value – Minimum Value

= 17 – 4

= 13

**MCQs Statistics Class 11 Question 8.**

If Mean = Median = Mode, then it is

(a) Symmetric distribution

(b) Asymmetric distribution

(c) Both symmetric and asymmetric distribution

(d) None of these

## Answer

Answer: (a) Symmetric distribution

Hint:

In a symmetric distribution,

Mean = Median = Mode

**Class 11 Maths Chapter 15 MCQ With Answers Question 9.**

If the difference of mode and median of a data is 24, then the difference of median and mean is

(a) 12

(b) 24

(c) 8

(d) 36

## Answer

Answer: (a) 12

Hint:

Given the difference of mode and median of a data is 24

⇒ Mode – Median = 24

⇒ Mode = Median + 24

Now, Mode = 3×Median – 2×Mean

⇒ Median + 24 = 3×Median – 2×Mean

⇒ 24 = 3×Median – 2×Mean – Median

⇒ 24 = 2×Median – 2×Mean

⇒ Median – Mean = 24/2

⇒ Median – Mean = 12

**Statistics Class 11 MCQ Questions Question 10.**

If r is the correlation coefficient, then

(a) |r| ≤ 1

(b) r ≤ 1

(c) |r| ≥ 1

(d) r ≥ 1

## Answer

Answer: (a) |r| ≤ 1

Hint:

If r is the correlation coefficient, then |r| ≤ 1

**Class 11 Maths Ch 15 MCQ Question 11.**

If the varience of the data is 121 then the standard deviation of the data is

(a) 121

(b) 11

(c) 12

(d) 21

## Answer

Answer: (b) 11

Hint:

Given, varience of the data = 121

Now, the standard deviation of the data = √(121) = 11

**Ch 15 Maths Class 11 MCQ Question 12.**

If the mean of the following data is 20.6, then the value of p is

x = 10 15 p 25 35

f = 3 10 25 7 5

(a) 30

(b) 20

(c) 25

(d) 10

## Answer

Answer: (b) 20

Hint:

Mean = ∑ f_{ i}× x_{ i} /∑ f_{ i}

⇒ 20.6 = (10 × 3 + 15 × 10 + p × 25 + 25 × 7 + 35 × 5)/(3 + 10 + 25 + 7 + 5)

⇒ 20.6 = (30 + 150 + 25p + 175 + 175)/50

⇒ 20.6 = (530 + 25p)/50

⇒ 530 + 25p = 20.6 × 50

⇒ 530 + 25p = 1030

⇒ 25p = 1030 – 530

⇒ 25p = 500

⇒ p = 500/25

⇒ p = 20

So, the value of p is 20

**MCQs On Statistics Class 11 Question 13.**

If the mean of first n natural numbers is 5n/9, then n =

(a) 5

(b) 4

(c) 9

(d) 10

## Answer

Answer: (c) 9

Hint:

Given mean of first n natural number is 5n/9

⇒ (n+1)/2 = 5n/9

⇒ n + 1 = (5n×2)/9

⇒ n + 1 = 10n/9

⇒ 9(n + 1) = 10n

⇒ 9n + 9 = 10n

⇒ 10n – 9n = 9

⇒ n = 9

**MCQ On Statistics Class 11 Maths Question 14.**

If one of the observation is zero then geometric mean is

(a) (Sum of observation)/n

(b) (Multiplication of all observations)^{n}

(c) (Multiplication of all observations)^{1/n}

(d) 0

## Answer

Answer: (d) 0

Hint:

Let the observations are 0, a, b, c, ……… up to n

Now, geometric mean = (0 × a × b × c × ……… up to n)^{1/n}

= 0

So, geometric mean is 0

**MCQ For Statistics Class 11 Question 15.**

Which one is measure of dispersion method

(a) Renge

(b) Quartile deviation

(c) Mean deviation

(d) all of the above

## Answer

Answer: (d) all of the above

Hint:

Range, Quartile deviation, Mean deviation all are the measure of dispersions method.

**MCQs Of Statistics Class 11 Question 16.**

If a variable takes discrete values x + 4, x – 7/2, x – 5/2, x – 3, x – 2, x + 1/2, x – 1/2, x + 5 (x > 0), then the median is

(a) x – 5/4

(b) x – 1/2

(c) x – 2

(d) x + 5/4

## Answer

Answer: (a) x – 5/4

Hint:

Given, discrete values x + 4, x – 7/2, x – 5/2, x – 3, x – 2, x + 1/2, x – 1/2, x + 5

Now, arrange them in ascending order, we get

x – 7/2, x – 3, x – 5/2, x – 2, x – 1/2, x + 1/2, x + 4, x + 5

Total observations = 8

Now, median = AM of 4th and 5th observations

= AM of (x – 2) and (x – 1/2) observations

= (x – 2 + x – 1/2)/2

= (2x – 5/2)/2

= x – 5/4

**Statistics Class 11 MCQs Question 17.**

If covariance between two variables is 0, then the correlation coefficient between them is

(a) nothing can be said

(b) 0

(c) positive

(d) negative

## Answer

Answer: (b) 0

Hint:

The relationship between the correlation coefficient and covariance for two variables as shown below:

r_{(x, y)} = COV(x, y)/{s_{x} × s_{y}}

r_{(x, y)} = correlation of the variables x and y

COV(x, y) = covariance of the variables x and y

s_{x} = sample standard deviation of the random variable x

s_{x} = sample standard deviation of the random variable y

Now given COV(x, y) = 0

Then r_{(x, y)} = 0

**Statistics MCQs Class 11 Question 18.**

The mean of a group of 100 observations was found to be 20. Later on, it was found that three observations were incorrect, which was recorded as 21, 21 and 18. Then the mean if the incorrect observations are omitted is

(a) 18

(b) 20

(c) 22

(d) 24

## Answer

Answer: (b) 20

Hint:

Given mean of 100 observations is 20

Now

∑ xi/100 = 20 (1 = i = 100)

⇒ ∑xi = 100×20

⇒ ∑xi = 2000

3 observations 21, 21 and 18 are recorded in-correctly.

So ∑xi = 2000 – 21 – 21 – 18

⇒ ∑xi = 2000 – 60

⇒ ∑xi = 1940

Now new mean is

∑ xi/100 = 1940/97 = 20

So, the new mean is 20

**Statistics Class 11 MCQ Maths Question 19.**

Varience is independent of change of

(a) origin only

(b) scale only

(c) origin and scale both

(d) None of these

## Answer

Answer: (a) origin only

Hint:

Varience is independent of change of origin only.

**MCQ Of Chapter 15 Maths Class 11 Question 20.**

Let x_{1}, x_{2}, x_{3}, ……… , x_{n}, be n observations and X be the arithmetic mean. Then formula for the standard deviation is given by

(a) ∑(x_{i} – mean)²

(b) ∑(x_{i} – mean)2 /n

(c) √{∑(x_{i} – mean)²/n}

(d) None of these

## Answer

Answer: (c) √{∑(x_{i} – mean)²/n}

Hint:

Given, x_{1}, x_{2}, x_{3}, ………. , x_{n} be n observations and X be the arithmetic mean.

Now standard deviation = √{∑(x_{i} – mean)²/n}

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