### CBSE Class 11 Maths Notes Chapter 15 Statistics

**Measure of Dispersion**

The dispersion is the measure of variations in the values of the variable. It measures the degree of scatteredness of the observation in a distribution around the central value.

**Range**

The measure of dispersion which is easiest to understand and easiest to calculate is the range.

Range is defined as the difference between two extreme observation of the distribution.

Range of distribution = Largest observation – Smallest observation.

**Mean Deviation**

Mean deviation for ungrouped data

For n observations x_{1}, x_{2}, x_{3},…, x_{n}, the mean deviation about their mean x¯ is given by

Mean deviation about their median M is given by

Mean deviation for discrete frequency distribution

Let the given data consist of discrete observations x_{1}, x_{2}, x_{3},……., x_{n} occurring with frequencies f_{1}, f_{2}, f_{3},……., f_{n} respectively in case

Mean deviation about their Median M is given by

Mean deviation for continuous frequency distribution

where x_{i} are the mid-points of the classes, x¯ and M are respectively, the mean and median of the distribution.

**Variance**

Variance is the arithmetic mean of the square of the deviation about mean x¯.

Let x_{1}, x_{2}, ……x_{n} be n observations with x¯ as the mean, then the variance denoted by σ^{2}, is given by

**Standard deviation**

If σ^{2} is the variance, then σ is called the standard deviation is given by

Standard deviation of a discrete frequency distribution is given by

Standard deviation of a continuous frequency distribution is given by

**Coefficient of Variation**

In order to compare two or more frequency distributions, we compare their coefficient of variations. The coefficient of variation is defined as

Note: The distribution having a greater coefficient of variation has more variability around the central value, then the distribution having a smaller value of the coefficient 0f variation.

### Statistics Class 11 MCQs Questions with Answers

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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.

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

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

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

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.

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}