Statistics(11th And 12th > Mathematics ) Questions and answers for exam Preparation

Question 1.

The range of the observations in 3, 8, 4, 9, 16, 19 is

16
19
8
7

Explanation:-

Answer: Option A. -> 16
:
A

Range = Highest observation - Lowest observation
= 19 − 3
= 16

Question 2.

The mean of 10 numbers is 12.5; the mean of the first six is 15 and the last five is 10. The sixth number is

15
12
18
none of these

Explanation:-

Answer: Option A. -> 15
:
A

L e t t h e m e a n o f t h e l a s t f o u r b e A_{2} . T h e n b y t h e f o r m u l a f o r c o m b i n e d m e a n, 12.5 = \frac{6 \times 15 + 4 \times A_{2}}{6 + 4}; o r 125 = 90 + 4 A_{2}; ∴ A_{2} = \frac{35}{4} L e t t h e s i x t h n u m b e r = x; t h e n t a k i n g t h e s i x t h n u m b e r a s a c o l l e c t i o n, t h e c o m b i n e d m e a n o f t h i s c o l l e c t i o n a n d t h e c o l l e c t i o n o f t h e l a s t f i v e i s 10, b y q u e s t i o n . ∴ B y d e f i n i t i o n o f c o m b i n e d m e a n 10 = \frac{1 \times x + 4 \times \frac{35}{4}}{1 + 4}; 50 = x + 35; ∴ x = 15 ∴ s i x t h n u m b e r = 15

Question 3.

If the coefficient of variation and standard deviation of a distribution are 2 and 0.4 respectively, then mean of the distribution is

40
20
10
8

Explanation:-

Answer: Option B. -> 20
:
B

$C . V = 2; σ = 0.4 (g i v e n) C . V = 100 \times \frac{s . d}{m e a n} 2 = \frac{100 \times 0.4}{m e a n} M e a n = \frac{40}{2} = 20$

Question 4.

If the mean of the set of number $x_{1}, x_{2} . . . x_{n}$ is $¯ x$ ,then the mean of the number $x_{i} + 2 i, 1 \leq\leq n$ is

$¯ x + 2 n$
$¯ x + n + 1$
$¯ x + 2$
$¯ x + n$

Explanation:-

Answer: Option B. ->

¯ x + n + 1

:
B

¯ x = \frac{\sum_{i = 1}^{n} x_{i}}{n} \Rightarrow \sum_{i = 1}^{n} x_{i} = n ¯ x ∴ \frac{\sum_{i}^{n} (x_{i} + 2 i)}{n} = \frac{\sum_{i = 1}^{n} x_{i} + 2 (1 + 2 + . . . + n)}{n} = \frac{n ¯ x + 2 \frac{n (n + 1)}{2}}{n} = ¯ x + (n + 1)

Question 5.

Mean deviation of the series a, a + d, a + 2d, a + 2nd from its mean is

$\frac{(n + 1) d}{(2 n + 1)}$
$\frac{(n d)}{(2 n + 1)}$
$\frac{n (n + 1) d}{(2 n + 1)}$
$\frac{2 (n + 1) d}{n (n + 1)}$

Explanation:-

Answer: Option C. ->

\frac{n (n + 1) d}{(2 n + 1)}

:
C

¯ x = \frac{\frac{2 n + 1}{2} (a + a + 2 n d))}{(2 n + 1)} = a + n d \sum | x - ¯ x | = 2 d (1 + 2 + . . . . . + n) = n (n + 1) d ∴ M . D = \frac{n (n + 1) d}{(2 n + 1)}

Question 6.

The means of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2, and 6, then the other two are:

2 and 9
3 and 8
4 and 7
5 and 6

Explanation:-

Answer: Option C. -> 4 and 7
:
C

Let the other two numbers be
$α a n d β .$
$¯ x = 4, N = 5 a n d \frac{\sum (x - ¯ x)^{2}}{N} = 5.2 \Rightarrow \sum (x - ¯ x)^{2} = (5.2) 5 ∴ \sum (x - ¯ x)^{2} = 26 ∴ (1 - 4)^{2} + (2 - 4)^{2} + (6 - 4)^{2} + (α - 4)^{2} + (β - 4)^{2} = 26 ∴ (α - 4)^{2} + (β - 4)^{2} = 9 A l s o, \frac{1 + 2 + 6 + α + β}{5} = 4 ∴ α + β = 20 - 9 = 11 C l e a r l y 4, 7 o n l y s a t i s f y t h e a b o v e e q u a t i o n i n α, β . H e n c e r e q u i r e d n u m b e r s a r e 4, 7.$

Question 7.

The mean deviation from the mean for the set of observations – 1, 0, 4 is

less than 3
less than 1
greater than 2.5
greater than 4.9

Explanation:-

Answer: Option A. -> less than 3
:
A

$¯ x = \frac{- 1 + 0 + 4}{3} = 1. ∴ Mean Deviation = \frac{1}{3} [| - 1 - 1 | + | 0 - 1 | + | 4 - 1 |] = 2$

Question 8.

If the S.D of a set of observations is 4 and if each observation is divided by 4, the S.D of the new set of observations will be

4
3
2
1

Explanation:-

Answer: Option D. -> 1
:
D

We know that if we multiply all numbers of a set by a constant "k" then the mean of these numbers will also be a multiple of the old mean.
And the variance of these numbers will be multiple of " $k^{2}$ ".
Since standard deviation is nothing but the square root of variance, s.d will be multiple of "k" of the old standard deviation.
We are given the s.d was 4.
So after multiplying each observation by (1/4), the s.d will also be multiplied by (1/4), thus giving the value = 4 $\times \frac{1}{4}$ = 1

Question 9.

If the mean of 3, 4, x, 7, 10 is 6, then the value of x is

4
5
6
7

Explanation:-

Answer: Option C. -> 6
:
C

6 = \frac{3 + 4 + x + 7 + 10}{5}

\Rightarrow 30 = 24 + x

\Rightarrow x = 6

Question 10.

The A.M. of a set of 50 numbers is 38. If two numbers of the set, namely 55 and 45 are discarded, the A.M. of the remaining set of numbers is

36
36.5
37.5
38.5

Explanation:-

Answer: Option C. -> 37.5
:
C

$w e h a v e, \frac{\sum x_{i}}{50} = 38, ∴ \sum x_{i} = 1900 N e w v a l u e o f \sum x_{i} = 1900 - 55 - 45 = 1800 a n d n = 48 ∴ n e w m e a n = \frac{1800}{48} = \frac{450}{12} = \frac{225}{6} = 37.5.$