Statistics — Page 5 | JEE Mathematics

Q41

If

\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} = n

and

\sum\limits_{i = 1}^n {{{\left( {{x_i} - a} \right)}^2}} = na

(n, a > 1) then the standard deviation of n observations x1 , x2 , ..., xn is :

A

a

– 1

B

n\sqrt {a - 1}

C

\sqrt {n\left( {a - 1} \right)}

D

\sqrt {a - 1}

Correct Answer

Option D

Solution

S.D =

\sqrt {{{\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} } \over n} - {{\left( {{{\sum\limits_{i = 1}^n {\left( {{x_i} - a} \right)} } \over n}} \right)}^2}}

=

\sqrt {{{na} \over n} - {{\left( {{n \over n}} \right)}^2}}

=

\sqrt {a - 1}

Q42

If the mean and the standard deviation of the data 3, 5, 7, a, b are 5 and 2 respectively, then a and b are the roots of the equation :

A x2 – 20x + 18 = 0

B 2x2 – 20x + 19 = 0

C x2 – 10x + 18 = 0

D x2 – 10x + 19 = 0

Correct Answer

Option D

Solution

Mean =

{{3 + 5 + 7 + a + b} \over 5}

= 5 $\Rightarrow$

a

+ b = 10 Variance =

{{{3^2} + {5^2} + {7^2} + {a^2} + {b^2}} \over 5}

- (5)2 = 4 $\Rightarrow$

{{a^2} + {b^2}}

= 62 $\Rightarrow$

{\left( {a + b} \right)^2} - 2ab

= 62 $\Rightarrow$

ab

= 19 So

a

and b are the roots of the equation x2 – 10x + 19 = 0

Q43

The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :

A 2

B 3

C 1

D 4

Correct Answer

Option A

Solution

\overline x = {{2 + 4 + + 10 + 12 + 14 + x + y} \over 7} = 8

x + y = 14 ....(i)

{(\sigma )^2} = {{\sum {{{({x_i})}^2}} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2}

$\Rightarrow$

16 = {{4 + 16 + 100 + 144 + 196 + {x^2} + {y^2}} \over 2} - {8^2}

$\Rightarrow$

16 + 64 = {{460 + {x^2} + {y^2}} \over 7}

$\Rightarrow$ 560 = 460 + x2 + y2 $\Rightarrow$ x2 + y2 = 100 ......(ii) Clearly by (i) and (ii), (x + y)2 - 2xy = 100 $\Rightarrow$ (14)2 - 2xy = 100 $\Rightarrow$ 2xy = 96 $\Rightarrow$ xy = 48 Now, |x - y| =

\sqrt {{{\left( {x + y} \right)}^2} - 4xy}

=

\sqrt {196 - 192}

= 2

Q44

The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :

A 5

B 3

C 7

D 9

Correct Answer

Option C

Solution

Let the two remaining observations be x and y. $\because$

\bar x = 10 = {{5 + 7 + 10 + 12 + 14 + 15 + x + y} \over 8}

\Rightarrow x + y = 17

....(1) $\because$

{\mathop {\rm var}} (x) = 13.5 = {{25 + 49 + 100 + 144 + 196 + 225 + {x^2} + {y^2}} \over 8} - {(10)^2}

\Rightarrow {x^2} + {y^2} = 169

....(2) From (1) and (2) (x, y) = (12, 5) or (5, 12) So

\left| {x - y} \right| = 7

Q45

Let xi (1

\le

i

\le

10) be ten observations of a random variable X. If

\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3

and

\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9

where 0

\ne

p

\in

R, then the standard deviation of these observations is :

A

{7 \over {10}}

B

{9 \over {10}}

C

{4 \over 5}

D

\sqrt {{3 \over 5}}

Correct Answer

Option B

Solution

Standard deviation =

\sqrt {Variance}

= \sqrt {{{\sum {x_1^2} } \over n} - {{(\overline x )}^2}}

= \sqrt {{{\sum\limits_{i = 1}^{10} {{{({x_i} - p)}^2}} } \over {10}} - {{\left( {{{\sum\limits_{i = 1}^{10} {({x_i} - p)} } \over {10}}} \right)}^2}}

[ Standard deviation is free from shifting of origin.]

= \sqrt {{9 \over {10}} - {{\left( {{3 \over {10}}} \right)}^2}}

= \sqrt {{9 \over {10}} - {9 \over {100}}}

= \sqrt {{{90 - 9} \over {100}}}

= \sqrt {{{81} \over {100}}}

= {9 \over {10}}

Q46

If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and

{{37} \over 4}

respectively, then (a

-

b)2 is equal to :

A 24

B 12

C 32

D 16

Correct Answer

Option D

Solution

Mean =

{{6 + 10 + 7 + 13 + a + 12 + b + 12} \over 8} = 9

60 + a + b = 72 a + b = 12 .....(1) variance

= {{\sum {x_i^2} } \over n} - {\left( {{{\sum {x_i^{}} } \over n}} \right)^2} = {{37} \over 4}

\sum {x_i^2} = {6^2} + {10^2} + {7^2} + {13^2} + {a^2} + {b^2} + {12^2} + {12^2} = {a^2} + {b^2} + 642

{{{a^2} + {b^2} + 642} \over 8} - {(9)^2} = {{37} \over 4}

{{{a^2} + {b^2}} \over 8} + {{321} \over 4} - 81 = {{37} \over 4}

{{{a^2} + {b^2}} \over 8} = 81 + {{37} \over 4} - {{321} \over 4}

{{{a^2} + {b^2}} \over 8} = 81 - 71

$\therefore$ a2 + b2 + 2ab = 144 80 + 2ab = 144 $\therefore$ 2ab = 64 $\therefore$

{(a - b)^2} = {a^2} + {b^2} - 2ab = 80 - 64 = 16

Q47

Let the observations xi (1

\le

i

\le

10) satisfy the equations,

\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)}

= 10 and

\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}}

= 40. If

\mu

and

\lambda

are the mean and the variance of the observations, x1 – 3, x2 – 3, ...., x10 – 3, then the ordered pair (

\mu

,

\lambda

) is equal to :

A (6, 6)

B (3, 3)

C (3, 6)

D (6, 3)

Correct Answer

Option B

Solution

\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)}

= 10 $\Rightarrow$ x1 + x2 + .... + x10 = 60 ....(1)

\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}}

= 40 $\Rightarrow$ (

x_1^2 + x_2^2 + ... + x_{10}^2

) + 25 $\times$ 10 - 10( x1 + x2 + .... + x10) = 40 $\Rightarrow$

x_1^2 + x_2^2 + ... + x_{10}^2

= 390 .....(2) From question, $\mu$ =

{{\left( {{x_1} - 3} \right) + \left( {{x_2} - 3} \right) + ... + \left( {{x_{10}} - 3} \right)} \over {10}}

=

{{60 - 3 \times 10} \over {10}}

= 3 And $\lambda$ = variance =

{{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - 3} \right)}^2}} } \over {10}}

- $\mu$ 2 =

{{\left( {x_1^2 + x_2^2 + ... + x_{10}^2} \right) + 90 - 6\left( {\sum {{x_i}} } \right)} \over {10}}

- 9 =

{{390 + 90 - 360} \over {10}}

- 9 = 12 - 9 = 3

Q48

The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is

A 3.98

B 3.99

C 4.01

D 4.02

Correct Answer

Option B

Solution

Let 20 observation be x1 , x2 ,....., x20 Mean = x1 + x2 +, .....+ x20 20 = 10 $\Rightarrow$ x1 + x2 +, .....+ x20 = 200 Variance =

{{\sum\limits_{i = 1}^{i = n} {x_i^2} } \over n} - {\left( {\overline x } \right)^2}

$\Rightarrow$ 4 =

{{x_1^2 + x_2^2 + ... + x_{20}^2} \over {20}}

- 102 $\Rightarrow$

{x_1^2 + x_2^2 + ... + x_{20}^2}

= 2080 Also x1 + x2 +, .....+ x20 - 9 + 11 = 202 new variance will be =

{{x_1^2 + x_2^2 + ... + x_{20}^2 - 81 + 121} \over {20}}

-

{\left( {{{202} \over {20}}} \right)^2}

= 3.99

Q49

The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 resepectively. Each of these 10 observations is multiplied by p and then reduced by q, where p

\ne

0 and q

\ne

0. If the new mean and new s.d. become half of their original values, then q is equal to

A 10

B -20

C -10

D -5

Correct Answer

Option B

Solution

Let observations are x1, x2, ...., x10 Here mean = 20 and standard deviation(S.D) = 2 When each of these 10 observations is multiplied by p then new observations are px1, px2, ....., px10 and new mean = 20p and new standard deviation(S.D) = 2|p| Now when Reduced by q then new observations are px1 - q, px2 - q, ....., px10 - q and new mean = 20p - q and new standard deviation(S.D) = 2|p| Given 20p - q =

{{20} \over 2}

= 10 and 2|p| =

{2 \over 2}

= 1 $\Rightarrow$ p = $\pm$

{1 \over 2}

If p =

{1 \over 2}

then q = 0 (not possible as given q $\ne$ 0) If p = -

{1 \over 2}

then q = -20

Q50

Let in a series of 2n observations, half of them are equal to a and remaining half are equal to

-

a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a2 + b2 is equal to :

A 425

B 250

C 925

D 650

Correct Answer

Option A

Solution

Given series (a, a, a, ........ n times), ( $-$ a, $-$ a, $-$ a, ...... n times) Now

\overline x

=

{{\sum {{x_i}} } \over {2n}} = 0

as, xi $\to$ xi + b then

\overline x

$\to$

\overline x

+ b So,

\overline x

+ b = 5 $\Rightarrow$ b = 5 No change in S.D. due to change in origin Standard deviation ( $\sigma$ ) =

\sqrt {{{\sum\limits_{i = 1}^{2n} {{{({x_i} - \overline x )}^2}} } \over {2n}}}

= \sqrt {{{\sum\limits_{i = 1}^{2n} {x_i^2} } \over {2n}}} = \sqrt {{{2n{a^2}} \over {2n}}} = \sqrt {{a^2}}

$\therefore$

20 = \sqrt {{a^2}} \Rightarrow a = 20

$\therefore$ a2 + b2 = 425