Statistics — Page 4 | JEE Mathematics

Q31

The mean of set of 30 observations is 75. If each observation is multiplied by a non-zero number

\lambda

and then each of them is decreased by 25, their mean remains the same. Then

\lambda

is equal to :

A

{1 \over 3}

B

{2 \over 3}

C

{4 \over 3}

D

{10 \over 3}

Correct Answer

Option C

Solution

As mean is a linear operation, so if each observation is multiplied by $\lambda$ and decreased by 25 then the mean becomes 75 $\lambda$ $-$ 25.

According to the question, 75 $\lambda$ $-$ 25 = 75 $\Rightarrow$ $\lambda$ =

{4 \over 3}

.

Q32

The mean and the standard deviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

A 0

B 1

C 2

D 4

Correct Answer

Option C

Solution

Here mean =

\overline x

= 9 $\Rightarrow$

\overline x

=

{{\sum {{x_i}} } \over n}

= 9 $\Rightarrow$

{\sum {{x_i}} }

= 9 $\times$ 5 = 45 Now, standard deviation = 0

\therefore\,\,\,

all the five terms are same i.e.; 9 Now for changed observation

{\overline x _{new}}

=

{{36 + {x_5}} \over 5} = 10

$\Rightarrow$ x5 = 14

\therefore\,\,\,

$\sigma$ new =

\sqrt {{{\sum {{{\left( {{x_i} - {{\overline x }_{new}}} \right)}^2}} } \over n}}

=

\sqrt {{{4{{\left( {9 - 10} \right)}^2} + {{\left( {14 - 10} \right)}^2}} \over 5}}

= 2

Q33

If the standard deviation of the numbers –1, 0, 1, k is

\sqrt 5

where k > 0, then k is equal to

A 2

\sqrt 6

B

\sqrt 6

C

2\sqrt {{{5} \over 6}}

D

2\sqrt {{{10} \over 3}}

Correct Answer

Option A

Solution

standard deviation =

\sqrt 5

$\therefore$ Variance =

{\left( {\sqrt 5 } \right)^2}

= 5 Also variance =

{{\sum {x_i^2} } \over N} - {\mu ^2}

Where $\mu$ = Mean =

{{ - 1 + 0 + 1 + k} \over 4}

=

{k \over 4}

$\therefore$ Variance =

{{{{\left( { - 1} \right)}^2} + 0 + {1^2} + {k^2}} \over 4}

-

{{{k^2}} \over {16}}

$\Rightarrow$ 5 =

{{2 + {k^2}} \over 4}

-

{{{k^2}} \over {16}}

$\Rightarrow$ 8 + 3k2 = 80 $\Rightarrow$ k2 = 24 = 2

\sqrt 6

Q34

If the data x1, x2,......., x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is :

A

\sqrt 2

B 2

C 2

\sqrt 2

D 4

Correct Answer

Option B

Solution

{\sigma ^2} = {{\sum {x_i^2} } \over {10}} - {\left( {{{\sum {{x_i}} } \over {10}}} \right)^2} \to (i)

Now x1 + x2 + x3 + x4 = 44 & x5 + x6 + ......... + x10 = 96 Hence

{\sigma ^2}

=

{{2000} \over {10}} - {\left( {{{140} \over {10}}} \right)^2}

= 200 - 196 = 4 Hence $\sigma$ = 2

Q35

If both the mean and the standard deviation of 50 observations x1, x2,..., x50 are equal to 16, then the mean of (x1 – 4)2 , (x2 – 4)2 ,....., (x50 – 4)2 is :

A 400

B 480

C 380

D 525

Correct Answer

Option A

Solution

Mean(\mu ) = {{\sum {{x_i}} } \over {50}} = 16

$\therefore$

\sum {{x_i}} = 16 \times 50

S.D.\left( \sigma \right) = \sqrt {{{\sum {{x_i}^2} } \over {50}} - {{\left( \mu \right)}^2}} = 16

\Rightarrow {{\sum {{x_i}^2} } \over {50}} = 256 \times 2

Required mean =

{{\sum {{{\left( {{x_i} - 4} \right)}^2}} } \over {50}}

\Rightarrow {{\sum {{x_i}^2} + 16 \times 50 - 8\sum {{x_i}} } \over {50}}

$\Rightarrow$ 256 $\times$ 2 + 16 - 8 $\times$ 16 $\Rightarrow$ 400

Q36

The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then

{y \over x}

is equal to

A

{7 \over 2}

B

{8 \over 3}

C

{9 \over 4}

D

{7 \over 3}

Correct Answer

Option D

Solution

Given ten numbers are 10, 22, 26, 29, 34, x, 42, 67, 70, y. As the numbers are in increasing order so Mediun =

{{34 + x} \over 2}

= 35 $\Rightarrow$ x = 36 Also given mean = 42 $\Rightarrow$

{{10 + 22 + 26 + 29 + 34 + x + 42 + 67 + 70 + y} \over {10}}

= 42 $\Rightarrow$

{{300 + x + y} \over {10}}

= 42 $\Rightarrow$ 300 + x + y = 420 $\Rightarrow$ 336 + y = 420 $\Rightarrow$ y = 84 $\therefore$

{y \over x} = {{84} \over {36}}

=

{7 \over 3}

Q37

A student scores the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is

A

100 \over {\sqrt 3}

B

10 \over {\sqrt 3}

C

10 \over3

D

100 \over3

Correct Answer

Option B

Solution

Let the score in the sixth test = x Given, Mean (

\overline x

) = 48 $\Rightarrow$

{{45 + 54 + 41 + 57 + 43 + x} \over 6}

= 48 $\Rightarrow$ x = 48 Standard deviation (SD) =

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

=

\sqrt {\begin{array}{ll}{\left( {45 - 48} \right)^2} + {\left( {54 - 48} \right)^2} \\ + {\left( {41 - 48} \right)^2} + {\left( {57 - 48} \right)^2} \\ + {\left( {43 - 48} \right)^2} + {\left( {48 - 48} \right)^2} \end{array} \over 6}

=

\sqrt {{{9 + 36 + 49 + 81 + 25} \over 6}}

=

\sqrt {{{200} \over 6}}

=

\sqrt {{{100} \over 3}}

=

{{10} \over {\sqrt 3 }}

Q38

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :

A 40

B 48

C 49

D 45

Correct Answer

Option B

Solution

Given mean ( $\mu$ ) = 8 variance (

{\sigma ^2}

) = 16 No of observations (N) = 7 Let the two unknown observation = x and y We know,

{\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}

= 16 $\Rightarrow$

{{{2^2} + {4^2} + {{10}^2} + {{12}^2} + {{14}^2} + {x^2} + {y^2}} \over 7} - {\left( 8 \right)^2}

= 16 $\Rightarrow$ x2 + y2 = 100 ........(1) We know, $\mu$ =

{{\sum {{x_i}} } \over N}

= 8 $\Rightarrow$

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

= 8 $\Rightarrow$ x + y = 14 ...........(

2) As (x + y)2 = x2 + y2 + 2xy $\Rightarrow$ (14)2 = 100 + 2xy $\Rightarrow$ 196 = 100 + 2xy $\Rightarrow$ xy = 48

Q39

If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is :

A 31

B 50

C 51

D 30

Correct Answer

Option A

Solution

\sum\limits_{i = 1}^{50} {\left( {{x_i} - 30} \right) = 50}

\sum {{x_i}} = 50 \times 30 = 50

\sum {{x_i}} = 50 + 50 + 30

Mean

= \overline x = {{\sum {{x_i}} } \over n} = {{50 \times 30 + 50} \over {50}}

= 30 + 1 = 31

Q40

Let X = {x

\in

N : 1

\le

x

\le

17} and Y = {ax + b: x

\in

X and a, b

\in

R, a > 0}. If mean and variance of elements of Y are 17 and 216 respectively then a + b is equal to :

A 7

B 9

C -7

D -27

Correct Answer

Option C

Solution

Mean of X =

{{\sum\limits_{x = 1}^{17} x } \over {17}}

=

{{17 \times 18} \over {17 \times 2}}

= 9 Mean of Y =

{{\sum\limits_{x = 1}^{17} {\left( {ax + b} \right)} } \over {17}}

= 17 $\Rightarrow$

a{{\sum\limits_{x = 1}^{17} x } \over {17}} + b

= 17 $\Rightarrow$ 9a + b = 17 ....(1) Given Var(Y) = 216 $\Rightarrow$

{{\sum\limits_{x = 1}^{17} {{{\left( {ax + b} \right)}^2}} } \over {17}} - {\left( {17} \right)^2}

= 216 $\Rightarrow$

{\sum\limits_{x = 1}^{17} {{{\left( {ax + b} \right)}^2}} }

= 8585 $\Rightarrow$ (a + b)2 + (2a + b)2 +....+ (17a + b)2 = 8585 $\Rightarrow$ 105a2 + b2 + 18ab = 505 ....(

2) From equation (1) & (2) a = 3 & b = -10 $\therefore$ a + b = –7