Statistics — Page 6 | JEE Mathematics

Q51

The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are :

A 10, 11

B 3, 18

C 8, 13

D 1, 20

Correct Answer

Option A

Solution

Let other two numbers be a, (21 $-$ a) Now,

10.25 = {{(4 + 16 + 25 + 49 + {a^2} + {{(21 - a)}^2}} \over 6} - {(6.5)^2}

(Using formula for variance)

\Rightarrow 6(10.25) + 6{(6.5)^2} = 94 + {a^2} + {(21 - a)^2}

\Rightarrow {a^2} + {(21 - a)^2} = 221

$\therefore$ a = 10 and (21 $-$ a) = 21 $-$ 10 = 11 So, remaining two observations are 10, 11.

$\Rightarrow$ Option (1) is correct.

Q52

If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and

{{20} \over 3}

, respectively, then the value of | a

-

b | is equal to :

A 9

B 11

C 7

D 1

Correct Answer

Option D

Solution

10 = {{7 + 10 + 11 + 15 + a + b} \over 6}

$\Rightarrow$ a + b = 17 ..... (i)

{{20} \over 3} = {{{7^2} + {{10}^2} + {{11}^2} + {{15}^2} + {a^2} + {b^2}} \over 6} - {10^2}

a2 + b2 = 145 ...... (ii) Solve (i) and (ii) a = 9, b = 8 or a = 8, b = 9 | a $-$ b | = 1

Q53

Let the mean and variance of the frequency distribution

\begin{array}{lll}{x:} & {{x_1} = 2} & {{x_2} = 6} & {{x_3} = 8} & {{x_4} = 9} \\ {f:} & 4 & 4 & \alpha & \beta \end{array}

be 6 and 6.8 respectively. If x3 is changed from 8 to 7, then the mean for the new data will be :

A 4

B 5

C

{{17} \over 3}

D

{{16} \over 3}

Correct Answer

Option C

Solution

Given 32 + 8 $\alpha$ + 9 $\beta$ = (8 + $\alpha$ + $\beta$ ) $\times$ 6 $\Rightarrow$ 2 $\alpha$ + 3 $\beta$ = 16 ..... (i) Also, 4 $\times$ 16 + 4 $\times$ $\alpha$ + 9 $\beta$ = (8 + $\alpha$ + $\beta$ ) $\times$ 6.8 $\Rightarrow$ 640 + 40 $\alpha$ + 90 $\beta$ = 544 + 68 $\alpha$ + 68 $\beta$ $\Rightarrow$ 28 $\alpha$ $-$ 22 $\beta$ = 96 $\Rightarrow$ 14 $\alpha$ $-$ 11 $\beta$ = 48 ..... (ii) from (i) & (ii) $\alpha$ = 5 & $\beta$ = 2 So, new mean =

{{32 + 35 + 18} \over {15}} = {{85} \over {15}} = {{17} \over 3}

Q54

The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation

\sqrt {13.44}

, then the standard deviation of the second sample is :

A 8

B 6

C 4

D 5

Correct Answer

Option C

Solution

n1 = 100 m = 250

\overline X

1 = 15

\overline X

= 15.6 V1(x) = 9 Var(x) = 13.44

{\sigma ^2} = {{{n_1}\sigma _1^2 + {n_2}\sigma _2^2} \over {{n_1} + {n_2}}} + {{{n_1}{n_2}} \over {{{({n_1} + {n_2})}^2}}}{({\overline x _1} - {\overline x _2})^2}

n2 = 150,

{\overline x _2}

= 16, V2(x) = $\sigma$ 2

13.44 = {{100 \times 9 + 150 \times \sigma _2^2} \over {250}} + {{100 \times 150} \over {{{(250)}^2}}} \times 1

$\Rightarrow$

{\sigma _2}^2

= 16 $\Rightarrow$ $\sigma$ 2 = 4

Q55

The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. if

\alpha

and

\sqrt \beta

are the mean and standard deviation respectively for correct data, then (

\alpha

,

\beta

) is :

A (11, 26)

B (10.5, 25)

C (11, 25)

D (10.5, 26)

Correct Answer

Option D

Solution

Given : Mean

(\overline x ) = {{\sum {{x_i}} } \over {20}} = 10

or

\Sigma

xi = 200 (incorrect) or 200 $-$ 25 + 35 = 210 =

\Sigma

xi (Correct) Now correct

\overline x = {{210} \over {20}} = 10.5

again given S.D = 2.5 ( $\sigma$ )

{\sigma ^2} = {{\sum {{x_i}^2} } \over {20}} - {(10)^2} = {(2.5)^2}

or

\sum {{x_i}^2} = 2125

(incorrect) or

\sum {{x_i}^2} = 2125 - {25^2} + {35^2}

= 2725 (correct) $\therefore$ correct

{\sigma ^2} = {{2725} \over {20}} - {(10.5)^2}

{{\sigma } ^2}

= 26 or $\sigma$ = 26 $\therefore$ $\alpha$ = 10.5, $\beta$ = 26

Q56

The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :

A

{{92} \over 5}

B

{{134} \over 5}

C

{{536} \over {25}}

D

{{112} \over 5}

Correct Answer

Option C

Solution

Let 8, 16, x1, x2, x3, x4, x5 be the observations. Now,

{{{x_1} + {x_2} + .... + {x_5} + 14} \over 7} = 8

\Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 42}

.... (1) Also,

{{x_1^2 + x_2^2 + ...x_5^2 + {8^2} + {6^2}} \over 7} - 64 = 16

\Rightarrow \sum\limits_{i = 1}^5 {x_i^2 = 560 - 100 = 460}

..... (2) So variance of x1, x2, ......., x5

= {{460} \over 5} - {\left( {{{42} \over 5}} \right)^2} = {{2300 - 1764} \over {25}} = {{536} \over {25}}

Q57

Let the mean and the variance of 5 observations x1, x2, x3, x4, x5 be

{24 \over 5}

and

{194 \over 25}

respectively. If the mean and variance of the first 4 observation are

{7 \over 2}

and a respectively, then (4a + x5) is equal to:

A 13

B 15

C 17

D 18

Correct Answer

Option B

Solution

Mean

(\overline x ) = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}

Given,

{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = {{24} \over 5}

\Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 24

...... (1) Now, Mean of first 4 observation

= {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4}

Given,

= {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4} = {7 \over 2}

\Rightarrow {x_1} + {x_2} + {x_3} + {x_4} = 14

...... (2) From equation (1) and (2), we get

14 + {x_5} = 24

\Rightarrow {x_5} = 10

Now, variance of first 5 observation

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

= {{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2}

Given,

{{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2} = {{194} \over {24}}

\Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 5\left( {{{194} \over {25}} + {{576} \over {25}}} \right)

\Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 154

\Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + {(10)^2} = 154

\Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 = 54

Now, variance of first 4 observation

= {{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2}

Given,

{{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2} = a

\Rightarrow {{54} \over 4} - {{49} \over 4} = a

\Rightarrow a = {5 \over 4}

$\therefore$

4a + {x_5}

= 4 \times {5 \over 4} + 10 = 15

Q58

The number of values of a

\in

N such that the variance of 3, 7, 12, a, 43

-

a is a natural number is :

A 0

B 2

C 5

D infinite

Correct Answer

Option A

Solution

Given, 5 numbers are 3, 7, 12, a, 43 $-$ a $\therefore$ Mean

(\overline x ) = {{3 + 7 + 12 + a + 43 - a} \over 5}

= {{65} \over 5}

= 13

We know, Variance

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

= {{{3^2} + {7^2} + {{12}^2} + {a^2} + {{(43 - a)}^2}} \over 5} - {(13)^2}

= {{9 + 49 + 144 + {a^2} + 1849 + {a^2} - 86a} \over 5} - 169

= {{2{a^2} - 86a + 2051} \over 5} - 169

= {{2{a^2} - 86a + 2051 - 845} \over 5}

= {{2{a^2} - 86a + 1206} \over 5}

= {2 \over 5}({a^2} - 43a + 603)

Variance will be natural number if

{a^2} - 43a + 603

in multiple of 5. $\therefore$

{a^2} - 43a + 603 = 5n

\Rightarrow {a^2} - 43a + 603 - 5n = 0

..... (1) $\therefore$

a = {{43\, \pm \,\sqrt D } \over 2}

Now "a" will be natural number if (1) D is a perfect square and (2)

43\, \pm \,\sqrt D

is multiple of 2 From equation (1),

D = {( - 43)^2} - 4\,.\,1\,.\,(603 - 5n)

= 1849 - 2412 + 20n

= 20n - 563

For any value of n, unit digit of 20n is always 0.

Then 20n $-$ 563 will give a number whose unit digit is 7.

For perfect square numbers ex : 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49, 82 = 64, 92 = 81, 102 = 100 So, unit digit is either 1, 4, 6, 5, 6, 9, 0 it can't be 7.

So D can't be perfect square.

Q59

The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x

A 162

B 320

C 674

D 420

Correct Answer

Option B

Solution

Mean

= {{4 + 5 + 6 + 6 + 7 + 8 + x + y} \over 8} = 6

$\therefore$

x + y = 12

..... (i) And variance

= {{{2^2} + {1^2} + {0^2} + {0^2} + {1^2} + {2^2} + {{(x - 6)}^2} + {{(y - 6)}^2}} \over 8}

= {9 \over 4}

$\therefore$

{(x - 6)^2} + {(y - 6)^2} = 8

..... (ii) From (i) and (ii) x = 4 and y = 8 $\therefore$

{x^4} + {y^2} = 320

Q60

The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to :

A 60

B 55

C 50

D 45

Correct Answer

Option A

Solution

$\because$

\overline x = 6 = {{a + b + 8 + 5 + 10} \over 5} \Rightarrow a + b = 7

...... (i) And

{\sigma ^2} = {{{a^2} + {b^2} + {8^2} + {5^2} + {{10}^2}} \over 5} - {6^2} = 6.8

\Rightarrow {a^2} + {b^2} = 25

..... (ii) From (i) and (ii) (a, b) = (3, 4) or (4, 3) Now mean deviation about mean

M = {1 \over 5}(3 + 2 + 2 + 1 + 4) = {{12} \over 5}

\Rightarrow 25M = 60