Statistics — Page 10 | JEE Mathematics

Q91

The frequency distribution of the age of students in a class of 40 students is given below. .tg .tg Age 15 16 17 18 19 20 No of Students 5 8 5 12

x

y

If the mean deviation about the median is 1.25, then

4x+5y

is equal to :

A 43

B 46

C 44

D 47

Correct Answer

Option C

Solution

.tg .tg Age No. of Students CF 15 5 5 16 8 13 17 5 18 18 12 30 19

x

30+x

20

y

30+x+y

\begin{aligned} & 30+x+y=40 \\ & x+y=10 \end{aligned}

Median

=\left(\frac{n+1}{2}\right)^{\text{th }}

observation

=\frac{40+1}{2}=\frac{41}{2}

Median

=18

Mean deviation about median

\begin{aligned} & 5.3+8.2+5.1+12.0+x \cdot 1+y \cdot 2=1.25 \times 40 \\ & 15+16+5+x+2 y=50 \\ & x+2 y=14 \\ & \quad x+y=10 \\ & \Rightarrow \quad x=4 \\ & \Rightarrow \quad y=6 \\ & \begin{aligned} 4 x+5 y & =24+20 \\ & =44 \end{aligned} \end{aligned}

Q92

A data consists of n observations : x1, x2, . . . . . . ., xn. If

\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n

and

\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,

then the standard deviation of this data is :

A 2

B

\sqrt 5

C 5

D

\sqrt 7

Correct Answer

Option B

Solution

\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n

\Rightarrow \sum\limits_{i = 1}^n {x_i^2} + 2\sum\limits_{i = 1}^n {{x_i}} + n = 9n\,\,\,\,\,...\,(1)

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

\Rightarrow \sum\limits_{i = 1}^n {x_i^2} - 2\sum\limits_{i = 1}^n {{x_i}} + n = 5n\,\,\,\,\,...\,(2)

Performing (1) + (2), we get

2\sum\limits_{i = 1}^n {x_i^2} + 2n = 14n

\sum\limits_{i = 1}^n {x_i^2} = 6n

Performing (1) $-$ (2), we get

\Rightarrow 4\sum\limits_{i = 1}^n {{x_i}} = 4n

$\Rightarrow$

\Rightarrow \sum\limits_{i = 1}^n {{x_i}} = n

S.D( $\sigma$ )

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

$\sigma$

= \sqrt {{{6n} \over n} - \left( 1 \right)}

$\sigma$

= \sqrt 5

Q93

For the frequency distribution : Variate (x) : x1 x2 x3 .... x15 Frequency (f) : f1 f2 f3 ...... f15 where 0 < x1 < x2 < x3 < ... < x15 = 10 and

\sum\limits_{i = 1}^{15} {{f_i}}

> 0, the standard deviation cannot be :

A 6

B 1

C 4

D 2

Correct Answer

Option A

Solution

If variate varries from m to M then variance

{\sigma ^2} \le {1 \over 4}{\left( {M - m} \right)^2}

(M = upper bound of value of any random variable, m = Lower bound of value of any random variable) Here M = 10 and m = 0 $\therefore$

{\sigma ^2} \le {1 \over 4}{\left( {10 - 0} \right)^2}

$\Rightarrow$

{\sigma ^2} \le 25

$\Rightarrow$

- 5 \le \sigma \le 5

$\therefore$ $\sigma$ $\ne$ 6

Q94

If the mean deviation about median for the numbers 3, 5, 7, 2k, 12, 16, 21, 24, arranged in the ascending order, is 6 then the median is :

A 11.5

B 10.5

C 12

D 11

Correct Answer

Option D

Solution

Median

= {{2k + 12} \over 2} = k + 6

Mean deviation

= \sum {{{|{x_i} - M|} \over n} = 6}

\Rightarrow {{(k + 3) + (k + 1) + (k - 1) + (6 - k) + (6 - k) + (10 - k) + (15 - k) + (18 - k)} \over 8}

$\therefore$

{{58 - 2k} \over 8} = 6

k = 5

Median

= {{2 \times 5 + 12} \over 2} = 11

Q95

If the mean deviation of the numbers 1, 1 + d, ..., 1 +100d from their mean is 255, then a value of d is :

A 10.1

B 20.2

C 10

D 5.05

Correct Answer

Option A

Solution

Given numbers are, 1, 1 + d, 1 + 2d . . . . . 1 + 100d $\therefore$ Total 101 number are present. $\therefore$ n = 101 $\therefore$ mean

\left( {\overline x } \right)

=

{{1 + \left( {1 + d} \right) + ......\left( {1 + 100d} \right)} \over {101}}

=

{1 \over {101}} \times {{101} \over 2}

[1 + (1 + 100d)] = 1 + 50d $\therefore$ Mean deviation from mean =

{1 \over {101}}

\left[ {\left| {1 - \left( {1 + 50d} \right)} \right| + \left| {\left( {1 + d} \right) - \left( {1 + 50d} \right)} \right|} \right.

\left. { + ...... + \left| {\left( {1 + 100d} \right) - \left( {1 + 50d} \right)} \right|} \right]

=

{{2\left| d \right|} \over {101}}

( 1 + 2 + 3 + . . . . . + 50) =

{{2\left| d \right|} \over {101}} \times {{50 \times 51} \over 2}

=

{{2550} \over {101}}\left| d \right|

From question,

{{2550} \over {101}}\left| d \right|

= 255 $\Rightarrow$

\left| d \right|

= 10.1

Q96

The outcome of each of 30 items was observed; 10 items gave an outcome

{1 \over 2}

– d each, 10 items gave outcome

{1 \over 2}

each and the remaining 10 items gave outcome

{1 \over 2}

+ d each. If the variance of this outcome data is

{4 \over 3}

then |d| equals :

A

{2 \over 3}

B

{{\sqrt 5 } \over 2}

C

{\sqrt 2 }

D 2

Correct Answer

Option C

Solution

Variance is independent of region. So we shift the given data by

{1 \over 2}

. so,

{{10{d^2} + 10 \times {0^2} + 10{d^2}} \over {30}} - {\left( 0 \right)^2} = {4 \over 3}

$\Rightarrow$ d2 $=$ 2 $\Rightarrow$

\left| d \right| = \sqrt 2