Statistics — Page 9 | JEE Mathematics

Q81

For a statistical data

\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}

of 10 values, a student obtained the mean as 5.5 and

\sum_{i=1}^{10} x_i^2=371

. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is

A 5

B 7

C 9

D 4

Correct Answer

Option B

Solution

\begin{aligned} & \text{ Mean } \overline{\mathrm{x}}=5.5 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=5.5 \times 10=55 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2=371 \\ & \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text{new }}=55-(4+5)+(6+8)=60 \\ & \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text{new }}=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430 \\ & \\ & \text{ Variance } \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\ & \\ & \sigma^2=\frac{430}{10}-\left(\frac{60}{10}\right)^2 \\ & \sigma^2=43-36 \\ & \sigma^2=7 \end{aligned}

Q82

If the mean and the variance of

6,4, a, 8, b, 12,10,13

are 9 and 9.25 respectively, then

a+b+a b

is equal to :

A 103

B 106

C 100

D 105

Correct Answer

Option A

Solution

Let’s set up two equations from the given mean and variance: Mean = 9 ⇒ total sum = 8·9 = 72 Known values sum to 6+4+8+12+10+13 = 53, so

a + b = 72 - 53 = 19.

Population variance = 9.25 ⇒

\frac{\sum x_i^2}{8} - 9^2 = 9.25\quad\Longrightarrow\quad \sum x_i^2 = 8\,(81 + 9.25) = 722.

Known squares sum to 6²+4²+8²+12²+10²+13² = 529, so

a^2 + b^2 = 722 - 529 = 193.

Now use

(a + b)^2 = a^2 + 2ab + b^2 \quad\Longrightarrow\quad 19^2 = 193 + 2ab

so

361 = 193 + 2ab\quad\Longrightarrow\quad ab = 84.

Finally,

a + b + ab = 19 + 84 = 103.

Answer: 103 (Option A).

Q83

Let the mean and the standard deviation of the observation

2,3,3,4,5,7, a, b

be 4 and

\sqrt{2}

respectively. Then the mean deviation about the mode of these observations is :

A

\dfrac{1}{2}

B

\dfrac{3}{4}

C 1

D 2

Correct Answer

Option C

Solution

\begin{aligned} &\begin{aligned} & \frac{2+3+3+4+5+7+a+b}{8}=4 \\ & \Rightarrow a+b=8 \\ & (\sqrt{2})^2=\frac{2^2+3^2+3^2+4^2+5^2+7^2+a^2+b^2}{8}-16 \\ & 112+a^2+b^2=18 \times 8 \\ & \Rightarrow a^2+b^2=32 \\ & \Rightarrow a=b=4 \end{aligned}\\ &\text{ Now numbers be }\\ &2,3,3,4,4,4,5,7 \end{aligned}

Mode $=4$ Mean deviation about mode :

\begin{aligned} & \frac{|2-4|+|3-4|+|3-4|+0+0+0+|5-4|+|4-7|}{8} \\ = & \frac{2+1+1+1+3}{8}=\frac{8}{8}=1 \end{aligned}

Q84

Let the Mean and Variance of five observations

x_1=1, x_2=3, x_3=a, x_4=7

and

x_5=\mathrm{b}, a>\mathrm{b}

, be 5 and 10 respectively. Then the Variance of the observations

n+x_n, n=1,2, \ldots, 5

is

A 17

B 16

C 16.4

D 17.4

Correct Answer

Option B

Solution

First, find the values of $a$ and $b$ using the given conditions: Mean Condition: $5 = \dfrac{1 + 3 + a + 7 + b}{5}$ This implies: $a + b = 14$ Variance Condition: $\dfrac{1 + 9 + a^2 + 49 + b^2}{5} - (5)^2 = 10$ Simplifying, we get: $a^2 + b^2 = 116$ Using $a + b = 14$ , we have: $b = 14 - a$ Substitute $b$ into the equation for $a^2 + b^2$ : $a^2 + (14 - a)^2 = 116$ Expand and simplify: $a^2 + 196 - 28a + a^2 = 116$ $2a^2 - 28a + 80 = 0$ Simplify further: $a^2 - 14a + 40 = 0$ Solve the quadratic equation: $(a - 10)(a - 4) = 0$ Thus, $a = 10$ and $b = 4$ (since $a > b$ ).

New Observations: The transformed observations are: $1+1=2, \quad 3+2=5, \quad 10+3=13, \quad 7+4=11, \quad 4+5=9$ Calculate the Variance of the New Observations: First, find the mean of the new data: $\text{Mean} = \dfrac{2 + 5 + 13 + 11 + 9}{5} = 8$ Calculate the variance: $\text{Variance} = \dfrac{(2-8)^2 + (5-8)^2 + (13-8)^2 + (11-8)^2 + (9-8)^2}{5}$ $= \dfrac{36 + 9 + 25 + 9 + 1}{5}$ $= \dfrac{80}{5} = 16$ Therefore, the variance of the transformed observations is 16.

Q85

Let

\mu

be the mean and

\sigma

be the standard deviation of the distribution .tg .tg

{x_i}

0 1 2 3 4 5

{f_i}

k + 2

2k

{k^2} - 1

{k^2} - 1

{k^2} + 1

k - 3

where

\sum f_{i}=62

. If

[x]

denotes the greatest integer

\leq x

, then

\left[\mu^{2}+\sigma^{2}\right]

is equal to :

A 9

B 8

C 6

D 7

Correct Answer

Option B

Solution

We have, $\Sigma f_i=62$

\begin{aligned} & \left.(K+2)+2 K+\left(K^2-1\right)\right)+\left(K^2-1\right)+\left(K^2+1\right)+(K-3)=62 \\\\ & \Rightarrow 3 K^2+4 K-64=0 \\\\ & \Rightarrow (3 K+16)(K-4)=0 \\\\ & \Rightarrow K=4 \quad \end{aligned}

\left(\because k=\frac{-16}{3} \text{ is not possible }\right)

\begin{array}{|r|c|c|c|} \\ x_i & f_i & f_i x_i & f_i x_i^2 \\ \\ 0 & 6 & 0 & 0 \\ 1 & 8 & 8 & 8 \\ 2 & 15 & 30 & 60 \\ 3 & 15 & 45 & 135 \\ 4 & 17 & 68 & 272 \\ 5 & 1 & 5 & 25 \\ \\ \text{ Total } & 62 & 156 & 500 \\ \\ \end{array}

\mu=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{0+8+30+45+68+5}{62}=\frac{156}{62}

\begin{aligned} \sigma^2= & \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\\\ = & \frac{1 \times 8+4 \times 15+9 \times 15+16 \times 17+25 \times 1}{62}-\left(\frac{156}{62}\right)^2 \\\\ & =\frac{500}{62}-\left(\frac{156}{62}\right)^2=\frac{500}{62}-\mu^2 \end{aligned}

\begin{aligned} & \therefore \sigma^2+\mu^2=\frac{500}{62} \\\\ & \text{ Hence, }\left[\sigma^2+\mu^2\right]=8 \end{aligned}

Q86

Let M denote the median of the following frequency distribution .tg .tg Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 Frequency 3 9 10 8 6 Then 20M is equal to :

A 104

B 52

C 208

D 416

Correct Answer

Option C

Solution

.tg .tg Class Frequency Cumulative frequency 0-4 3 3 4-8 9 12 8-12 10 22 12-16 8 30 16-20 6 36

\begin{aligned} & \mathrm{M}=1+\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{\mathrm{f}}\right) \mathrm{h} \\ & \mathrm{M}=8+\frac{18-12}{10} \times 4 \\ & \mathrm{M}=10.4 \\ & 20 \mathrm{M}=208 \end{aligned}

Q87

The mean of 5 observations is 5 and their variance is 124. If three of the observations are 1, 2 and 6 ; then the mean deviation from the mean of the data is :

A 2.4

B 2.8

C 2.5

D 2.6

Correct Answer

Option B

Solution

Let 5 observations are x1, x2, x3, x4, x5 given, x1 = 1, x2 = 2, x3 = 6 Mean = 5 $\therefore$ Mean

\left( {\overline x } \right)

=

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

= 5 $\Rightarrow$ 1 + 2 + 6 + x4 + x5 = 25 $\therefore$ x4 + x5 = 16 $\Rightarrow$ (x4 $-$ 5) + (x5 $-$ 5) + 10 = 16 $\Rightarrow$ (x4 $-$ 5) + (x5 $-$ 5) = 6 $\therefore$ Mean deviation about mean, =

{{\sum {\left| {{x_i} - \overline x } \right|} } \over n}

=

{{\left| {1 - 5} \right| + \left| {2 - 5} \right| + \left| {6 - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|} \over 5}

=

{{4 + 3 + 1 + 6} \over 5}

=

{{14} \over 5}

= 2.8

Q88

If the variance of the frequency distribution .tg .tg

x

c

2c

3c

4c

5c

6c

f

2 1 1 1 1 1 is 160, then the value of

c\in N

is

A 5

B 8

C 6

D 7

Correct Answer

Option D

Solution

.tg .tg

x_i

f(x_i)

x(f(x)

x^2f(x)

C 2 2C 2C

^2

2C 1 2C 4C

^2

3C 1 3C 9C

^2

4C 1 4C 16C

^2

5C 1 5C 25C

^2

6C 1 6C 36C

^2

\begin{aligned} & \sigma^2=E\left(x^2\right)-[E(x)], \sum f\left(x_i\right)=7 \\ & E(x)=\sum x f(x)=22 C \\ & E\left(x^2\right)=\sum x^2 f(x)=92 C^2 \end{aligned}

\begin{aligned} & \sigma^2=160=\frac{92 C^2}{7}-\left(\frac{22 C}{7}\right)^2 \\ & \Rightarrow C= \pm 7 \text{ but } C \in N \\ & \Rightarrow C=7 \end{aligned}

Q89

Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If

\mu

and

\sigma^2

represent mean and variance of X, respectively, then

10(\mu^2+\sigma^2)

is equal to :

A 20

B 30

C 250

D 25

Correct Answer

Option A

Solution

3 rotten apples are mixed with 7 good apples.

$\therefore$ Total apples = 10 Among those 10 apples 4 are chosen randomly. .tg .tg

{x_i}

{p_i}

{p_i}{x_i}

{p_i}{({x_i})^2}

0

{{{}^7{C_4}} \over {{}^{10}{C_4}}} = {{35} \over {210}}

0 0 1

{{{}^3{C_1} \times {}^7{C_3}} \over {{}^{10}{C_4}}} = {{105} \over {210}}

{{105} \over {210}}

{{105} \over {210}}

2

{{{}^3{C_2} \times {}^7{C_2}} \over {{}^{10}{C_4}}} = {{63} \over {210}}

{{126} \over {210}}

{{252} \over {210}}

3

{{{}^3{C_3} \times {}^7{C_1}} \over {{}^{10}{C_4}}} = {7 \over {210}}

{{21} \over {210}}

{{63} \over {210}}

{x_i}

= Number of rotten apples drawn.

{p_i}

= Probability of rotten apple. We know, Mean

(\mu ) = \sum {{p_i}{x_i}}

= 0 + {{105} \over {210}} + {{126} \over {210}} + {{21} \over {210}}

= {{252} \over {210}} = {6 \over 5}

Also, Variance

({\sigma ^2}) = \left( {\sum {{p_i}{{({x_i})}^2}} } \right) - {\mu ^2}

= {{105} \over {210}} + {{252} \over {210}} + {{63} \over {210}} - {{36} \over {25}}

= {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {{36} \over {25}} = {{14} \over {25}}

$\therefore$

10(\mu^2 + {\sigma ^2})

= 10\left( {({6 \over 5})^2 + {{14} \over {25}}} \right)

= 10\left( {{{36 + 14} \over {25}}} \right)

= 10 \times {{50} \over {25}} = 20

Q90

If for some x

\in

R, the frequency distribution of the marks obtained by 20 students in a test is : .tg .tg Marks 2 3 5 7 Frequency (x + 1)2 2x - 5 x2 - 3x x then the mean of the marks is

A 3.0

B 2.8

C 2.5

D 3.2

Correct Answer

Option B

Solution

Number of students

\Rightarrow {\left( {x + 1} \right)^2} + (2x - 5) + \left( {{x^2} - 3x} \right) + x = 20

\Rightarrow 2{x^2} + 2x - 4 = 20

\Rightarrow {x^2} + x - 12 = 0

\Rightarrow (x + 4)(x - 3) = 0

x = 3

.tg .tg Marks 2 3 5 7 No. of students 16 1 0 3 Average marks =

{{32 + 3 + 21} \over {20}} = {{56} \over {20}} = 2.8