Statistics — Page 7 | JEE Mathematics

Q61

The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :

A 10

B 36

C 43

D 60

Correct Answer

Option C

Solution

Given

\overline x = 15,\,\sigma = 2 \Rightarrow {\sigma ^2} = 4

$\therefore$

{x_2} + {x_2} + \,\,.....\,\, + \,\,{x_{50}} = 15 \times 50 = 750

4 = {{x_1^2 + x_2^2 + \,\,.....\,\, + \,\,x_{50}^2} \over {50}} - 225

$\therefore$

x_1^2 + x_2^2 + \,\,.....\,\, + \,\,x_{50}^2 = 50 \times 229

Let a be the correct observation and b is the incorrect observation then

a + b = 70

and

16 = {{750 - b + a} \over {50}}

$\therefore$

a - b = 50 \Rightarrow a = 60,\,b = 10

$\therefore$ Correct variance

= {{50 \times 229 + {{60}^2} - {{10}^2}} \over {50}} - 256 = 43

Q62

Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and

\alpha(>

0 ), and the mean and standard deviation of marks of class

B

of

n

students be respectively 55 and 30

-\alpha

. If the mean and variance of the marks of the combined class of

100+\mathrm{n}

studants are respectively 50 and 350 , then the sum of variances of classes

A

and

B

is :

A 450

B 900

C 650

D 500

Correct Answer

Option D

Solution

\begin{array}{cll} \quad \mathbf{A} & \mathbf{B} & \mathbf{A}+\mathbf{B} \\ \overline{\mathrm{x}}_{1}=40 & \overline{\mathrm{x}}_{2}=55 & \overline{\mathrm{x}}=50 \\ \sigma_{1}=\alpha & \sigma_{2}=30-\alpha & \sigma^{2}=350 \\ \mathrm{n}_{1}=100 & \mathrm{n}_{2}=\mathrm{n} & 100+\mathrm{n} \\\\ \overline{\mathrm{x}}=\frac{100 \times 40+55 \mathrm{n}}{100+\mathrm{n}} & \end{array}

\begin{aligned} & 5000+50 \mathrm{n}=4000+55 \mathrm{n} \\\\ & 1000=5 \mathrm{n} \\\\ & \mathrm{n}=200 \\\\ & \sigma_{1}^{2}=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{100}-40^{2} \\\\ & \sigma_{2}^{2}=\frac{\sum \mathrm{x}_{\mathrm{j}}^{2}}{100}-55^{2} \\\\ & 350=\sigma^{2}=\frac{\sum \mathrm{x}_{\mathrm{i}}{ }^{2}+\sum \mathrm{x}_{\mathrm{j}}^{2}}{300}-(\overline{\mathrm{x}})^{2} \\\\ & 350=\frac{\left(1600+\alpha^{2}\right) \times 100+\left[(30-\alpha)^{2}+3025\right] \times 200}{300}-(50)^{2} \\\\ & 2850 \times 3=\alpha^{2}+2(30-\alpha)^{2}+1600+6050 \\\\ & 8550=\alpha^{2}+2(30-\alpha)^{2}+7650 \\\\ & \alpha^{2}+2(30-\alpha)^{2}=900 \\\\ & \alpha^{2}-40 \alpha+300=0 \\\\ & \alpha=10,30 \\\\ & \sigma_{1}^{2}+\sigma_{2}^{2}=10^{2}+20^{2}=500 \end{aligned}

Q63

The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :

A 1792

B 1216

C 1456

D 1072

Correct Answer

Option D

Solution

Let observation $1,3,5, a, b$

\begin{aligned} & \text{ Mean } = \frac{9+a+b}{5}=5 \\\\ & \text{ Variance } = \frac{a^{2}+b^{2}+35}{5}-25=8 \\\\ & \Rightarrow a+b=16 \text{ and } a^{2}+b^{2}=130 \\\\ & \therefore a \text{ and } b \text{ are } 7 \text{ and } 9 \\\\ & \therefore a^{3}+b^{3}=7^{3}+9^{3}=1072 \end{aligned}

Q64

Let

S

be the set of all values of

a_1

for which the mean deviation about the mean of 100 consecutive positive integers

a_1, a_2, a_3, \ldots ., a_{100}

is 25 . Then

S

is :

A

\{9\}

B

\phi

C

\{99\}

D N

Correct Answer

Option D

Solution

Let

{a_1} = a \Rightarrow {a_2} = a + 1,\,.....\,{a_{100}} = a + 90

\mu = {{{{100a + (99 \times 100)} \over 2}} \over {100}} = a + {{99} \over 2}

M.D

= {{\sum {|{a_1} - \mu |} } \over {100}} = {{{{\left( {{{99} \over 2} + {{97} \over 2}\, + \,...\, + \,{1 \over 2}} \right)}^2}} \over {100}} = {{2500} \over {100}} = 25

$\therefore$

a \to z

Q65

The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :

A 3.92

B 4.08

C 3.96

D 4.04

Correct Answer

Option C

Solution

$\bar{x}=10 ~\&~ \sigma^{2}=4$ , No. of students $=N$ (let)

\therefore \quad \frac{\sum x_{i}}{N}=10 ~\&~ \frac{\sum x_{i}^{2}}{N}-(10)^{2}=4

Now if one of $x_{i}$ is changed from 8 to 12 we have New mean $\dfrac{\sum x_{i}+4}{N}=10+\dfrac{4}{N}=10.2$ $\Rightarrow N=20$ and $\sigma_{\text{new }}^{2}=\dfrac{\sum x_{i}^{2}-(8)^{2}+(12)^{2}}{20}-(10 \cdot 2)^{2}$

\begin{aligned} & =\frac{\sum x_{i}^{2}}{20}+\frac{144-64}{20}-(10 \cdot 2)^{2} \\\\ & =104+4-(10 \cdot 2)^{2} \\\\ & =108-104.04=3.96 \end{aligned}

Q66

Let the six numbers

\mathrm{a_1,a_2,a_3,a_4,a_5,a_6}

, be in A.P. and

\mathrm{a_1+a_3=10}

. If the mean of these six numbers is

\dfrac{19}{2}

and their variance is

\sigma^2

, then 8

\sigma^2

is equal to :

A 220

B 210

C 105

D 200

Correct Answer

Option B

Solution

{a_1},{a_2},{a_3},{a_4},{a_5},{a_6}

are in AP. Let

{a_1} = a

{a_2} = a + d

{a_3} = a + 2d

{a_4} = a + 3d

{a_5} = a + 4d

{a_6} = a + 5d

Now Mean of

{a_1},{a_2},{a_3},{a_4},{a_5}

and

{a_6}

is

= {{{a_1} + {a_2} + {a_3} + {a_4} + {a_5} + {a_6}} \over 6} = {{19} \over 2}

\Rightarrow {{6a + 15d} \over 6} = {{19} \over 2}

\Rightarrow 2a + 5d = 19

...... (1) Also, given,

{a_1} + {a_3} = 10

\Rightarrow a + a + 2d = 10

\Rightarrow 2a + 2d = 10

\Rightarrow a + d = 5

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

a = 2

and

d = 3

$\therefore$ AP is 2, 5, 8, 11, 14, 17 Now, Variance

({\sigma ^2}) = {{\sum {x_i^2} } \over 6} - {\left( {\overline x } \right)^2}

= {{{2^2} + {5^2} + {8^2} + {{11}^2} + {{14}^2} + {{17}^2}} \over 6} - {\left( {{{19} \over 2}} \right)^2}

= {{105} \over 4}

$\therefore$

8{\sigma ^2} = 8 \times {{105} \over 4} = 210

Q67

The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is :

A 11

B 12

C 13

D 14

Correct Answer

Option C

Solution

1. Calculate the sum of the original observations:

\frac{x_1+x_2+\ldots+x_9+50}{10}=20

x_1+x_2+\ldots+x_9=150

2. Calculate the sum of the squares of the original observations using the original variance:

\sigma^2 = 8^2 = 64

64 = \frac{x_1^2+x_2^2+\ldots+x_9^2+2500}{10} - 400

x_1^2+x_2^2+\ldots+x_9^2 = 2140

3. Calculate the new mean after correcting the error:

\text{New mean} = \frac{150+40}{10} = 19

4. Calculate the new variance using the corrected sum of observations and the corrected sum of squares of observations:

\text{New } \sigma^2 = \frac{2140+1600}{10} - (19)^2

$\Rightarrow$

\sigma^2 = 13

The correct variance after correcting the error is 13 (Option C).

Q68

Let the mean of 6 observations

1,2,4,5, \mathrm{x}

and

\mathrm{y}

be 5 and their variance be 10 . Then their mean deviation about the mean is equal to :

A

\dfrac{10}{3}

B

\dfrac{8}{3}

C

\dfrac{7}{3}

D 3

Correct Answer

Option B

Solution

Given that the mean of the observations ${x, y, 1, 2, 4, 5}$ is 5, we get the equation: $x + y + 1 + 2 + 4 + 5 = 6 \cdot 5 \Rightarrow x + y = 18$ ..........

$(1)$ We are also given that the variance of the observations is 10.

Using the formula for variance, we have: $V = \dfrac{\Sigma x_i^2}{n} - \bar{x}^2$ , where $n$ is the number of observations, and $\bar{x}$ is the mean.

Substituting the given values in the variance formula, we get: $10 = \dfrac{\Sigma x_i^2}{6} - 5^2 = \dfrac{\Sigma x_i^2}{6} - 25 \Rightarrow \Sigma x_i^2 = 210$ .

Here, $\Sigma x_i^2$ is the sum of the squares of all the observations, thus: $x^2 + y^2 + 1 + 4 + 16 + 25 = 210 \Rightarrow x^2 + y^2 = 164$ .........

$(2)$ By solving the system of equations $(1)$ and $(2)$ , we get $x = 8$ and $y = 10$ .

Now, the mean deviation about the mean ( $\bar{x} = 5$ ) is calculated by taking the average of the absolute differences of each observation from the mean: $MD(5) = \dfrac{1}{6} [|1-5| + |2-5| + |4-5| + |5-5| + |8-5| + |10-5|] = \dfrac{16}{6} = \dfrac{8}{3}$ .

So, the mean deviation about the mean is $\dfrac{8}{3}$ , and the correct answer is Option B, $\dfrac{8}{3}$ .

Q69

Let the mean and variance of 12 observations be

\dfrac{9}{2}

and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is

\dfrac{m}{n}

, where

\mathrm{m}

and

\mathrm{n}

are coprime, then

\mathrm{m}+\mathrm{n}

is equal to :

A 317

B 316

C 314

D 315

Correct Answer

Option A

Solution

\begin{aligned} & \text{ Since, Mean }=\frac{9}{2} \\\\ & \Rightarrow \Sigma x=\frac{9}{2} \times 12=54 \end{aligned}

Also, variance $=4$

\begin{aligned} & \Rightarrow \frac{\sum x^2}{12}=\left[\frac{\sum x_i}{12}\right]^2=4 \\\\ & \Rightarrow \frac{\sum x^2}{12}=4+\frac{81}{4}=\frac{97}{4} \\\\ & \Rightarrow \sum x^2=291 \end{aligned}

\begin{aligned} & \sum x^{\prime}=54-(9+10)+7+14 \\\\ & =54-19+21=56 \\\\ & \text{ and } \sum x^2=291-(81+100)+49+196 \\\\ & =291-181+49+196=355 \end{aligned}

\begin{aligned} & \text{ So, } \sigma_{\text{new }}^2=\frac{\sum x_{\text{new }}^2}{12}-\left(\frac{\sum x_{\text{new }}}{12}\right)^2 \\\\ & =\frac{355}{12}-\left(\frac{56}{12}\right)^2 \\\\ & =\frac{4260-3136}{144}=\frac{1124}{144}=\frac{281}{36} \\\\ & =\frac{m}{n} \\\\ & \Rightarrow m=281, n=36 \\\\ & \Rightarrow m+n=281+36=317 \end{aligned}

Q70

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and

\sigma^{2}

respectively. If the variance of all the 30 numbers in the two sets is 13 , then

\sigma^{2}

is equal to :

A 12

B 11

C 10

D 9

Correct Answer

Option C

Solution

We know that if $n_1, n_2$ are the sizes, $\bar{X}_1, \bar{X}_2$ are the means and $\sigma_1, \sigma_2$ are the standard deviation of the series, then the combine variance of the series.

\begin{array}{ll} & \sigma^2=\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 \cdot n_2}{\left(n_1+n_2\right)^2}\left(\bar{X}_1-\bar{X}_2\right)^2 \\\\ &\Rightarrow 13=\frac{15 \times 14+15 \times \sigma^2}{15+15}+\frac{15 \times 15}{(15+15)^2}(12-14)^2 \\\\ &\Rightarrow 13=\frac{14+\sigma^2}{2}+\frac{1}{4} \times 4 \\\\ &\Rightarrow 14+\sigma^2=2 \times 12 \\\\ &\Rightarrow \sigma^2=10 \end{array}