Statistics — Page 8 | JEE Mathematics

Q71

Consider 10 observations

x_1, x_2, \ldots, x_{10}

such that

\sum\limits_{i=1}^{10}\left(x_i-\alpha\right)=2

and

\sum\limits_{i=1}^{10}\left(x_i-\beta\right)^2=40

, where

\alpha, \beta

are positive integers. Let the mean and the variance of the observations be

\dfrac{6}{5}

and

\dfrac{84}{25}

respectively. Then

\dfrac{\beta}{\alpha}

is equal to :

A 2

B 1

C

\dfrac{5}{2}

D

\dfrac{3}{2}

Correct Answer

Option A

Solution

We have given $\bar{x}($ mean $)=\dfrac{6}{5}$

\begin{aligned} & \text{ Variance }=\frac{84}{25} \\\\ & \sum_{i=1}^{10}\left(x_i-\alpha\right)=2 \\\\ & \Rightarrow x_1+x_2+\ldots+x_{10}-10 \alpha=2 \\\\ & \Rightarrow \frac{x_1+x_2+\ldots+x_{10}}{10}-\alpha=\frac{2}{10} \\\\ & \Rightarrow \frac{6}{5}-\alpha=\frac{2}{10} \\\\ & \Rightarrow \alpha=1 \end{aligned}

$\begin{aligned} & \text{ and } \sum_{i=1}^{10}\left(x_i-\beta\right)^2=40 \\\\ & \left(x_1-\beta\right)^2+\left(x_2-\beta\right)^2+\ldots+\left(x_{10}-\beta\right)^2=40 \\\\ & x_1^2+x_2^2+\ldots+x_{10}^2+10 \beta^2-2 \beta\left(x_1+x_2+\ldots+x_{10}\right)=40 \\\\ & \Rightarrow \dfrac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}+\beta^2-\dfrac{2 \beta\left(x_1+x_2+\ldots+x_{10}\right)}{10}=4 \\\\ & \Rightarrow \dfrac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}-\dfrac{36}{25}+\dfrac{36}{25}+\beta^2-2 \beta \times \dfrac{6}{5}=4\end{aligned}$ $\left[\right.$ Variance $\left.=\dfrac{\sum_{i=1}^n x_i^2}{n}-(\bar{x})^2\right]$ $\begin{aligned} & \Rightarrow \dfrac{84}{25}+\dfrac{36}{25}+\beta^2-\dfrac{12 \beta}{5}-4=0 \\\\ & \Rightarrow \dfrac{120}{25}+\beta^2-\dfrac{12 \beta}{5}-4=0 \\\\ & \Rightarrow 25 \beta^2-60 \beta+20=0 \\\\ & \Rightarrow 5 \beta^2-12 \beta+4=0 \\\\ & \Rightarrow \beta=2, \dfrac{2}{5}\end{aligned}$ Take $\beta=2$

\frac{\beta}{\alpha}=\frac{2}{1}=2

Q72

Let the median and the mean deviation about the median of 7 observation

170,125,230,190,210

, a, b be 170 and

\dfrac{205}{7}

respectively. Then the mean deviation about the mean of these 7 observations is :

A 31

B 28

C 30

D 32

Correct Answer

Option C

Solution

\text{ Median }=170 \Rightarrow 125, \mathrm{a}, \mathrm{b}, 170,190,210,230

Mean deviation about Median $=$

\begin{aligned} & \frac{0+45+60+20+40+170-a+170-b}{7}=\frac{205}{7} \\\\ & \Rightarrow \mathrm{a}+\mathrm{b}=300 \\\\ & \text{ Mean }=\frac{170+125+230+190+210+a+b}{7}=175 \end{aligned}

Mean deviation About mean $=$

\frac{50+175-a+175-b+5+15+35+55}{7}=30

Q73

Let

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

be 10 observations such that

\sum\limits_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50

and

\sum\limits_{\forall \mathrm{k} < \mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100

. Then the standard deviation of

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

is equal to :

A 5

B

\sqrt{115}

C 10

D

\sqrt{5}

Correct Answer

Option D

Solution

\begin{aligned} & \sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50 \\ & \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50 \quad \text{.... (i)}\\ & \sum_{\forall \mathrm{k}

\begin{aligned} & =\sqrt{\frac{\sum \mathrm{a}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{a}_{\mathrm{i}}}{10}\right)^2}=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2} \\ & =\sqrt{30-25}=\sqrt{5} \end{aligned}$$

Q74

Let the mean and the variance of 6 observations

a, b, 68,44,48,60

be

55

and

194

, respectively. If

a>b

, then

a+3 b

is

A 180

B 210

C 190

D 200

Correct Answer

Option A

Solution

\begin{aligned} & a, b, 68,44,48,60 \\ & \text{ Mean }=55 \quad a>b \\ & \text{ Variance }=194 \quad a+3 b \\ & \frac{a+b+68+44+48+60}{6}=55 \\ & \Rightarrow 220+a+b=330 \\ & \therefore a+b=110 \ldots . .(1) \end{aligned}

Also,

\begin{aligned} & \sum \frac{\left(x_i-\bar{x}\right)^2}{n}=194 \\ & \Rightarrow(a-55)^2+(b-55)^2+(68-55)^2+(44-55)^2 \\ & +(48-55)^2+(60-55)^2=194 \times 6 \\ & \Rightarrow(a-55)^2+(b-55)^2+169+121+49+25=1164 \\ & \Rightarrow(a-55)^2+(b-55)^2=1164-364=800 \\ & a^2+3025-110 a+b^2+3025-110 b=800 \\ & \Rightarrow a^2+b^2=800-6050+12100 \\ & a^2+b^2=6850 \ldots \ldots . .(2) \end{aligned}

Solve (1) & (2);

\begin{aligned} & a=75, b=35 \\ & \therefore a+3 b=75+3(35)=75+105=180 \end{aligned}

Q75

The mean and standard deviation of 100 observations are 40 and 5.1 , respectively. By mistake one observation is taken as 50 instead of 40 . If the correct mean and the correct standard deviation are

\mu

and

\sigma

respectively, then

10(\mu+\sigma)

is equal to

A 447

B 445

C 449

D 451

Correct Answer

Option C

Solution

\begin{aligned} &\text{ Let the observations be } x_1, x_2, \ldots, x_{99}, 50\\ &\begin{aligned} & \text{ Mean }=\frac{x_1+x_2+\ldots+x_9+50}{100}=40 \\ & \Rightarrow x_1+x_2+\ldots+x_{99}=4000-50 \\ & \Rightarrow x_1+x_2+\ldots+x_{99}=3950 \\ & \text{ Current Mean }=\frac{3950+40}{100} \\ & \mu=\frac{399}{10}=39.9 \\ & (\text{ S.D })^2=\sum_{i=1}^{99} \frac{\left(x_i\right)^2+2500}{100}-(40)^2 \\ & \sum_{i=1}^{99} x_i^2=160101 \\ & (\text{ Correct S.D })^2=\frac{160101+1600}{100}-\left(\frac{399}{10}\right)^2 \\ & \sigma=5 \\ & 10(\mu+\sigma)=10(39.9+5)=449 \end{aligned} \end{aligned}

Q76

If the mean and variance of five observations are

\dfrac{24}{5}

and

\dfrac{194}{25}

respectively and the mean of the first four observations is

\dfrac{7}{2}

, then the variance of the first four observations in equal to

A

\dfrac{5}{4}

B

\dfrac{4}{5}

C

\dfrac{105}{4}

D

\dfrac{77}{12}

Correct Answer

Option A

Solution

\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}

Let first four observation be

\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4

Here,

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

. ..... (1) Also,

\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4}{4}=\frac{7}{2}

\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4=14

Now from eqn -1

\mathrm{x}_5=10

Now,

\sigma^2=\frac{194}{25}

\begin{aligned} & \frac{\mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2+\mathrm{x}_5^2}{5}-\frac{576}{25}=\frac{194}{25} \\ & \Rightarrow \mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2=54 \end{aligned}

Now, variance of first 4 observations

\begin{aligned} \operatorname{Var} & =\frac{\sum\limits_{i=1}^4 x_i^2}{4}-\left(\frac{\sum\limits_{i=1}^4 x_i}{4}\right)^2 \\ & =\frac{54}{4}-\frac{49}{4}=\frac{5}{4} \end{aligned}

Q77

Let

\alpha, \beta \in \mathbf{R}

. Let the mean and the variance of 6 observations

-3,4,7,-6, \alpha, \beta

be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

A

\dfrac{16}{3}

B

\dfrac{11}{3}

C

\dfrac{14}{3}

D

\dfrac{13}{3}

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Mean }=\frac{-3+4+7+(-6)+\alpha+\beta}{6}=2 \\ & \Rightarrow \alpha+\beta=10 \\ & \text{ Variance }=\frac{\sum x_i^2}{n}-\left(\frac{\bar{x}}{n}\right)^2=23 \\ & \Rightarrow \sum x_i^2=27 \times 6 \\ & \Rightarrow 9+16+49+36+\alpha^2+\beta^2=162 \\ & \Rightarrow \alpha^2+\beta^2=52 \end{aligned}

We get $\alpha$ and $\beta$ as 4 and 6 So, mean deviation about mean

\begin{aligned} & =\frac{|-3-2|+|4-2|+|7-2|+|-6-2|+|4-2|+|6-2|}{6} \\ & =\frac{5+2+5+8+2+4}{6} \\ & =\frac{13}{3} \end{aligned}

Q78

The mean and standard deviation of 20 observations are found to be 10 and 2 , respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

A 1.94

B

\sqrt{3.96}

C

\sqrt{3.86}

D 1.8

Correct Answer

Option B

Solution

To find the correct standard deviation, we first need to adjust the mean and sum of squares of the observations based on the correction of the erroneous entry.

The original erroneous observation was 8, and the correct observation is 12.

The mean and standard deviation of the 20 observations before correction were 10 and 2, respectively.

The sum of all 20 original observations ( $S_{\text{original}}$ ) can be calculated from the mean formula: $\text{Mean} = \dfrac{\text{Sum of all observations}}{\text{Number of observations}}$ $10 = \dfrac{S_{\text{original}}}{20}$ $S_{\text{original}} = 10 \times 20 = 200$ The corrected sum of observations ( $S_{\text{correct}}$ ) will replace the incorrect observation (8) with the correct one (12): $S_{\text{correct}} = S_{\text{original}} - 8 + 12 = 200 - 8 + 12 = 204$ The corrected mean ( $\mu_{\text{correct}}$ ) is: $\mu_{\text{correct}} = \dfrac{S_{\text{correct}}}{20} = \dfrac{204}{20} = 10.2$ To find the corrected standard deviation, we need the sum of squares of the deviations from the mean for both the original and corrected data.

The original sum of squares ( $SS_{\text{original}}$ ) is calculated from the original standard deviation formula, where $\sigma = 2$ : $\sigma^2 = \dfrac{SS}{n}$ $4 = \dfrac{SS_{\text{original}}}{20}$ $SS_{\text{original}} = 4 \times 20 = 80$ To calculate the corrected sum of squares ( $SS_{\text{correct}}$ ), we need to adjust $SS_{\text{original}}$ by removing the square of the deviation of the incorrect observation and adding the square of the deviation of the correct observation: $SS_{\text{correct}} = SS_{\text{original}} - (8 - 10)^2 + (12 - 10.2)^2$ $SS_{\text{correct}} = 80 - (-2)^2 + (1.8)^2$ $SS_{\text{correct}} = 80 - 4 + 3.24 = 79.24$ Finally, the corrected standard deviation ( $\sigma_{\text{correct}}$ ) is: $\sigma_{\text{correct}} = \sqrt{\dfrac{SS_{\text{correct}}}{20}}$ $\sigma_{\text{correct}} = \sqrt{\dfrac{79.24}{20}}$ $\sigma_{\text{correct}} = \sqrt{3.962}$ The corrected standard deviation is closest to the value given in Option B, which is $\sqrt{3.96}$

Q79

Let

x_1, x_2, ..., x_{10}

be ten observations such that

\sum\limits_{i=1}^{10} (x_i - 2) = 30

,

\sum\limits_{i=1}^{10} (x_i - \beta)^2 = 98

,

\beta > 2

, and their variance is

\dfrac{4}{5}

. If

\mu

and

\sigma^2

are respectively the mean and the variance of

2(x_1 - 1) + 4\beta, 2(x_2 - 1) + 4\beta, ..., 2(x_{10} - 1) + 4\beta

, then

\dfrac{\beta\mu}{\sigma^2}

is equal to :

A 100

B 90

C 120

D 110

Correct Answer

Option A

Solution

\begin{aligned} & \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=50, \quad \therefore \text{ mean }=5 \\ & \text{ Variance }=\frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\ & \frac{4}{5}=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{10}-25 \\ & \Rightarrow \sum \mathrm{x}_{\mathrm{i}}^2=258 \quad\text{.... (1)}\\ & \text{ Now } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\beta\right)^2=98 \\ & \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}^2-2 \beta \cdot \mathrm{x}_{\mathrm{i}}+\beta^2\right)=98 \\ & 258-2 \beta(50)+10 \beta^2=98 \\ & (\beta-8)(\beta-2)=0 \\ & \beta=\text{ or } \beta=2 \quad(\text{ as } \beta>2) \\ & \therefore \beta=8\quad\text{..... (2)} \end{aligned}

Now as per the question

\begin{aligned} &2\left(\mathrm{x}_1-1\right)+4 \beta, 2\left(\mathrm{x}_2-1\right)+4 \beta, \ldots .2\left(\mathrm{x}_{10}-1\right)+4 \beta\\ &\text{ can be simplified to }\\ &2 \mathrm{x}_1+30,2 \mathrm{x}_2+30, \ldots . .2 \mathrm{x}_{10}+30 \text{ using eq. (2) }\\ &\mu=2(5)+30=40\quad\text{..... (3)}\\ &\sigma^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5}\\ &\because \frac{\beta \mu}{\sigma^2}=\frac{8 \times 40}{16 / 5}=100 \end{aligned}

Q80

Marks obtains by all the students of class 12 are presented in a freqency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18 , then the total number of students is :

A 52

B 44

C 40

D 48

Correct Answer

Option B

Solution

The median for grouped data is given by:

\text{Median} = L + \left(\frac{\frac{n}{2} - CF}{f}\right) \times h

where

L

is the lower limit (or boundary) of the median class.

CF

is the cumulative frequency of all classes preceding the median class.

f

is the frequency of the median class.

h

is the class width.

n

is the total number of students. Given: Median

= 14

Median class interval is

12-18

, so

L = 12

and the class width

h = 18 - 12 = 6

. Frequency of median class

f = 12

. Cumulative frequency below the median class

= 18

(i.e.,

CF = 18

). Plugging these into the formula:

14 = 12 + \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6

Step 1: Subtract 12 from both sides:

2 = \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6

Step 2: Simplify the multiplication factor:

\left(\frac{6}{12}\right) = \frac{1}{2}

So the equation becomes:

2 = \frac{1}{2}\left(\frac{n}{2} - 18\right)

Step 3: Multiply both sides by 2 to remove the fraction:

4 = \frac{n}{2} - 18

Step 4: Solve for

\frac{n}{2}

:

\frac{n}{2} = 4 + 18 = 22

Step 5: Multiply both sides by 2 to find

n

:

n = 44

Thus, the total number of students is

44

.