JEE Main Previous Year Statistics Questions and Solutions

Q 1 :
Let α, β ∈ R. Let the mean and the variance of 6 observations -3, 4, 7, -6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is [2024]

16/3
13/3
11/3
14/3

1

2

3

4

(2)

$Mean = \frac{- 3 + 4 + 7 + (- 6) + α + β}{6} = 2$

$\Rightarrow α + β = 10$ ...(i)

$Variance = \frac{\sum x_{i}^{2}}{n} - {(\frac{\sum x_{i}}{n})}^{2} = 23$

$\Rightarrow \sum x_{i}^{2} = 27 \times 6$ $(∵ \bar{x} = \frac{\sum_{x_{i}}}{n} = 2)$

$\Rightarrow 9 + 16 + 49 + 36 + α^{2} + β^{2} = 162$

$\Rightarrow α^{2} + β^{2} = 52$ ...(ii)

Solving (i) and (ii), we get $α = 4$ and $β = 6$

So, mean deviation about mean $= \frac{| - 3 - 2 | + | 4 - 2 | + | 7 - 2 | + | - 6 - 2 | + | 4 - 2 | + | 6 - 2 |}{6}$

$= \frac{5 + 2 + 5 + 8 + 2 + 4}{6}$ $= \frac{13}{3}$

Q 2 :
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 as 8 instead of 12. The correct standard deviation is [2024]

1.8
1.94
$\sqrt{3.86}$
$\sqrt{3.96}$

1

2

3

4

(4)

Incorrect mean, $\bar{x} = 10$

$\Rightarrow Incorrect \frac{\sum x_{i}}{20} = 10$

Incorrect sum, $\sum x_{i} = 200$

$Correct sum = 200 - 8 + 12 = 204$

$Correct mean = \frac{204}{20} = 10.2$

Incorrect S.D. = 2

Incorrect variance = 4

$\Rightarrow \frac{\sum x_{i}^{2}}{20} - {(\frac{\sum x_{i}}{20})}^{2} = 4 \Rightarrow \frac{\sum x_{i}^{2}}{20} - 100 = 4$

$\Rightarrow \sum x_{i}^{2} = 2080$

Incorrect $\sum x_{i}^{2} = 2080$

Correct $\sum x_{i}^{2} = 2080 - 8^{2} + 12^{2} = 2160$

Correct Variance $= \frac{2160}{20} - {(10.2)}^{2} = 108 - 104.04 = 3.96$

Correct S.D. = $\sqrt{3.96}$

Q 3 :
The frequency distribution of the age of students in a class of 40 students is given below. [2024]

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:

43
44
46
47

1

2

3

4

(2)

Age	No. of Students	C.F.
15	5	5
16	8	13
17	5	18
18	12	30
19	$x$	$30 + x$
20	$y$	$30 + x + y$

$30 + x + y = 40$ (Given)

$\Rightarrow x + y = 10$ ...(i)

Median of the given data = 18

$Mean deviation about median = \frac{\sum f_{i} | x_{i} - 18 |}{\sum f_{i}}$

$\sum f_{i} | x_{i} - 18 | = 5 \times 3 + 8 \times 2 + 5 \times 1 + 12 \times 0 + x \times 1 + y \times 2$

$= 15 + 16 + 5 + x + 2 y = 36 + x + 2 y$

$\Rightarrow \frac{36 + x + 2 y}{40} = 1.25$ (Given)

$\Rightarrow x + 2 y = 14$ ...(ii)

On solving (i) and (ii), we get

$x = 6, y = 4$

Now,

$4 x + 5 y = 4 \times 6 + 5 \times 4 = 24 + 20 = 44$

Q 4 :
If the variance of the frequency distribution

$x$    $c$    $2 c$ $3 c$ $4 c$    $5 c$ $6 c$

$y$ 2 1 1 1 1 1

is 160, then the value of $c \in N$ is   [2024]

6
8
5
7

1

2

3

4

(4)

We have, $\bar{x} = \frac{2 c + 2 c + 3 c + 4 c + 5 c + 6 c}{2 + 1 + 1 + 1 + 1 + 1}$ $= \frac{22 c}{7}$

$Var (X) = \frac{c^{2} (2 + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2})}{7} - {(\frac{22 c}{7})}^{2}$

$= \frac{92 c^{2}}{7} - \frac{484 c^{2}}{49} = \frac{(644 - 484) c^{2}}{49} = \frac{160 c^{2}}{49}$

$\Rightarrow 160 = \frac{160 c^{2}}{49}$

$\Rightarrow c^{2} = 49$

$\Rightarrow c = 7$ [ $∵ c \in N$ ]

Q 5 :
Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is: [2024]

32
30
28
31

1

2

3

4

(2)

The ascending order of 7 observations are 125, a, b, 170, 190, 210, 230

$M.D. (Med) = \frac{1}{7} \sum_{i = 1}^{7} | x_{i} - Med | = \frac{205}{7}$

$\Rightarrow | 125 - 170 | + | a - 170 | + | b - 170 | + | 170 - 170 | + | 190 - 170 | + | 210 - 170 | + | 230 - 170 | = 205$

$[∵ Median = 170]$

$\Rightarrow a + b = 300$ $. . . (i)$

$∴ Mean = \frac{125 + 300 + 170 + 190 + 210 + 230}{7} = \frac{1225}{7} = 175$ (By (i))

$∴ M.D. (Mean) = \frac{1}{7} \sum_{i = 1}^{7} | x_{i} - Mean |$ $= \frac{1}{7} (50 + 350 - 300 + 5 + 15 + 35 + 55)$

$= \frac{210}{7} = 30$

Q 6 :
Consider 10 observations $x_{1}, x_{2}, . . ., x_{10}$ such that $\sum_{i = 1}^{10} (x_{i} - α) = 2$ and $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 40$ , where $α, β$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. Then $\frac{β}{α}$ is equal to : [2024]

2
1
$\frac{5}{2}$
$\frac{3}{2}$

1

2

3

4

(1)

We have, $\frac{\sum_{i = 1}^{10} x_{i}}{n} = \frac{6}{5} \Rightarrow \sum_{i = 1}^{10} x_{i} = 12$ $[∵ n = 10]$

Also, $\sum_{i = 1}^{10} (x_{i} - α) = 2 \Rightarrow \sum_{i = 1}^{10} x_{i} - 10 α = 2 \Rightarrow α = 1$

Now, $\frac{\sum_{i = 1}^{10} x_{i}^{2}}{n} - {(\bar{x})}^{2} = \frac{84}{25} \Rightarrow \frac{\sum_{i = 1}^{10} x_{i}^{2}}{10} = \frac{84}{25} + \frac{36}{25} = \frac{120}{25}$

$\Rightarrow \sum_{i = 1}^{10} x_{i}^{2} = 48$

Also, $\sum_{i = 1}^{10} {(x_{i} - β)}^{2} = 40 \Rightarrow \sum_{i = 1}^{10} x_{i}^{2} + 10 β^{2} - 2 β \sum_{i = 1}^{10} x_{i} = 40$

$\Rightarrow 48 + 10 β^{2} - 24 β = 40 \Rightarrow 5 β^{2} - 12 β + 4 = 0$

$\Rightarrow (5 β - 2) \Rightarrow (β - 2) = 0 \Rightarrow β = \frac{2}{5} or β = 2$

If $β = 2$ , then $\frac{β}{α} = 2$

If $β = \frac{2}{5}$ , then $\frac{β}{α} = \frac{2}{5}$

Q 7 :
Let $a_{1}, a_{2}, \dots, a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100 .$ Then the standard deviation of $a_{1}, a_{2}, \dots, a_{10}$ is equal to [2024]

5
$\sqrt{115}$
$\sqrt{5}$
10

1

2

3

4

(3)

Given, $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100$

$σ^{2} = \frac{\sum a_{k}^{2}}{10} -$ $({Mean)}^{2}$ $= \frac{\sum a_{k}^{2}}{10} - {(\frac{\sum a_{k}}{10})}^{2} = \frac{\sum a_{k}^{2}}{10} - {(\frac{50}{10})}^{2}$

$= \frac{\sum a_{k}^{2}}{10} - 25$ ...(i)

Also, ${(\sum a_{k})}^{2} = \sum a_{k}^{2} + 2 \sum_{\forall k < j} a_{k} a_{j}$

$\Rightarrow 2500 = \sum a_{k}^{2} + 2 (1100) \Rightarrow \sum a_{k}^{2} = 2500 - 2200 = 300$

From (i), we get

$σ^{2} = \frac{300}{10} - 25 = 30 - 25 = 5 \Rightarrow σ = \sqrt{5}$

Q 8 :
If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of the first four observations is $\frac{7}{2}$ , then the variance of the first four observations is equal to: [2024]

$\frac{77}{12}$
$\frac{105}{4}$
$\frac{5}{4}$
$\frac{4}{5}$

1

2

3

4

(3)

Let $a, b, c, d$ and $e$ be the five observations.

$∴$ $Mean of 5 observations, \bar{x} = \frac{24}{5}$

$\Rightarrow a + b + c + d + e = 24$ ...(i)

$Mean of four observations, {\bar{x}}_{1} = \frac{7}{2}$

$\Rightarrow a + b + c + d = 14$ ...(ii)

From (i) and (ii), we have $e = 10$

Now, variance of 5 observations $= \frac{194}{25}$

$\Rightarrow \frac{\sum x_{i}^{2}}{5} - {(\bar{x})}^{2} = \frac{194}{25}$

$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} + e^{2} = 5 (\frac{194}{25} + \frac{576}{25}) = 154$

$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} = 154 - 100 (∵ e = 10)$

$\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} = 54$

$∴ Variance of 4 observations = \frac{a^{2} + b^{2} + c^{2} + d^{2}}{4} - {({\bar{x}}_{1})}^{2}$

$= \frac{54}{4} - \frac{49}{4} = \frac{5}{4}$

Q 9 :
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 + 3b is: [2024]

210
190
200
180

1

2

3

4

(4)

Given, mean $(\bar{x}) = 55$

$∴$ $\frac{a + b + 68 + 44 + 48 + 60}{6} = 55 \Rightarrow a + b = 110$ ...(i)

Given, variance $(σ^{2}) = 194$

$∴$ $\frac{a^{2} + b^{2} + {(68)}^{2} + {(44)}^{2} + {(48)}^{2} + {(60)}^{2}}{6} - {(55)}^{2} = 194$

$\Rightarrow a^{2} + b^{2} = 6850$ ...(ii)

From (i) and (ii), we get

$a = 75$ and $b = 35$

$∴$ $a + 3 b = 75 + 3 (35) = 180$

Q 10 :
If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α,$ $β,$ 60 where $α > β$ , are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to _____ . [2024]

0%

(6344)

Mean = 56

$Variance = \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - {(\bar{x})}^{2}$

$\Rightarrow 66.2 = \frac{1}{10} [{(65)}^{2} + {(68)}^{2} + {(58)}^{2} + {(44)}^{2} + {(48)}^{2} + {(45)}^{2} + (60)$ ²+α²+β²+(60)²]-(56)²

$\Rightarrow 3202.2 \times 10 = 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + α^{2} + β^{2} + 3600$

$\Rightarrow α^{2} + β^{2} = 6344$

Topic Question Set

Q 1 :
Let α, β ∈ R. Let the mean and the variance of 6 observations -3, 4, 7, -6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is [2024]

Q 2 :
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 as 8 instead of 12. The correct standard deviation is [2024]

Q 3 :
The frequency distribution of the age of students in a class of 40 students is given below. [2024]

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:

Q 4 :
If the variance of the frequency distribution

$x$    $c$    $2 c$ $3 c$ $4 c$    $5 c$ $6 c$

$y$ 2 1 1 1 1 1

is 160, then the value of $c \in N$ is   [2024]

Q 5 :
Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is: [2024]

Q 7 :
Let $a_{1}, a_{2}, \dots, a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100 .$ Then the standard deviation of $a_{1}, a_{2}, \dots, a_{10}$ is equal to [2024]

Q 8 :
If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of the first four observations is $\frac{7}{2}$ , then the variance of the first four observations is equal to: [2024]

Q 9 :
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 + 3b is: [2024]

Q 10 :
If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α,$ $β,$ 60 where $α > β$ , are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to _____ . [2024]

0%

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 1 : Let α, β ∈ R. Let the mean and the variance of 6 observations -3, 4, 7, -6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is [2024]

Q 2 : 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 as 8 instead of 12. The correct standard deviation is [2024]

Q 3 : The frequency distribution of the age of students in a class of 40 students is given below. [2024] 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:

Q 4 : If the variance of the frequency distribution x c 2c 3c 4c 5c 6c y 2 1 1 1 1 1 is 160, then the value of c∈N is [2024]

Q 5 : Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and 2057 respectively. Then the mean deviation about the mean of these 7 observations is: [2024]

Q 6 : Consider 10 observations x1,x2, ..., x10 such that ∑i=110(xi-α)=2 and ∑i=110(xi-β)2=40, where α,β are positive integers. Let the mean and the variance of the observations be 65 and 8425 respectively. Then βα is equal to : [2024]

Q 7 : Let a1,a2, …, a10 be 10 observations such that ∑k=110ak=50 and ∑∀k<jak·aj=1100. Then the standard deviation of a1,a2, …, a10 is equal to [2024]

Q 8 : If the mean and variance of five observations are 245 and 19425 respectively and the mean of the first four observations is 72, then the variance of the first four observations is equal to: [2024]

Q 9 : 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 + 3b is: [2024]

Q 10 : If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, α,β, 60 where α>β, are 56 and 66.2 respectively, then α2+β2 is equal to _____ . [2024]

0%

Want Us to Email you About Special Offers & Updates?

Q 1 :
Let α, β ∈ R. Let the mean and the variance of 6 observations -3, 4, 7, -6, α, β be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is [2024]

Q 2 :
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 as 8 instead of 12. The correct standard deviation is [2024]

Q 3 :
The frequency distribution of the age of students in a class of 40 students is given below. [2024]

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:

Q 4 :
If the variance of the frequency distribution

$x$ $c$ $2 c$ $3 c$ $4 c$ $5 c$ $6 c$

$y$ 2 1 1 1 1 1

is 160, then the value of $c \in N$ is [2024]

Q 5 :
Let the median and the mean deviation about the median of 7 observations 170, 125, 230, 190, 210, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is: [2024]

Q 7 :
Let $a_{1}, a_{2}, \dots, a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\sum_{\forall k < j} a_{k} \cdot a_{j} = 1100 .$ Then the standard deviation of $a_{1}, a_{2}, \dots, a_{10}$ is equal to [2024]

Q 8 :
If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of the first four observations is $\frac{7}{2}$ , then the variance of the first four observations is equal to: [2024]

Q 9 :
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 + 3b is: [2024]

Q 10 :
If the mean and variance of the data 65, 68, 58, 44, 48, 45, 60, $α,$ $β,$ 60 where $α > β$ , are 56 and 66.2 respectively, then $α^{2} + β^{2}$ is equal to _____ . [2024]