JEE Main Previous Year Statistics Questions and Solutions

Q 41 :

If the mean of the frequency distribution

Class : 0–10 10–20 20–30 30–40 40–50

Frequency : 2 3 $x$ 5 4

is 28, then its variance is _____ . [2023]

0%

(151)

$x_{i}$	$f_{i}$	$x_{i} \cdot f_{i}$
5	2	10
15	3	45
25	$x$	$25 x$
35	5	175
45	4	180
	$n = 14 + x$	$\sum x_{i} f_{i} = 410 + 25 x$

$Mean, \bar{x} = \frac{410 + 25 x}{14 + x}$ $\Rightarrow 28 = \frac{410 + 25 x}{14 + x}$

$\Rightarrow 392 + 28 x = 410 + 25 x \Rightarrow 3 x = 18$

$\Rightarrow x = 6$

$∴$ $n = 20$

$Variance (σ^{2}) = \frac{\sum f_{i} x_{i}^{2}}{n} - {(28)}^{2}$

$= \frac{2 \times 5^{2} + 3 \times 15^{2} + 6 \times 25^{2} + 5 \times 35^{2} + 4 \times 45^{2}}{20} - {(28)}^{2}$

$= \frac{(2 \times 25) + (3 \times 225) + (6 \times 625) + (5 \times 1225) + (4 \times 2025)}{20} - 784$

$= 935 - 784 = 151$

Q 42 :

Let the positive numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{m}{n}$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_{3} + a_{4} + a_{5} = 14$ , then $m + n$ is equal to _______ . [2023]

0%

(211)

$Let \frac{a}{r^{2}}, \frac{a}{r}, a, a r, a r^{2} be positive numbers in G.P.$

$\Rightarrow \frac{a}{r^{2}} + \frac{a}{r} + a + a r + a r^{2} = 5 \times \frac{31}{10}$ ...(i)

$and$ $\frac{r^{2}}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{a r} + \frac{1}{a r^{2}} = 5 \times \frac{31}{40}$ ...(ii)

$Divide (i) by (ii), we get$

$\frac{a (\frac{1}{r^{2}} + \frac{1}{r} + 1 + r + r^{2})}{\frac{1}{a} (r^{2} + r + 1 + \frac{1}{r} + \frac{1}{r^{2}})} = \frac{5 \times \frac{31}{10}}{5 \times \frac{31}{40}}$

$\Rightarrow a^{2} = 4$

$\Rightarrow a = 2$ $(∵ a cannot be negative)$

$From (i), a (\frac{1}{r^{2}} + \frac{1}{r} + 1 + r + r^{2}) = 5 \times \frac{31}{10}$

$\Rightarrow {(r + \frac{1}{r})}^{2} + (r + \frac{1}{r}) = \frac{31}{4} + 1$

$\Rightarrow 4 t^{2} + 4 t - 35 = 0$ $[Let, r + \frac{1}{r} = t]$

$\Rightarrow t = \frac{5}{2} \Rightarrow r = 2$

$∴$ $Numbers are = \frac{1}{2}, 1, 2, 4, 8$

$∴$ $σ^{2} = \frac{\sum x^{2}}{N} - {(\frac{\sum x}{N})}^{2} = \frac{\frac{1}{4} + 1 + 4 + 16 + 64}{5} - {(\frac{31}{10})}^{2}$

$= \frac{341}{20} - \frac{961}{100} = \frac{186}{25}$ $∴$ $m + n = 186 + 25 = 211$

Q 43 :

If the variance of the frequency distribution

$x_{𝑖}$ 2 3 4 5 6 7 8

Frequency $f_{i}$ 3 6 16 $α$ 9 5 6

is 3, then $α$ is equal to _______. [2023]

0%

(5)

$x_{i}$	$f_{i}$	$d_{i} = x_{i} - 5$	$d_{i}^{2}$	$f_{i} d_{i}^{2}$	$f_{i} d_{i}$
2	3	- 3	9	27	- 9
3	6	- 2	4	24	- 12
4	16	- 1	1	16	- 16
5	$α$	0	0	0	0
6	9	1	1	9	9
7	5	2	4	20	10
8	6	3	9	54	18
Total	$45 + α$			150	0

$σ^{2} = \frac{\sum f_{i} d_{i}^{2}}{\sum f_{i}} - {(\frac{\sum f_{i} d_{i}}{\sum f_{i}})}^{2} = \frac{150}{45 + α} - {(\frac{0}{45 + α})}^{2}$

$\Rightarrow 3 = \frac{150}{45 + α} (∵ variance = 3)$

$\Rightarrow 135 + 3 α = 150 \Rightarrow 3 α = 15 \Rightarrow α = 5$

Q 44 :

Let the mean and variance of 8 numbers $- 10, - 7, - 1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$ , respectively. Then the mean of 4 numbers $x, y, x + y + 1,$ $| x - y |$ is: [2026]

9
12
10
11

1

2

3

4

(4)

Q 45 :

Let the mean and variance of 7 observations 2, 4, 10, $x$ , 12, 14, $y$ , $x > y$ , be 8 and 16 respectively. Two numbers are chosen from {1, 2, 3, $x - 4$ , $y$ , 5} one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is: [2026]

$\frac{4}{5}$
$\frac{1}{3}$
$\frac{3}{5}$
$\frac{2}{5}$

1

2

3

4

(1)

$Mean (\bar{x}) = 8 (Given)$

$\Rightarrow \frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8$

$\Rightarrow x + y = 14 . . . (1)$

$Variance (σ^{2}) = 16 (Given)$

$\Rightarrow 16 = \frac{2^{2} + 4^{2} + 10^{2} + x^{2} + 12^{2} + 14^{2} + y^{2}}{7} - 8^{2}$

$\Rightarrow x^{2} + y^{2} = 100 . . . (2)$

$∵$ ${(x + y)}^{2} = x^{2} + y^{2} + 2 x y$

$\Rightarrow x y = 48 (sum is 14, product is 48)$

$Since problem states x > y$

$∴$ $x = 8 and y = 6$

$Now set X = {1, 2, 3, 4, 6, 5}$

Now we choose two numbers one after another without replacement

Total outcomes $= 6 \times 5 = 30$

We want the probability that the smaller number among the two is less than 4

$P (smaller < 4) = 1 - P (smaller \geq 4)$

$= 1 - \frac{6}{30} = \frac{4}{5}$

Q 46 :

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observation $α$ in this data is replaced by $β,$ then the mean and variance become 10.1 and 1.99, respectively. Then $α + β$ equals: [2026]

5
15
10
20

1

2

3

4

(4)

Q 47 :

Let $X = {x \in ℕ : 1 \leq x \leq 19}$ and for some $a, b \in ℝ, Y = {a x + b : x \in X} .$ If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of $b$ is [2026]

100
20
80
60

1

2

3

4

(4)

Q 48 :

A random variable X takes values 0, 1, 2, 3 with probabilities $\frac{2 a + 1}{30}, \frac{8 a - 1}{30}, \frac{4 a + 1}{30}, b$ respectively,

where $a, b \in R$ . Let $μ a n d σ$ respectively be the mean and standard deviation of X such that $σ^{2} + μ^{2} = 2 .$

Then $\frac{a}{b}$ is equal to : [2026]

3
30
60
12

1

2

3

4

(4)

Q 49 :

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are 2,3,5,10,11,13,15,21, then the mean deviation about the median of all the 10 observations is [2026]

6
7
4
5

1

2

3

4

(4)

Topic Question Set

Q 41 :

If the mean of the frequency distribution

Class : 0–10 10–20 20–30 30–40 40–50

Frequency : 2 3 $x$ 5 4

is 28, then its variance is _____ . [2023]

0%

0%

Q 43 :

If the variance of the frequency distribution

$x_{𝑖}$ 2 3 4 5 6 7 8

Frequency $f_{i}$ 3 6 16 $α$ 9 5 6

is 3, then $α$ is equal to _______. [2023]

0%

Q 44 :

Let the mean and variance of 8 numbers $- 10, - 7, - 1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$ , respectively. Then the mean of 4 numbers $x, y, x + y + 1,$ $| x - y |$ is: [2026]

Q 46 :

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observation $α$ in this data is replaced by $β,$ then the mean and variance become 10.1 and 1.99, respectively. Then $α + β$ equals: [2026]

Q 47 :

Let $X = {x \in ℕ : 1 \leq x \leq 19}$ and for some $a, b \in ℝ, Y = {a x + b : x \in X} .$ If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of $b$ is [2026]

Q 49 :

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are 2,3,5,10,11,13,15,21, then the mean deviation about the median of all the 10 observations is [2026]

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Class :	0–10	10–20	20–30	30–40	40–50
Frequency :	2	3	$x$	5	4