JEE Main Previous Year Statistics Questions and Solutions

Q 31 :

Let $9 = x_{1} < x_{2} < \dots < x_{7}$ be in an A.P. with common difference $d$ . If the standard deviation of $x_{1}, x_{2}, \dots, x_{7}$ is $4$ and the mean is $\bar{x}$ , then $\bar{x} + x_{6}$ is equal to [2023]

$18 (1 + \frac{1}{\sqrt{3}})$
$2 (9 + \frac{8}{\sqrt{7}})$
$34$
$25$

1

2

3

4

(3)

$Mean = \bar{x} = \frac{\sum_{i = 1}^{7} x_{i}}{7} = \frac{\frac{7}{2} [2 a + 6 d]}{7} = a + 3 d = x_{4}$

$Variance = \frac{\sum_{i = 1}^{7} {(x_{i} - \bar{x})}^{2}}{7} = {(4)}^{2} \Rightarrow \frac{\sum_{i = 1}^{7} {(x_{i} - x_{4})}^{2}}{7} = 16$

$\Rightarrow \frac{{(3 d)}^{2} + {(2 d)}^{2} + d^{2} + 0 + d^{2} + {(2 d)}^{2} + {(3 d)}^{2}}{7} = 16$

$\Rightarrow \frac{28 d^{2}}{7} = 16 \Rightarrow d = 2 \Rightarrow \bar{x} = 9 + 3 (2) = 15$

$x_{6} = a + 5 d = 9 + 5 (2) = 19$ $∴$ $\bar{x} + x_{6} = 15 + 19 = 34$

Q 32 :

Let the six numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ be in A.P. and $a_{1} + a_{3} = 10$ . If the mean of these six numbers is $\frac{19}{2}$ and their variance is $σ^{2}$ , then $8 σ^{2}$ is equal to [2023]

105
210
200
220

1

2

3

4

(2)

$Given a_{1}, a_{2}, a_{3}, \dots, a_{6} are in A.P.$

$Sum of these terms in A.P. = \frac{6}{2} [2 a + 5 d] = \frac{19}{2} \times 6$

$\Rightarrow 2 a + 5 d = 19 ...(i)$

$Since a_{1} + a_{3} = 10 \Rightarrow a + a + 2 d = 10$

$\Rightarrow 2 a + 2 d = 10 ...(ii)$

$Solving (i) and (ii), we get a = 2, d = 3$

$So, 2, 5, 8, 11, 14, 17 are the terms.$

$Variance, σ^{2} = \frac{2^{2} + 5^{2} + 8^{2} + 11^{2} + 14^{2} + 17^{2}}{6} - {(\frac{19}{2})}^{2}$

$= \frac{699}{6} - \frac{361}{4} = \frac{466 - 361}{4} = \frac{105}{4}$ $\Rightarrow 8 σ^{2} = \frac{8 \times 105}{4} = 210$

Q 33 :

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: [2023]

3.92
3.96
4.04
4.08

1

2

3

4

(2)

$Mean, μ = 10$

$Variance, σ^{2} = 4$

$Let the number of observations be n$

$Sum of observations \sum x_{i} = 10 n$

$When marks of one student increased from 8 to 12$

$\sum x_{i} = 10 n - 8 + 12 = 10 n + 4$

$New mean = \frac{10 n + 4}{n} = 10.2$ $\Rightarrow n = 20;$ $4 = \frac{\sum x_{i}^{2}}{20} - 10^{2}$

$Incorrect \sum x_{i}^{2} = 104 \times 20 = 2080$ $⇒Correct \sum x_{i}^{2} = 2080 - 8^{2} + 12^{2} = 2160$

$New σ^{2} = \frac{2160}{20} - 10 . 2^{2} = 3.96$

Q 34 :

Let $S$ be the set of all values of $a_{1}$ for which the mean deviation about the mean of 100 consecutive positive integers $a_{1}, a_{2}, a_{3}, \dots, a_{100}$ is $25$ . Then $S$ is [2023]

${99}$
$ϕ$
$N$
${9}$

1

2

3

4

(3)

$Let a_{1} be a natural number.$

$The 100 consecutive positive numbers are a_{1}, a_{1} + 1, a_{1} + 2, a_{1} + 3, \dots, a_{1} + 99$

$Mean (\bar{x}) = \frac{a_{1} + a_{1} + 1 + a_{1} + 2 + \dots + a_{1} + 99}{100} = a_{1} + \frac{99}{2}$

$Mean deviation about the mean = \frac{\sum_{i = 1}^{100} | x_{_{i}}^{2} - \bar{x} |}{100}$

$= \frac{\frac{99}{2} + \frac{97}{2} + \frac{95}{2} + \dots + \frac{99}{2}}{100} = 25$

$Hence, a_{1} is a natural number.$

Q 35 :

Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$ , then $3 x - 2 y$ is equal to _____. [2023]

0%

(25)

We have, $x, y, 10, 12, 6, 12, 4, 8$

Mean, $\bar{x} = \frac{x + y + 10 + 12 + 6 + 12 + 4 + 8}{8} = 9$

$\Rightarrow x + y = 20 \dots (i)$

Now, Variance $= \frac{x^{2} + y^{2} + {(10)}^{2} + {(12)}^{2} + {(6)}^{2} + {(12)}^{2} + {(4)}^{2} + {(8)}^{2}}{8} - 81$

$\Rightarrow 9.25 = \frac{x^{2} + y^{2} + 504}{8} - 81$ $\Rightarrow x^{2} + y^{2} = 218 \dots (i i)$

Now, ${(x + y)}^{2} = x^{2} + y^{2} + 2 x y$

$\Rightarrow {(20)}^{2} = 218 + 2 x y$ $\Rightarrow 2 x y = 182$

Now, ${(x - y)}^{2} = 218 - 182 = 36$ $\Rightarrow x - y = \pm 6 \dots (i i i)$

From (i) and (iii), we get $x = 13, y = 7$ $(∵ x > y)$

So, $3 x - 2 y = 39 - 14 = 25$

Q 36 :

Let the mean of the data

$x$ 1 3 5 7 9

Frequency $(f)$ 4 24 28 $α$ 8

be 5. If $m$ and $σ^{2}$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 α}{m + σ^{2}}$ is equal to ________ . [2023]

0%

(8)

Given, $\bar{x} = 5$

$\Rightarrow \frac{\sum x_{i} f_{i}}{\sum f_{i}} = \frac{4 + 72 + 140 + 7 α + 72}{64 + α} = 5$

$\Rightarrow 288 + 7 α = 320 + 5 α \Rightarrow 2 α = 32 \Rightarrow α = 16$

Now, $M.D. (\bar{x}) = \frac{\sum f_{i} | x_{i} - \bar{x} |}{\sum f_{i}}$

$= \frac{4 \times 4 + 24 \times 2 + 28 \times 0 + 16 \times 2 + 8 \times 4}{80} = \frac{128}{80} = \frac{8}{5} \Rightarrow m = \frac{8}{5}$

Variance $(σ^{2}) = \frac{\sum f_{i} | x_{i} - \bar{x} |^{2}}{\sum f_{i}}$

$= \frac{4 \times 16 + 24 \times 4 + 28 \times 0 + 16 \times 4 + 8 \times 16}{80} = \frac{352}{80} = \frac{22}{5}$

$∴$ $\frac{3 α}{m + σ^{2}} = \frac{3 \times 16}{\frac{8}{5} + \frac{22}{5}} = \frac{48}{6} = 8$

Q 37 :

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is __________. [2023]

0%

(269)

Given, Mean $\bar{X} = 50, σ = 12$

$\bar{X} = 50 - \frac{50}{10} = 45$

$\frac{\sum x_{i}^{2}}{10} - 50^{2} = 144$ $[σ^{2} = \frac{\sum x_{i}^{2}}{n} - {(\bar{x})}^{2}]$

$\frac{\sum x_{i}^{2}}{10} = 2500 + 144$

$\frac{\sum x_{i}^{2}}{10} = 2500 + 144 + \frac{400 + 625 - 45^{2} - 50^{2}}{10} = 2294$

$Now correct variance: σ_{T}^{2} = 2294 - {(45)}^{2} = 2294 - 2025 = 269$

Q 38 :

Let $X = {11, 12, 13, \dots, 40, 41}$ and $Y = {61, 62, 63, \dots, 90, 91}$ be the two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $σ^{2}$ is the variance of all the observations in $X \cup Y$ , then $| \bar{x} + \bar{y} - σ^{2} |$ is equal to _______ . [2023]

0%

(603)

$Mean \bar{x} = \frac{\sum_{i = 11}^{41} i}{31} = \frac{11 + 41}{2} = 26$

$\bar{y} = \frac{\sum_{j = 61}^{91} j}{31} = \frac{61 + 91}{2} = 76$

$Combined mean μ = \frac{31 \times 26 + 31 \times 76}{31 + 31} = \frac{26 + 76}{2} = 51$

$σ^{2} = \frac{1}{62} (\sum_{i = 1}^{31} {(x_{i} - μ)}^{2} + \sum_{i = 1}^{31} {(y_{i} - μ)}^{2}) = 705$

$∴$ $| \bar{x} + \bar{y} - σ^{2} |$ $=$ $| 26 + 76 - 705 |$ $= 603$

Q 39 :

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observations, then $a + 3 b - 5$ is equal to ______. [2023]

0%

(37)

$Mean = 7, Variance = 16$

Let the observations be $x_{1}, x_{2}, \dots, x_{7}$ .

Now, $\frac{x_{1} + x_{2} + \dots + x_{7}}{7} = 8,$ $x_{1} + x_{2} + \dots + x_{7} = 56$

When one observation $14$ is omitted, the mean is

$a = \frac{x_{1} + x_{2} + \dots + x_{7} - 14}{6} \Rightarrow a = \frac{56 - 14}{6} = 7$

$Variance = \frac{\sum_{i = 1}^{7} {(x_{i})}^{2}}{7} - 64 = 16 \Rightarrow \sum_{i = 1}^{7} {(x_{i})}^{2} = 560$

When one observation is omitted, then variance

$b = \frac{\sum_{i = 1}^{6} {(x_{i})}^{2}}{6} - 49 \Rightarrow b = \frac{560 - {(14)}^{2}}{6} - 49 = \frac{364}{6} - 49 = \frac{35}{3}$

So, $a + 3 b - 5 = 7 + 3 \times \frac{35}{3} - 5 = 37$

Q 40 :

If the mean and variance of the frequency distribution

$x_{i}$ 2 4 6 8 10 12 14 16

$f_{i}$ 4 4 $α$ 15 8 $β$ 4 5

are 9 and 15.08 respectively, then the value of $α^{2} + β^{2} - α β$ is ________ . [2023]

0%

(25)

$x_{i}$	$f_{i}$	$x_{i} f_{i}$	$f_{i} x_{i}^{2}$
2	4	8	16
4	4	16	64
6	$α$	$6 α$	$36 α$
8	15	120	960
10	8	80	800
12	$β$	$12 β$	$144 β$
14	$4$	56	784
16	5	80	1280
Total	$40 + α + β$	$360 + 6 α + 12 β$	$3904 + 36 α + 144 β$

$Mean (\bar{x}) = \frac{\sum f_{i} x_{i}}{\sum f_{i}} \Rightarrow \frac{360 + 6 α + 12 β}{40 + α + β} = 9$

$\Rightarrow 360 + 6 α + 12 β = 360 + 9 α + 9 β$

$\Rightarrow 3 α = 3 β \Rightarrow α = β$ ... $(i)$

$Also, variance σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{\sum f_{i}} - {(\frac{\sum f_{i} x_{i}}{\sum f_{i}})}^{2}$

$\Rightarrow \frac{3904 + 36 α + 144 β}{40 + α + β} - {(\bar{x})}^{2} = 15.08$

$\Rightarrow \frac{3904 + 180 α}{40 + 2 α} - {(9)}^{2} = 15.08$ $\Rightarrow α = 5$

From (i), $β = 5$

$∴$ $α^{2} + β^{2} - α β = 25 + 25 - 25 = 25$

Topic Question Set