JEE Main Previous Year Statistics Questions and Solutions

Q 11 :
Let $a, b, c \in N$ and $a < b < c .$ Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, $a, b, c$ be 18, 4, and $\frac{136}{5}$ , respectively. Then $2 a + b - c$ is equal to _______ . [2024]

0%

(33)

$a, b, c \in N$ and $a < b < c$

Since, mean = 18 $\Rightarrow \frac{9 + 25 + a + b + c}{5} = 18$

$\Rightarrow 34 + a + b + c = 90 \Rightarrow a + b + c = 56$

Mean deviation about the mean is 4

$\Rightarrow \frac{| 9 - 18 | + | 25 - 18 | + | a - 18 | + | b - 18 | + | c - 18 |}{5} = 4$

$\Rightarrow 9 + 7 + | a - 18 | + | b - 18 | + | c - 18 | = 20$

$\Rightarrow | a - 18 | + | b - 18 | + | c - 18 | = 4$

Variance $= \frac{136}{5}$

$\Rightarrow \frac{\sum | x_{i} - \bar{x} |^{2}}{n} = \frac{136}{5}$

$\Rightarrow \frac{81 + 49 + | a - 18 |^{2} + | b - 18 |^{2} + | c - 18 |^{2}}{5} = \frac{136}{5}$

$\Rightarrow | a - 18 |^{2} + | b - 18 |^{2} + | c - 18 |^{2} = 136 - 130 = 6$

Possible values are $| a - 18 |^{2} = 1, | b - 18 |^{2} = 1$ and $| c - 18 |^{2} = 4$

$∵$ $a < b < c$

$∴$ $18 - a = 1, b - 18 = 1$ and $c - 18 = 2$

$\Rightarrow a = 17, b = 19$ and $c = 20$

$Hence, 2 a + b - c = 34 + 19 - 20 = 33$

Q 12 :
The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking, it was found that an observation was read as 10 in place of 12. If $μ$ and $σ^{2}$ denote the mean and variance of the correct observations respectively, then $15 (μ + μ^{2} + σ^{2})$ is equal to _________ . [2024]

0%

(2521)

We have, mean = 12, standard deviation = 3

Let $x_{1}, x_{2}, \dots, x_{15}$ be 15 observations

$∴$ $n = 15$

$\frac{x_{1} + x_{2} + \dots + x_{14} + 10}{15} = 12$ $\Rightarrow x_{1} + x_{2} + \dots + x_{14} = 170$

Now, correct mean = $μ$

$\frac{x_{1} + x_{2} + \dots + x_{14} + 12}{15} = μ \Rightarrow \frac{170 + 12}{15} = μ \Rightarrow μ = \frac{182}{15}$

Standard deviation $= \frac{1}{15} \sqrt{15 \sum_{i = 1}^{15} x_{i}^{2} - {(\sum_{i = 1}^{15} x_{i})}^{2}}$

$\Rightarrow 3 = \frac{1}{15} \sqrt{15 \times \sum_{i = 1}^{15} x_{i}^{2} - {(180)}^{2}}$ $\Rightarrow 45 = \sqrt{15 \times \sum_{i = 1}^{15} x_{i}^{2} - 32400}$

$\Rightarrow 2025 = 15 \sum_{i = 1}^{15} x_{i}^{2} - 32400$

$\Rightarrow 2025 + 32400 = 15 (x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} + 100)$

$\Rightarrow 2295 = x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} + 100$

$\Rightarrow x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} = 2195$

Correct Variance, $σ^{2} = \frac{1}{n^{2}} (15 \times \sum_{i = 1}^{15} x_{i}^{2} - {(\sum_{i = 1}^{n} x_{i})}^{2})$

$= \frac{1}{{(15)}^{2}} (15 \times (x_{1}^{2} + x_{2}^{2} + \dots + x_{14}^{2} + 144) - {(182)}^{2})$

$= \frac{1}{225} (15 \times (2339) - 33124)$ $= \frac{1}{225} (35085 - 33124)$

$σ^{2} = \frac{1961}{225}$

$∴$ $15 (μ + μ^{2} + σ^{2}) = 15 (\frac{182}{15} + {(\frac{182}{15})}^{2} + \frac{1961}{225})$

$= 15 (\frac{2730 + 33124 + 1961}{225})$ $= \frac{37815}{15} = 2521$

Q 13 :
The variance $σ^{2}$ of the data

$x_{i}$ 0 1 5 6 10 12 17

$f_{i}$ 3 2 3 2 6 3 3

is ________ . [2024]

0%

(29)

Mean $(\bar{x})$ $= \frac{\sum f_{i} x_{i}}{\sum f_{i}}$

$= \frac{0 \times 3 + 1 \times 2 + 5 \times 3 + 6 \times 2 + 10 \times 6 + 12 \times 3 + 17 \times 3}{3 + 2 + 3 + 2 + 6 + 3 + 3}$

$= \frac{176}{22} = 8$

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

$= \frac{3 {(0 - 8)}^{2} + 2 {(1 - 8)}^{2} + 3 {(5 - 8)}^{2} + 2 {(6 - 8)}^{2} + 6 {(10 - 8)}^{2} + 3 {(12 - 8)}^{2} + 3 {(17 - 8)}^{2}}{22}$

$= \frac{3 \times 64 + 2 \times 49 + 3 \times 9 + 2 \times 4 + 6 \times 4 + 3 \times 16 + 3 \times 81}{22}$

$= \frac{640}{22} = 29.09 \approx 29$

Q 14 :
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to : [2025]

106
103
100
105

1

2

3

4

(2)

Mean, $\bar{x} = 6 + 4 + a + 8 + b + 12 + 10 + 13$

$\Rightarrow 9 = \frac{53 + a + b}{8} \Rightarrow a + b = 72 - 53 = 19$

$∴$ a + b = 19 ... (i)

Variance, $σ^{2} = \frac{{(6 - 9)}^{2} + {(4 - 9)}^{2} + {(a - 9)}^{2} + {(8 - 9)}^{2} + {(b - 9)}^{2} + {(12 - 9)}^{2} + {(10 - 9)}^{2} + {(13 - 9)}^{2}}{8}$

$\Rightarrow 9.25 \times 8 = 9 + 25 + 1 + 9 + 1 + 16 + {(a - 9)}^{2} + {(b - 9)}^{2}$

$\Rightarrow 74 - 61 = {(a - 9)}^{2} + {(19 - a - 9)}^{2}$ [Using (i)]

$\Rightarrow 13 = {(a - 9)}^{2} + {(10 - a)}^{2}$

$\Rightarrow a^{2} + 81 - 18 a + a^{2} + 100 - 20 a = 13$

$\Rightarrow 2 a^{2} - 38 a = 13 - 181$

$\Rightarrow a^{2} - 19 a + 84 = 0$

$\Rightarrow (a - 12) (a - 7) = 0$

$\Rightarrow a = 12 or 7 \Rightarrow b = 7 or 12$

$∴$ a + b + ab = 12 + 7 + 84 = 103.

Q 15 :
Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, . . ., 5$ is [2025]

17
16.4
16
17.4

1

2

3

4

(3)

We have, $\bar{x} = 5 and σ^{2} = 10$

Now, $\bar{x} = \frac{\sum_{}^{} x_{i}^{}}{n} = \frac{1 + 3 + a + 7 + b}{5} \Rightarrow \frac{11 + a + b}{5} = 5$

$\Rightarrow a + b = 14$

Also, $σ^{2} = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(\bar{x})}^{2}$

$\Rightarrow 10 = \frac{1^{2} + 3^{2} + a^{2} + 7^{2} + b^{2}}{5} - 25$

$\Rightarrow a^{2} + b^{2} = 116$

Since, a > b $\Rightarrow$ a = 10 and b = 4

Now, $n + x_{n}$ : 2, 5, 13, 11, 9

$∴ σ^{2} = \frac{2^{2} + 5^{2} + 13^{2} + 11^{2} + 9^{2}}{5} - {(\frac{2 + 5 + 13 + 11 + 9}{5})}^{2} = 80 - 64 = 16$ .

Q 16 :
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

$\frac{3}{5}$
$\frac{2}{15}$
$\frac{28}{75}$
$\frac{11}{15}$

1

2

3

4

(3)

X	X=0	X=1	x=2
P(x)	$\frac{{}^{7}C_{2} \times {}^{3}C_{0}}{{}^{10}C_{2}} = \frac{7}{15}$	$\frac{{}^{7}C_{1} \times {}^{3}C_{1}}{{}^{10}C_{2}} = \frac{7}{15}$	$\frac{{}^{7}C_{0} \times {}^{3}C_{2}}{{}^{10}C_{2}} = \frac{1}{15}$

Mean, $E (X) = \sum_{}^{} X_{i} P (X_{i}) = 0 + \frac{7}{15} + \frac{2}{15} = \frac{3}{5}$

Now, variance $(X) = E (X^{2}) - (E {(X))}^{2} = \frac{28}{75}$

$= (0 + \frac{7}{15} + \frac{4}{15}) - {(\frac{3}{5})}^{2} = \frac{11}{15} - \frac{9}{25} = \frac{28}{75}$

Q 17 :
Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is : [2025]

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

1

2

3

4

(1)

Given observations are 2, 3, 3, 4, 5, 7, a, b.

Also, mean = 4 and S.D. = $\sqrt{2}$

$\Rightarrow \frac{24 + a + b}{8} = 4 \Rightarrow a + b = 8$

${(\sqrt{2})}^{2} = \frac{2^{2} + 3^{2} + 3^{2} + 4^{2} + 5^{2} + 7^{2} + a^{2} + b^{2}}{8} - 16$

$\Rightarrow 112 + a^{2} + b^{2} = 18 \times 8$

$\Rightarrow a^{2} + b^{2} = 32 \Rightarrow a = b = 4$

New numbers are 2, 3, 3, 4, 5, 7, 4, 4

Arrange the numbers in a ascending order

2, 3, 3, 4, 4, 4, 5, 7

So, mode = 4.

$∴$ Mean deviation about mode

$= \frac{2 + 1 + 1 + 0 + 1 + 3 + 0 + 0}{8} = 1$ .

Q 18 :
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 $μ$ and $σ$ respectively, then $10 (μ + σ)$ is equal to [2025]

447
445
451
449

1

2

3

4

(4)

We have, Mean = $(\bar{x}) = 40$

Mean $(\bar{x}) = \frac{\sum_{}^{} x_{i}^{}}{100}$

$\Rightarrow 40 \times 100 = \sum_{}^{} x_{i}^{} \Rightarrow \sum_{}^{} x_{i}^{} = 4000$

Correct mean, $μ = \frac{100 (40) - 50 + 40}{100}$

$= 40 - \frac{1}{10} = 39.9$

Also, S.D.( $σ$ ) = 5.1

$\Rightarrow {(5.1)}^{2} = \frac{\sum_{}^{} x_{i}^{2}}{100} - {(\bar{x})}^{2}$

$\Rightarrow \sum_{}^{} x_{i}^{2} = 100 \times {(40)}^{2} + 100 {(5.1)}^{2}$

$= 16 \times 10^{4} + {(5.1)}^{2} \times 100 = 162601$

Correct $σ^{2} = \frac{\sum_{}^{} x_{i}^{2} - 50^{2} + 40^{2}}{100} - {(μ)}^{2}$

$= 1617.01 - 1592.01 = 25 \Rightarrow σ = 5$

$∴ 10 (μ + σ) = 10 (39.9 + 5) = 10 \times 44.9 = 449$ .

Q 19 :
A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is : [2025]

64
48
51
32

1

2

3

4

(2)

When a coin is tossed three times, outcomes are:

{HHH, HHT, HTH, THH, THT, TTH, HTT, and TTT}

Outcomes when tails does not follows a head (only once):

{HHH, THH, TTT, TTH}

Outcomes when tail follows a head:

{HHT, HTT, HTH, THT}

$∴$ Probability distribution table is given by

$X_{i}$	0	1
$P (X_{i})$	1/2	1/2

Mean $(μ) = \sum_{}^{} X_{i}^{} P_{i} = \frac{1}{2}$

and variance $(σ^{2}) = \sum_{}^{} X_{i}^{2} P_{i} - μ^{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

$∴ 64 (μ + σ^{2}) = 64 (\frac{1}{2} + \frac{1}{4}) = 64 \times \frac{3}{4} = 48$ .

Q 20 :
For a statistical data $x_{1}, x_{2}, . . ., x_{10}$ of 10 values, a student obtained the mean as 5.5 and $\sum_{i = 1}^{10} x_{i}^{2} = 371$ . He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is [2025]

7
5
9
4

1

2

3

4

(1)

$\sum_{i = 1}^{10} x_{i}^{} = 5.5 \times 10 = 55$

Correct mean $= \frac{\sum_{i = 1}^{10} x_{i}^{} - 4 - 5 + 6 + 8}{10} = \frac{55 - 9 + 14}{10} = 6$

Corrected $\sum_{i = 1}^{10} x_{i}^{2} = 371 - 4^{2} - 5^{2} + 6^{2} + 8^{2} = 430$

Corrected variance $= \frac{Corrected \sum_{i = 1}^{10} x_{i}^{2}}{10} - {(Corrected \bar{x})}^{2}$

$= \frac{430}{10} - {(6)}^{2} = 43 - 36 = 7$

Topic Question Set

Q 11 :
Let $a, b, c \in N$ and $a < b < c .$ Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, $a, b, c$ be 18, 4, and $\frac{136}{5}$ , respectively. Then $2 a + b - c$ is equal to _______ . [2024]

0%

0%

Q 13 :
The variance $σ^{2}$ of the data

$x_{i}$ 0 1 5 6 10 12 17

$f_{i}$ 3 2 3 2 6 3 3

is ________ . [2024]

0%

Q 14 :
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to : [2025]

Q 15 :
Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, . . ., 5$ is [2025]

Q 16 :
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

Q 17 :
Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is : [2025]

Q 18 :
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 $μ$ and $σ$ respectively, then $10 (μ + σ)$ is equal to [2025]

Q 19 :
A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is : [2025]

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$x_{i}$	0	1	5	6	10	12	17
$f_{i}$	3	2	3	2	6	3	3

Topic Question Set

Q 11 : Let a,b,c∈N and a<b<c. Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, a,b,c be 18, 4, and 1365, respectively. Then 2a+b-c is equal to _______ . [2024]

0%

0%

Q 13 : The variance σ2 of the data xi 0 1 5 6 10 12 17 fi 3 2 3 2 6 3 3 is ________ . [2024]

0%

Q 14 : If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to : [2025]

Q 15 : Let the Mean and Variance of five observations x1=1,x2=3,x3=a,x4=7 and x5=b,a>b be 5 and 10 respectively. Then the Variance of the observations n+xn,n=1,2,...,5 is [2025]

Q 16 : A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

Q 17 : Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and 2 respectively. Then the mean deviation about the mode of these observations is : [2025]

Q 18 : 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 μ and σ respectively, then 10(μ+σ) is equal to [2025]

Q 19 : A coin is tossed three times. Let X denote the number of times a tail follows a head. If μ and σ2 denote the mean and variance of X, then the value of 64(μ+σ2) is : [2025]

Q 20 : For a statistical data x1,x2,...,x10 of 10 values, a student obtained the mean as 5.5 and ∑i=110xi2=371. He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is [2025]

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Q 11 :
Let $a, b, c \in N$ and $a < b < c .$ Let the mean, the mean deviation about the mean, and the variance of the 5 observations 9, 25, $a, b, c$ be 18, 4, and $\frac{136}{5}$ , respectively. Then $2 a + b - c$ is equal to _______ . [2024]

Q 13 :
The variance $σ^{2}$ of the data

$x_{i}$ 0 1 5 6 10 12 17

$f_{i}$ 3 2 3 2 6 3 3

is ________ . [2024]

Q 14 :
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then a + b + ab is equal to : [2025]

Q 15 :
Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, . . ., 5$ is [2025]

Q 16 :
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let X denote the number of defective pens. Then the variance of X is [2025]

Q 17 :
Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is : [2025]

Q 18 :
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 $μ$ and $σ$ respectively, then $10 (μ + σ)$ is equal to [2025]

Q 19 :
A coin is tossed three times. Let X denote the number of times a tail follows a head. If $μ$ and $σ^{2}$ denote the mean and variance of X, then the value of $64 (μ + σ^{2})$ is : [2025]