JEE Main Previous Year Statistics Questions and Solutions

Q 1 :
Let M denote the median of the following frequency distribution

Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20

Frequency 3 9 10 8 6

Then 20 M is equal to [2024]

416
52
208
104

1

2

3

4

(3)

Class	Frequency	Cumulative Frequency
0 - 4	3	3
4 - 8	9	12
8 - 12	10	22
12 - 16	8	30
16 - 20	6	36
	36

Where, $h = 4, l = 8, c . f . = 12, f = 10, n = 18$

$M = l + [\frac{\frac{n}{2} - c . f .}{f}] \times h = 8 + \frac{18 - 12}{10} \times 4$

$= 8 + \frac{6 \times 4}{10} = 10.4$

Now, $20 M = 20 \times 10.4 = 208$

Q 2 :
Marks obtains by all the students of class 12 are presented in a frequency 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 [2025]

48
52
40
44

1

2

3

4

(4)

In the given question, l = 12, cf = 18, median = 14, f = 12 and h = 6

$∴ Median = l + (\frac{\frac{N}{2} - c f}{f}) \times h$

$\Rightarrow 14 = 12 + (\frac{\frac{N}{2} - 18}{12}) \times 6$

$\Rightarrow 2 = \frac{\frac{N}{2} - 18}{2} \Rightarrow 4 = \frac{N}{2} - 18$

$\Rightarrow \frac{N}{2} = 22 \Rightarrow N = 44$

So, total number of students = 44.

Q 3 :
Let $x_{1}, x_{2}, \dots, x_{100}$ be in an arithmetic progression with $x_{1} = 2$ and their mean equal to 200. If $y_{i} = i (x_{i} - i),$ $1 \leq i \leq 100,$ then the mean of $y_{1}, y_{2}, \dots, y_{100}$ is [2023]

10101.50
10051.50
10049.50
10100

1

2

3

4

(3)

$Mean = 200 \Rightarrow \frac{\frac{100}{2} (2 \times 2 + 99 d)}{100} = 200$

$\Rightarrow 4 + 99 d = 400 \Rightarrow d = 4$

$y_{i} = i (x_{i} - i) = i (2 + (i - 1) 4 - i) = 3 i^{2} - 2 i$

$Mean = \frac{\sum y_{i}}{100} = \frac{1}{100} \sum_{i = 1}^{100} (3 i^{2} - 2 i)$

$= \frac{1}{100} {\frac{3 \times 100 \times 101 \times 201}{6} - \frac{2 \times 100 \times 101}{2}}$

$= 101 {\frac{201}{2} - 1} = 101 \times 99.5 = 10049.50$

Q 4 :
The mean of the coefficients of $x, x^{2}, \dots, x^{7}$ in the binomial expansion of ${(2 + x)}^{9}$ is __________. [2023]

0%

(2736)

${(2 + x)}^{9} = {}^{9}C_{0} 2^{9} + {}^{9}C_{1} 2^{8} x^{1} + {}^{9}C_{2} 2^{7} x^{2} + {}^{9}C_{3} 2^{6} x^{3} + {}^{9}C_{4} 2^{5} x^{4} + {}^{9}C_{5} 2^{4} x^{5} + {}^{9}C_{6} 2^{3} x^{6} + {}^{9}C_{7} 2^{2} x^{7} + {}^{9}C_{8} 2 x^{8} + {}^{9}C_{9} x^{9}$

$Coefficient of x = {}^{9}C_{1} 2^{8}$

$Coefficient of x^{2} = {}^{9}C_{2} 2^{7}, \dots, coefficient of x^{7} = {}^{9}C_{7} 2^{2}$

$Mean = \frac{{}^{9}C_{1} \cdot 2^{8} + {}^{9}C_{2} \cdot 2^{7} + \dots + {}^{9}C_{7} \cdot 2^{2}}{7}$

$= \frac{{(1 + 2)}^{9} - {}^{9}C_{0} \cdot 2^{9} - {}^{9}C_{8} \cdot 2^{1} - {}^{9}C_{9}}{7}$

$= \frac{3^{9} - 2^{9} - 18 - 1}{7} = \frac{19152}{7} = 2736$

Q 5 :
A fair $n (n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9},$ then $n$ is equal to _______ . [2023]

0%

(10)

$Mean = 1 \cdot \frac{n - 1}{n} + 2 \frac{1}{n} (\frac{n - 1}{n}) + 3 {(\frac{1}{n})}^{2} (\frac{n - 1}{n}) + \dots$

$∴$ $\frac{n}{9} = (\frac{n - 1}{n}) (1 + 2 (\frac{1}{n}) + 3 {(\frac{1}{n})}^{2} + \dots)$

$\Rightarrow \frac{n}{9} = (\frac{n - 1}{n}) {(1 - \frac{1}{n})}^{- 2} = (\frac{n - 1}{n}) \cdot \frac{n^{2}}{{(n - 1)}^{2}}$

$\Rightarrow \frac{n}{9} = \frac{n}{n - 1} \Rightarrow n = 10$

Topic Question Set

Q 1 :
Let M denote the median of the following frequency distribution

Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20

Frequency 3 9 10 8 6

Then 20 M is equal to [2024]

Q 3 :
Let $x_{1}, x_{2}, \dots, x_{100}$ be in an arithmetic progression with $x_{1} = 2$ and their mean equal to 200. If $y_{i} = i (x_{i} - i),$ $1 \leq i \leq 100,$ then the mean of $y_{1}, y_{2}, \dots, y_{100}$ is [2023]

Q 4 :
The mean of the coefficients of $x, x^{2}, \dots, x^{7}$ in the binomial expansion of ${(2 + x)}^{9}$ is __________. [2023]

0%

Q 5 :
A fair $n (n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9},$ then $n$ is equal to _______ . [2023]

0%

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Topic Question Set

Q 1 : Let M denote the median of the following frequency distribution Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 Frequency 3 9 10 8 6 Then 20 M is equal to [2024]

Q 3 : Let x1,x2,…,x100 be in an arithmetic progression with x1=2 and their mean equal to 200. If yi=i(xi-i), 1≤i≤100, then the mean of y1,y2,…,y100 is [2023]

Q 4 : The mean of the coefficients of x,x2,…,x7 in the binomial expansion of (2+x)9 is __________. [2023]

0%

Q 5 : A fair n(n>1) faces die is rolled repeatedly until a number less than n appears. If the mean of the number of tosses required is n9, then n is equal to _______ . [2023]

0%

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Q 1 :
Let M denote the median of the following frequency distribution

Class 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20

Frequency 3 9 10 8 6

Then 20 M is equal to [2024]

Q 3 :
Let $x_{1}, x_{2}, \dots, x_{100}$ be in an arithmetic progression with $x_{1} = 2$ and their mean equal to 200. If $y_{i} = i (x_{i} - i),$ $1 \leq i \leq 100,$ then the mean of $y_{1}, y_{2}, \dots, y_{100}$ is [2023]

Q 4 :
The mean of the coefficients of $x, x^{2}, \dots, x^{7}$ in the binomial expansion of ${(2 + x)}^{9}$ is __________. [2023]

Q 5 :
A fair $n (n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9},$ then $n$ is equal to _______ . [2023]