Statistics · JEE Main previous year questions

If the standard deviation of the numbers –1, 0, 1, k is $\sqrt{5}$ where k > 0, then k is equal to

1. A.$2\sqrt{6}$
2. B.$\sqrt{6}$
3. C.$2\sqrt{\frac{5}{6}}$
4. D.$2\sqrt{\frac{10}{3}}$

The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then $\frac{y}{x}$ is equal to

1. A.$\frac{7}{2}$
2. B.$\frac{8}{3}$
3. C.$\frac{9}{4}$
4. D.$\frac{7}{3}$

A student scores the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is

1. A.$\frac{100}{\sqrt{3}}$
2. B.$\frac{10}{\sqrt{3}}$
3. C.$\frac{10}{3}$
4. D.$\frac{100}{3}$

If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is :

1. A.31
2. B.50
3. C.51
4. D.30

The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4 ; then the absolute value of the difference of the other two observations, is :

1. A.1
2. B.7
3. C.3
4. D.5

5 students of a class have an average height $150\text{ cm}$ and variance $18{\text{ cm}}^2$. A new student, whose height is $156\text{ cm}$, joined them. The variance (in $\mathrm{c}{\mathrm{m}}^2$) of the height of these six students is :

1. A.16
2. B.22
3. C.20
4. D.18

The mean and the standard deviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

1. A.0
2. B.1
3. C.2
4. D.4

In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, $x+y=5$ and $x=4$. Then area of $\Delta \text{ ABC}$ (in sq. units) is :

1. A.12
2. B.4
3. C.5
4. D.9

The mean of set of 30 observations is 75. If each observation is multiplied by a non-zero number $\lambda$ and then each of them is decreased by 25, their mean remains the same. Then $\lambda$ is equal to :

1. A.$\frac{1}{3}$
2. B.$\frac{2}{3}$
3. C.$\frac{4}{3}$
4. D.$\frac{10}{3}$

If ${\sum }_{i=1}^9(x_i-5)=9$ and ${\sum }_{i=1}^9(x_i-5{)}^2=45$, then the standard deviation of the 9 items $x_1,x_2,\dots \dots \dots ,x_9$ is

1. A.$3$
2. B.$9$
3. C.$4$
4. D.$2$

The sum of 100 observations and the sum of their squares are 400 and 2475, respectively. Later on, three observations, 3, 4 and 5, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is :

1. A.8.25
2. B.8.50
3. C.8.00
4. D.9.00

The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is :

1. A.25
2. B.30
3. C.35
4. D.40

If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

1. A.$3a^2-26a+55=0$
2. B.$3a^2-32a+84=0$
3. C.$3a^2-34a+91=0$
4. D.$3a^2-23a+44=0$

The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is :

1. A.15.8
2. B.14.0
3. C.16.8
4. D.16.0

Let $PS$ be the median of the triangle with vertices $P(2,2),Q(6,-1)$ and $R(7,3)$. The equation of the line passing through $(1,-1)$ and parallel to PS is :

1. A.$4x+7y+3=0$
2. B.$2x-9y-11=0$
3. C.$4x-7y-11=0$
4. D.$2x+9y+7=0$

The variance of first 50 even natural numbers is

1. A.833
2. B.437
3. C.$\frac{437}{4}$
4. D.$\frac{833}{4}$

Statement 1: The variance of first $n$ odd natural numbers is $\frac{n^2-1}{3}$ Statement 2: The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{n(4n^2+1)}{3}$.

1. A.Statement 1 is true, Statement 2 is false.
2. B.Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.
3. C.Statement 1 is false, Statement 2 is true.
4. D.Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

If the mean of $4,7,2,8,6$ and $a$ is 7, then the mean deviation from the median of these observations is

1. A.8
2. B.5
3. C.1
4. D.3

The frequency distribution of daily working expenditure of families in a locality is as follows: Expenditure (in Rs.) ($x$) 0-50 50-100 100-150 150-200 200-250 No. of families ($f$) 24 33 37 $b$ 25 If the mode of the distribution is Rs. 140, then the value of $b$ is

1. A.34
2. B.31
3. C.26
4. D.36

For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is

1. A.$\frac{5}{2}$
2. B.$\frac{11}{2}$
3. C.$6$
4. D.$\frac{13}{2}$

If the mean deviation of numbers 1, 1 + d, 1 + 2d,........, 1 + 100d from their mean is 255, then d is equal to

1. A.20.0
2. B.10.1
3. C.20.2
4. D.10.0

Statement - 1 : The variance of first n even natural numbers is $\frac{n^2-1}{4}$ Statement - 2 : The sum of first n natural numbers is $\frac{n(n+1)}{2}$ and the sum of squares of first n natural numbers is $\frac{n(n+1)(2n+1)}{6}$

1. A.Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
2. B.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
3. C.Statement-1 is true, Statement-2 is false
4. D.Statement-1 is false, Statement-2 is true

Suppose a population A has 100 observations $101,102,\dots ,200$, and another population B has 100 observations $151,152,\dots ,250$. If $V_A$ and $V_B$ represent the variances of the two populations, respectively, then $\frac{V_A}{V_B}$ is

1. A.1
2. B.$\frac{9}{4}$
3. C.$\frac{4}{9}$
4. D.$\frac{2}{3}$

A random variable $X$ has Poisson distribution with mean $2$. Then $P(X>1.5)$ equals :

1. A.$\frac{2}{e^2}$
2. B.$0$
3. C.$1-\frac{3}{e^2}$
4. D.$\frac{3}{e^2}$

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately :

1. A.20.5
2. B.22.0
3. C.24.0
4. D.25.5