Variance and Standard Deviation

Interactive Feed

JEE Main 2012

LEVELBoard

Let $x_{1}, x_{2}, \dots, x_{n}$ be n observations, and let $\overset{x}{ˉ}$ be their arithmetic mean and $σ^{2}$ be the variance. Statement-1 : Variance of $2 x_{1}, 2 x_{2}, \dots, 2 x_{n}$ is $4 σ^{2}$ Statement-2 : Arithmetic mean $2 x_{1}, 2 x_{2}, \dots, 2 x_{n}$ is $4 \overset{x}{ˉ}$ .

(A)

Statement-1 is false, Statement-2 is true.

(B)

Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for Statement-1.

(C)

Statement-1 is true, statement-2 is true; statement-2 is not a correct explanation for Statement-1.

(D)

Statement-1 is true, statement-2 is false.

JEE Main 2009

LEVELJEE Main

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 $\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

(A)

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

(B)

Statement-1 is true, Statement-2 is false.

(C)

Statement-1 is false, Statement-2 is true.

(D)

Statement-1 is true, Statement-2 is true. Statement-2 is a correct explanation for Statement-1.

JEE Main 2008

LEVELJEE Main

The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b?

The standard deviation of 17 numbers is zero. Then

(A)

the numbers are in geometric progression with common ratio not equal to one.

(B)

eight numbers are positive, eight are negative and one is zero.

The marks obtained by 40 students are grouped in a frequency table in class intervals of 10 marks each. The mean and the variance obtained from this distribution are found to be 40 and 49 respectively. It was later discovered that two observations belonging to the class interval (21-30) were included in the class interval (31-40) by mistake. Find the mean and the variance after correcting the error.

JEE Advanced 1981

LEVELJEE Main

The mean square deviations of a set of observations $x_{1}, x_{2}, \dots, x_{n}$ about a point $c$ is defined to be $\frac{1}{n} \sum_{i = 1}^{n} (x_{i} - c)^{2}$ . The mean square deviations about $- 1$ and $+ 1$ of a set of observations are 7 and 3 respectively. Find the standard deviation of this set of observations.

JEE Main 2018 (Paper 1)

LEVELBoard

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 $x_{1}, x_{2}, \dots, x_{9}$ is :

JEE Main 2018 (15 April Evening)

LEVELBoard

If the mean of the data : 7, 8, 9, 7, 8, 7, $λ$ , 8 is 8, then the variance of this data is :-

JEE Main 2018 (16 April Shift 1)

LEVELJEE Main

The mean and the standard deviation(s.d.) of five observations are 9 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 :

JEE Main 2019 (9 January)

LEVELJEE Main

5 students of a class have an average height 150 cm and variance 18 cm². A new student, whose height is 156 cm, joined them. The variance (in cm²) of the height of these six students is:

JEE Main 2019 (10 January Shift 1)

LEVELJEE Main

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is :

JEE Main 2019 (9 January)

LEVELBoard

A data consists of $n$ observations: $x_{1}, x_{2}, ....., x_{n}$ . If $\sum_{i = 1}^{n} (x_{i} + 1)^{2} = 9 n$ and $\sum_{i = 1}^{n} (x_{i} - 1)^{2} = 5 n$ , then the standard deviation of this data is :

(A)

(B)

5

(C)

7

(D)

JEE Main 2019 (11 January)

LEVELJEE Main

The outcome of each of 30 items was observed; 10 items gave an outcome $\frac{1}{2} - d$ each, 10 items gave outcome $\frac{1}{2}$ each and the remaining 10 items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac{4}{3}$ then $∣ d ∣$ equals :

(A)

(B)

\frac{5}{2}

(C)

\frac{2}{3}

(D)

2

JEE Main 2019 (8 April Shift 1)

LEVELBoard

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :

(A)

40

(B)

49

(C)

48

(D)

45

JEE Main 2019 (08 April Shift 2)

LEVELBoard

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

(A)

10/ 3

(B)

100/ 3

(C)

100/3

(D)

10/3

JEE Main 2019 (09 April Shift 1)

LEVELJEE Main

If the standard deviation of the numbers $- 1, 0, 1, k$ is $5$ where $k > 0$ , then $k$ is equal to

(A)

2 \frac{10}{3}

(B)

26

(C)

4 \frac{5}{3}

(D)

6

JEE Main 2019 (10 April Shift 2)

LEVELJEE Main

If both the mean and the standard deviation of 50 observations $x_{1}, x_{2}, ..., x_{50}$ are equal to 16, then the mean of $(x_{1} - 4)^{2}, (x_{2} - 4)^{2}, ..., (x_{50} - 4)^{2}$ is :

JEE Main 2020 (7 January Shift 1)

LEVELBoard

If variance of first $n$ natural numbers is 10 and variance of first $m$ even natural numbers is 16, $n + m$ is equal to

JEE Main 2020 (7 January Shift 1)

LEVELBoard

If the mean and variance of eight numbers $3, 7, 9, 12, 13, 20, x$ and $y$ be 10 and 25 respectively then $x y$ is equal to ________.

JEE Main 2020 (8 January Shift 1)

LEVELBoard

The mean and standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by $p$ and then reduced by $q$ , where $p \neq = 0$ and $q \neq = 0$ . If the new mean and standard deviation become half of their original values, then $q$ is equal to:

(A)

- 20

(B)

- 5

(C)

10

(D)

- 10

JEE Main 2020 (9 January Shift 1)

LEVELBoard

Let the observations $x_{i} (1 \leq i \leq 10)$ satisfy the equations, $\sum_{i = 1}^{10} (x_{i} - 5) = 10$ and $\sum_{i = 1}^{10} (x_{i} - 5)^{2} = 40$ . If $μ$ and $λ$ are the mean and the variance of observations, $(x_{1} - 3), (x_{2} - 3) \dots (x_{10} - 3)$ , then the ordered pair $(μ, λ)$ is equal to:

JEE Main 2020 - 7 Jan (Evening)

LEVELBoard

If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, $x$ and $y$ be 10 and 25 respectively, then $x \cdot y$ is equal to....

JEE Main 2020 - 8 Jan (Morning)

LEVELJEE Main

The mean and the standard deviation ( $s, d$ ) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by $p$ and then reduced by $q$ , where $p \neq = 0$ and $q \neq = 0$ . If the new mean and new $s . d .$ become half of their original values, then $q$ is equal to :

(A)

- 10

(B)

- 20

(C)

- 5

(D)

10

JEE Main 2020 - 2 Sep (Morning)

LEVELJEE Main

Let $X = {x \in N : 1 \leq x \leq 17}$ and $Y = {a x + b : x \in X and a, b \in R, a > 0}$ . If mean and variance of elements of $Y$ are 17 and 216 respectively then $a + b$ is equal to :

(A)

7

(B)

- 27

(C)

9

(D)

- 7

JEE Main 2020 - 3 Sep (Morning)

LEVELJEE Main

For the frequency distribution: Variate ( $x$ ): $x_{1}, x_{2}, x_{3} \dots x_{15}$ Frequency ( $f$ ): $f_{1}, f_{2}, f_{3} \dots f_{15}$ where $0 < x_{1} < x_{2} < x_{3} < \dots < x_{15} = 10$ and $\sum_{i = 1}^{15} f_{i} > 0$ , the standard deviation cannot be:

(A)

1

(B)

6

(C)

2

(D)

4

The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b?

The standard deviation of 17 numbers is zero. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑i=19​(xi​−5)=9 and ∑i=19​(xi​−5)2=45, then the standard deviation of the 9 items x1​,x2​,…,x9​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean of the data : 7, 8, 9, 7, 8, 7, λ, 8 is 8, then the variance of this data is :-

The mean and the standard deviation(s.d.) of five observations are 9 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 :

5 students of a class have an average height 150 cm and variance 18 cm². A new student, whose height is 156 cm, joined them. The variance (in cm²) of the height of these six students is:

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A data consists of n observations: x1​,x2​,.....,xn​. If ∑i=1n​(xi​+1)2=9n and ∑i=1n​(xi​−1)2=5n, then the standard deviation of this data is :

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the standard deviation of the numbers −1,0,1,k is 5​ where k>0, then k is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If both the mean and the standard deviation of 50 observations x1​,x2​,...,x50​ are equal to 16, then the mean of (x1​−4)2,(x2​−4)2,...,(x50​−4)2 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If variance of first n natural numbers is 10 and variance of first m even natural numbers is 16, n+m is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean and variance of eight numbers 3,7,9,12,13,20,x and y be 10 and 25 respectively then xy is equal to ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively, then x⋅y is equal to....

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let X={x∈N:1≤x≤17} and Y={ax+b:x∈X and a,b∈R,a>0}. If mean and variance of elements of Y are 17 and 216 respectively then a+b is equal to :

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 $x_{1}, x_{2}, \dots, x_{9}$ is :

If the mean of the data : 7, 8, 9, 7, 8, 7, $λ$ , 8 is 8, then the variance of this data is :-

A data consists of $n$ observations: $x_{1}, x_{2}, ....., x_{n}$ . If $\sum_{i = 1}^{n} (x_{i} + 1)^{2} = 9 n$ and $\sum_{i = 1}^{n} (x_{i} - 1)^{2} = 5 n$ , then the standard deviation of this data is :

If the standard deviation of the numbers $- 1, 0, 1, k$ is $5$ where $k > 0$ , then $k$ is equal to

If both the mean and the standard deviation of 50 observations $x_{1}, x_{2}, ..., x_{50}$ are equal to 16, then the mean of $(x_{1} - 4)^{2}, (x_{2} - 4)^{2}, ..., (x_{50} - 4)^{2}$ is :

If variance of first $n$ natural numbers is 10 and variance of first $m$ even natural numbers is 16, $n + m$ is equal to

If the mean and variance of eight numbers $3, 7, 9, 12, 13, 20, x$ and $y$ be 10 and 25 respectively then $x y$ is equal to ________.

If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, $x$ and $y$ be 10 and 25 respectively, then $x \cdot y$ is equal to....

Let $X = {x \in N : 1 \leq x \leq 17}$ and $Y = {a x + b : x \in X and a, b \in R, a > 0}$ . If mean and variance of elements of $Y$ are 17 and 216 respectively then $a + b$ is equal to :