(D) Population $A$ consists of $100$ observations: $101, 102, \dots, 200$. The variance is $V_A$.Population $B$ consists of $100$ observations: $151,

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(D) Population $A$ consists of $100$ observations: $101, 102, \dots, 200$. The variance is $V_A$.Population $B$ consists of $100$ observations: $151, 152, \dots, 250$.We can write these observations as $(101 + 50), (102 + 50), \dots, (200 + 50)$.Since variance is independent of the change of origin,adding a constant $k$ to each observation does not change the variance.Therefore,$V_B = V_A$.Thus,$\frac{V_A}{V_B} = 1$.

Class 11 Mathematics Statistics Variance and Standard Deviation

Medium

Let population $A$ contain $100$ observations $101, 102, \dots, 200$ and another population $B$ contain $100$ observations $151, 152, \dots, 250$. If $V_A$ and $V_B$ represent the variances of the two populations respectively,then what is $V_A / V_B$?

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

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Class 11 Questions

Class 11

Mathematics Questions

Chapter

Statistics

Subtopic

Variance and Standard Deviation

Let $x_1, x_2, \dots, x_n$ be $n$ observations,$\bar{x}$ be their mean,and $\sigma^2$ be their variance.
Statement-$1$: The variance of $2x_1, 2x_2, \dots, 2x_n$ is $4\sigma^2$.
Statement-$2$: The mean of $2x_1, 2x_2, \dots, 2x_n$ is $4\bar{x}$.

Medium

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The mean and $S.D.$ of $1, 2, 3, 4, 5, 6$ are

Medium

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For a frequency distribution,the standard deviation is computed by:

Medium

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

Easy

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The means of two groups of size $200$ and $300$ are $25$ and $10$ respectively. Their standard deviations are $3$ and $4$ respectively. What is the variance of the combined sample of size $500$?

Difficult

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Let population $A$ contain $100$ observations $101, 102, \dots, 200$ and another population $B$ contain $100$ observations $151, 152, \dots, 250$. If $V_A$ and $V_B$ represent the variances of the two populations respectively,then what is $V_A / V_B$?

Similar Questions

Let $x_1, x_2, \dots, x_n$ be $n$ observations,$\bar{x}$ be their mean,and $\sigma^2$ be their variance.Statement-$1$: The variance of $2x_1, 2x_2, \dots, 2x_n$ is $4\sigma^2$.Statement-$2$: The mean of $2x_1, 2x_2, \dots, 2x_n$ is $4\bar{x}$.

The mean and $S.D.$ of $1, 2, 3, 4, 5, 6$ are

For a frequency distribution,the standard deviation is computed by:

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

The means of two groups of size $200$ and $300$ are $25$ and $10$ respectively. Their standard deviations are $3$ and $4$ respectively. What is the variance of the combined sample of size $500$?

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Let $x_1, x_2, \dots, x_n$ be $n$ observations,$\bar{x}$ be their mean,and $\sigma^2$ be their variance.
Statement-$1$: The variance of $2x_1, 2x_2, \dots, 2x_n$ is $4\sigma^2$.
Statement-$2$: The mean of $2x_1, 2x_2, \dots, 2x_n$ is $4\bar{x}$.

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