IITM BS Question 6 - Week 7 Practice

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IITM BS Question 6 - Week 7 Practice | Prasnya

00:00

Q6Comprehension0.5 Marks10 May 2026

Passage

Let

X

be a random variable with mean

\mu

and variance

\sigma^2

. Let

\bar{X}_1

and

\bar{X}_2

be the sample means of two independent random samples of sizes

n_1

and

n_2

, respectively, drawn from the distribution of

X

. Consider the estimator

\hat{\mu} = a\bar{X}_1 + b\bar{X}_2

, where

a

and

b

are constants such that

a + b = 1

. Based on the above data, answer the given subquestions.

Consider the derivation for

Var(\hat{\mu})

:\n

Var(\hat{\mu}) = Var(a\bar{X}_1 + b\bar{X}_2)

\nSince the two samples are independent,\n

\implies Var(\hat{\mu}) = \mathbf{A} Var(\bar{X}_1) + \mathbf{B} Var(\bar{X}_2) \quad \dots(1)

\nWhere,\n

Var(\bar{X}_1) = \frac{\mathbf{C}}{n_1}

Var(\bar{X}_2) = \frac{\sigma^2}{n_2}

\nNow, from equation (1):\n

Var(\hat{\mu}) = \sigma^2 \left( \frac{\mathbf{D}}{n_1} + \frac{\mathbf{B}}{n_2} \right)

\nOn substituting the values of

n_1, n_2, \sigma^2, a

and using the relation

a+b=1

, we get:\n

Var(\hat{\mu}) = \mathbf{E}

\n\n**Options list for the blanks:**\n(1)

b

\n(2)

\sigma^2

\n(3)

a

\n(4)

a^2

\n(5)

b^2

\n(6)

a^2

(or is it

\frac{a^2}{n_1}

? No, Option 6 is

a^2

)\n(7)

\frac{a}{n_1}

\n(8)

\frac{\sigma^2}{n_2}

\n(9)

\frac{b^2}{n_2}

\n(10)

\frac{\sigma^2}{n_1^2}

\n\nEnter the correct option number for

\mathbf{A}

Your answer

Press Enter to check.

00:00

10 May 2026

Statistics 2 · Week 7

Q6.Consider the derivation for

Var(\hat{\mu})

:\n

Var(\hat{\mu}) = Var(a\bar{X}_1 + b\bar{X}_2)

\nSince the two samples are independent,\n

\implies Var(\hat{\mu}) = \mathbf{A} Var(\bar{X}_1) + \mathbf{B} Var(\bar{X}_2) \quad \dots(1)

\nWhere,\n

Var(\bar{X}_1) = \frac{\mathbf{C}}{n_1}

Var(\bar{X}_2) = \frac{\sigma^2}{n_2}

\nNow, from equation (1):\n

Var(\hat{\mu}) = \sigma^2 \left( \frac{\mathbf{D}}{n_1} + \frac{\mathbf{B}}{n_2} \right)

\nOn substituting the values of

n_1, n_2, \sigma^2, a

and using the relation

a+b=1

, we get:\n

Var(\hat{\mu}) = \mathbf{E}

\n\n**Options list for the blanks:**\n(1)

b

\n(2)

\sigma^2

\n(3)

a

\n(4)

a^2

\n(5)

b^2

\n(6)

a^2

(or is it

\frac{a^2}{n_1}

? No, Option 6 is

a^2

)\n(7)

\frac{a}{n_1}

\n(8)

\frac{\sigma^2}{n_2}

\n(9)

\frac{b^2}{n_2}

\n(10)

\frac{\sigma^2}{n_1^2}

\n\nEnter the correct option number for

\mathbf{A}

.@passage

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00:00

Q6Comprehension0.5 Marks10 May 2026

Passage

Let

X

be a random variable with mean

\mu

and variance

\sigma^2

. Let

\bar{X}_1

and

\bar{X}_2

be the sample means of two independent random samples of sizes

n_1

and

n_2

, respectively, drawn from the distribution of

X

. Consider the estimator

\hat{\mu} = a\bar{X}_1 + b\bar{X}_2

, where

a

and

b

are constants such that

a + b = 1

. Based on the above data, answer the given subquestions.

Consider the derivation for

Var(\hat{\mu})

:\n

Var(\hat{\mu}) = Var(a\bar{X}_1 + b\bar{X}_2)

\nSince the two samples are independent,\n

\implies Var(\hat{\mu}) = \mathbf{A} Var(\bar{X}_1) + \mathbf{B} Var(\bar{X}_2) \quad \dots(1)

\nWhere,\n

Var(\bar{X}_1) = \frac{\mathbf{C}}{n_1}

Var(\bar{X}_2) = \frac{\sigma^2}{n_2}

\nNow, from equation (1):\n

Var(\hat{\mu}) = \sigma^2 \left( \frac{\mathbf{D}}{n_1} + \frac{\mathbf{B}}{n_2} \right)

\nOn substituting the values of

n_1, n_2, \sigma^2, a

and using the relation

a+b=1

, we get:\n

Var(\hat{\mu}) = \mathbf{E}

\n\n**Options list for the blanks:**\n(1)

b

\n(2)

\sigma^2

\n(3)

a

\n(4)

a^2

\n(5)

b^2

\n(6)

a^2

(or is it

\frac{a^2}{n_1}

? No, Option 6 is

a^2

)\n(7)

\frac{a}{n_1}

\n(8)

\frac{\sigma^2}{n_2}

\n(9)

\frac{b^2}{n_2}

\n(10)

\frac{\sigma^2}{n_1^2}

\n\nEnter the correct option number for

\mathbf{A}

Your answer

Press Enter to check.

00:00

10 May 2026

Statistics 2 · Week 7

Q6.Consider the derivation for

Var(\hat{\mu})

:\n

Var(\hat{\mu}) = Var(a\bar{X}_1 + b\bar{X}_2)

\nSince the two samples are independent,\n

\implies Var(\hat{\mu}) = \mathbf{A} Var(\bar{X}_1) + \mathbf{B} Var(\bar{X}_2) \quad \dots(1)

\nWhere,\n

Var(\bar{X}_1) = \frac{\mathbf{C}}{n_1}

Var(\bar{X}_2) = \frac{\sigma^2}{n_2}

\nNow, from equation (1):\n

Var(\hat{\mu}) = \sigma^2 \left( \frac{\mathbf{D}}{n_1} + \frac{\mathbf{B}}{n_2} \right)

\nOn substituting the values of

n_1, n_2, \sigma^2, a

and using the relation

a+b=1

, we get:\n

Var(\hat{\mu}) = \mathbf{E}

\n\n**Options list for the blanks:**\n(1)

b

\n(2)

\sigma^2

\n(3)

a

\n(4)

a^2

\n(5)

b^2

\n(6)

a^2

(or is it

\frac{a^2}{n_1}

? No, Option 6 is

a^2

)\n(7)

\frac{a}{n_1}

\n(8)

\frac{\sigma^2}{n_2}

\n(9)

\frac{b^2}{n_2}

\n(10)

\frac{\sigma^2}{n_1^2}

\n\nEnter the correct option number for

\mathbf{A}

.@passage

Your Answer

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