The mean and SD for a, b, and 2 are 3 and 1 respectively, the value of ab would be

Option a: 11.5. Solution: Given: Observations are a, b, 2; Mean = 3, SD = 1 Step 1: Use Mean condition. a + b + 2/3 = 3 a + b + 2 = 9 a + b = 7 quad s (1) Step 2: Use SD condition. sigma^2 = sum xi^2/n - x-bar^2 1^2 = a^2 + b^2 + 4/3 - 9 1 = a^2 + b^2 + 4/3 - 9 10 = a^2 + b^2 + 4/3 a^2 + b^2 + 4 = 30 a^2 + b^2 = 26 quad s (2) Step 3: Use the identity (a+b)^2 = a^2 + 2ab + b^2. 7^2 = 26 + 2ab 49 = 26 + 2ab 2ab = 23 ab = 11.5 So ab = 11.5, which is Option A. The given correctoption 'b' (5) is incorrect. Hence, Option A is the correct answer.

Central Tendency & DispersionMTP Nov 20Question 3166 of 473

All Questions

The mean and SD for $\displaystyle a, b,$ and $\displaystyle 2$ are $\displaystyle 3$ and $\displaystyle 1$ respectively, the value of $\displaystyle ab$ would be

Options

\displaystyle 11.5

\displaystyle 5

\displaystyle 12

\displaystyle 13

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Correct Answer

✅ Option a — $\displaystyle 11.5$

All Options:

A $\displaystyle 11.5$
B $\displaystyle 5$
C $\displaystyle 12$
D $\displaystyle 13$

Detailed Solution & Explanation

**Given:** Observations are

\displaystyle a, b, 2

; Mean = 3, SD = 1 **Step 1: Use Mean condition.**

\frac{a + b + 2}{3} = 3

a + b + 2 = 9

a + b = 7 \quad \cdots (1)

**Step 2: Use SD condition.**

\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2

1^2 = \frac{a^2 + b^2 + 4}{3} - 9

1 = \frac{a^2 + b^2 + 4}{3} - 9

10 = \frac{a^2 + b^2 + 4}{3}

a^2 + b^2 + 4 = 30

a^2 + b^2 = 26 \quad \cdots (2)

**Step 3: Use the identity

\displaystyle (a+b)^2 = a^2 + 2ab + b^2

.**

7^2 = 26 + 2ab

49 = 26 + 2ab

2ab = 23

ab = 11.5

\displaystyle ab = 11.5

, which is Option A. The given `correct_option` 'b' (5) is incorrect. Hence, **Option A** is the correct answer.

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The mean and SD for a,b,\displaystyle a, b,a,b, and 2\displaystyle 22 are 3\displaystyle 33 and 1\displaystyle 11 respectively, the value of ab\displaystyle abab would be