Measures of Central Tendency and DispersionMCQMTP Nov 20Question 3166 of 473
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The mean and SD for a,b,\displaystyle a, b, and 2\displaystyle 2 are 3\displaystyle 3 and 1\displaystyle 1 respectively, the value of ab\displaystyle ab would be

Options

A11.5\displaystyle 11.5
B5\displaystyle 5
C12\displaystyle 12
D13\displaystyle 13
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Correct Answer

Option a11.5\displaystyle 11.5

All Options:

  • A11.5\displaystyle 11.5
  • B5\displaystyle 5
  • C12\displaystyle 12
  • D13\displaystyle 13

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Detailed Solution & Explanation

**Given:** Observations are a,b,2\displaystyle a, b, 2; Mean = 3, SD = 1 **Step 1: Use Mean condition.** a+b+23=3\frac{a + b + 2}{3} = 3 a+b+2=9a + b + 2 = 9 a+b=7(1)a + b = 7 \quad \cdots (1) **Step 2: Use SD condition.** σ2=xi2nxˉ2\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 12=a2+b2+4391^2 = \frac{a^2 + b^2 + 4}{3} - 9 1=a2+b2+4391 = \frac{a^2 + b^2 + 4}{3} - 9 10=a2+b2+4310 = \frac{a^2 + b^2 + 4}{3} a2+b2+4=30a^2 + b^2 + 4 = 30 a2+b2=26(2)a^2 + b^2 = 26 \quad \cdots (2) **Step 3: Use the identity (a+b)2=a2+2ab+b2\displaystyle (a+b)^2 = a^2 + 2ab + b^2.** 72=26+2ab7^2 = 26 + 2ab 49=26+2ab49 = 26 + 2ab 2ab=232ab = 23 ab=11.5ab = 11.5 So 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.

About This Chapter: Measures of Central Tendency and Dispersion

Paper

Paper 3: Quantitative Aptitude

Weightage

12-15 Marks

Key Topics

Mean, Median, Mode, Range, Mean Deviation, Standard Deviation

The core foundation of Statistics. This chapter covers Mean (Arithmetic, Geometric, Harmonic), Median, Mode, and their properties. It also explores measures of spread like Range, Mean Deviation, Standard Deviation, and Quartile Deviation.

View Official ICAI Syllabus

Exam Strategy Tip

Do not just memorize formulas; ICAI loves asking about the mathematical properties (e.g., 'sum of deviations from the AM is always zero'). You can usually eliminate 2 options just by knowing the properties.

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