Central Tendency & DispersionPYQ Jun 23Question 2872 of 473
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A Professor has given assignment to students in a Statistics class. A student Jagan computes the arithmetic mean and standard deviation for a set of 100 observations as 50 and 5 respectively. Later on, Sonali points out to Jagan that he has made of mistake in taking one observation as 100 instead of 50. What would be the correct mean if the wrong observation is corrected?

Options

A50.5\displaystyle 50.5
B49.9\displaystyle 49.9
C49.5\displaystyle 49.5
D50.1\displaystyle 50.1
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Correct Answer

✅ Option c — 49.5\displaystyle 49.5

All Options:

  • A50.5\displaystyle 50.5
  • B49.9\displaystyle 49.9
  • C49.5\displaystyle 49.5
  • D50.1\displaystyle 50.1

Detailed Solution & Explanation

**Step 1: Find the wrong total sum.** Wrong sum=100×50=5000\text{Wrong sum} = 100 \times 50 = 5000 **Step 2: Compute the corrected sum.** Wrong observation: 100. Correct observation: 50. Correct sum=5000−100+50=4950\text{Correct sum} = 5000 - 100 + 50 = 4950 **Step 3: Compute the correct mean.** Correct mean=4950100=49.5\text{Correct mean} = \frac{4950}{100} = 49.5 Hence, **Option C** is the correct answer.

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