In calculating the mean and variance of...

MathematicsMeasures of DispersionJEE Advanced 1979Easy

In calculating the mean and variance of 10 readings, a student wrongly used the figure 52 for the correct figure of 25. He obtained the mean and variance as 45.0 and 16.0 respectively. Determine the correct mean and variance.

Answer:Correct mean 42.3 ; Correct variance 43.81

Visualized Solution (English)

Given Information

Number of readings $n = 10$
Incorrect Mean $\overset{x}{ˉ}_{o l d} = 45.0$
Incorrect Variance $σ_{o l d}^{2} = 16.0$
Incorrect value used $= 52$
Correct value to be used $= 25$

Calculating Incorrect Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum x_{o l d}

Formula for mean: $\overset{x}{ˉ} = \frac{\sum x}{n}$
Rearranging for sum: $\sum x = n \times \overset{x}{ˉ}$
Incorrect sum $\sum x_{o l d} = 10 \times 45.0 = 450$

Correcting the Sum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum x_{n e w}

Correct sum $\sum x_{n e w} = \sum x_{o l d} - (wrong value) + (correct value)$
$\sum x_{n e w} = 450 - 52 + 25$
$\sum x_{n e w} = 423$

Calculating Correct Mean .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \overset{x}{ˉ}_{n e w}

Correct mean $\overset{x}{ˉ}_{n e w} = \frac{\sum x _{n e w}}{n}$
$\overset{x}{ˉ}_{n e w} = \frac{423}{10}$
$\overset{x}{ˉ}_{n e w} = 42.3$

Formula for Variance

Variance formula: $σ^{2} = \frac{\sum x ^{2}}{n} - (\overset{x}{ˉ})^{2}$
We need to find $\sum x^{2}$ to correct it.

Finding Incorrect Sum of Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum x_{o l d}^{2}

$16.0 = \frac{\sum x _{o l d}^{2}}{10} - (45.0)^{2}$
$16.0 = \frac{\sum x _{o l d}^{2}}{10} - 2025$
$\frac{\sum x _{o l d}^{2}}{10} = 16 + 2025 = 2041$
$\sum x_{o l d}^{2} = 20410$

Correcting the Sum of Squares .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \sum x_{n e w}^{2}

Correct sum of squares $\sum x_{n e w}^{2} = \sum x_{o l d}^{2} - (52)^{2} + (25)^{2}$
$\sum x_{n e w}^{2} = 20410 - 2704 + 625$
$\sum x_{n e w}^{2} = 18331$

Calculating Correct Variance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } σ_{n e w}^{2}

Correct variance $σ_{n e w}^{2} = \frac{\sum x _{n e w}^{2}}{10} - (\overset{x}{ˉ}_{n e w})^{2}$
$σ_{n e w}^{2} = \frac{18331}{10} - (42.3)^{2}$
$σ_{n e w}^{2} = 1833.1 - 1789.29$
$σ_{n e w}^{2} = 43.81$

Summary and Takeaway

Correct Mean $= 42.3$
Correct Variance $= 43.81$
Key Takeaway: Always correct the fundamental sums ( $\sum x$ and $\sum x^{2}$ ) before recalculating statistical measures.

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