Question

The standard deviation of first 10 multiples of 4 is

• A. 7
• B. 8
• C. 11.5
  % Checkmark on this option in image
• D. 14

Hint:

For a set of numbers \(x_1, x_2, \dots, x_N\): Mean \(\bar{x} = \frac{\sum x_i}{N}\). Variance \(\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N} = \frac{\sum x_i^2}{N} - (\bar{x})^2\). Standard Deviation \(\sigma = \sqrt{\text{Variance}}\).
For the first N natural numbers (1 to N), Variance = \(\frac{N^2-1}{12}\).
If \(y_i = ax_i + b\), then \(\text{Var}(y) = a^2 \text{Var}(x)\) and \(\text{SD}(y) = |a| \text{SD}(x)\).

Answer 1

The Correct Option is C

The first 10 multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40. These can be written as \(4k\) for \(k=1, 2, \dots, 10\). Let \(x_k = 4k\). Mean (\(\bar{x}\)): \( \bar{x} = \frac{\sum_{k=1}^{10} 4k}{10} = \frac{4 \sum_{k=1}^{10} k}{10} \). We know \(\sum_{k=1}^{N} k = \frac{N(N+1)}{2}\). So for N=10, \(\sum_{k=1}^{10} k = \frac{10(11)}{2} = 55\). \( \bar{x} = \frac{4 \times 55}{10} = \frac{220}{10} = 22 \). Variance (\(\sigma^2\)): The variance of the first N natural numbers \(1, 2, \dots, N\) is \(\sigma_N^2 = \frac{N^2-1}{12}\). The numbers here are \(4 \times 1, 4 \times 2, \dots, 4 \times 10\). If \(y_i = c x_i\), then \(\text{Var}(y) = c^2 \text{Var}(x)\). Here, our numbers are \(4k\). Let \(X\) be the random variable taking values \(1,2,\dots,10\). \(\text{Var}(X) = \frac{10^2-1}{12} = \frac{99}{12} = \frac{33}{4} = 8.25\). The numbers are \(4X\). So, Variance(\(4X\)) = \(4^2 \text{Var}(X) = 16 \times \text{Var}(X)\). \( \sigma^2 = 16 \times \frac{33}{4} = 4 \times 33 = 132 \). Standard Deviation (\(\sigma\)): \( \sigma = \sqrt{\text{Variance}} = \sqrt{132} \). \( \sqrt{132} = \sqrt{4 \times 33} = 2\sqrt{33} \). Now, approximate \(\sqrt{33}\). We know \(5^2=25, 6^2=36\). So \(\sqrt{33}\) is between 5 and 6. \(5.5^2 = 30.25\). \(5.7^2 \approx (6-0.3)^2 = 36 - 3.6 + 0.09 = 32.49\). \(5.75^2 \approx (5.7+0.05)^2 \dots\) Let's estimate \(\sqrt{33}\). \( \sqrt{33} \approx 5.74456 \). So, \(\sigma = 2 \times 5.74456 \approx 11.48912\). This is approximately 11.5. Option (c) is 11.5. Alternative calculation for variance: \( \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N} \). \(x_i - \bar{x}\): (4-22)=-18, (8-22)=-14, (12-22)=-10, (16-22)=-6, (20-22)=-2, (24-22)=2, (28-22)=6, (32-22)=10, (36-22)=14, (40-22)=18. Squared deviations: \( (-18)^2 = 324 \) \( (-14)^2 = 196 \) \( (-10)^2 = 100 \) \( (-6)^2 = 36 \) \( (-2)^2 = 4 \) \( (2)^2 = 4 \) \( (6)^2 = 36 \) \( (10)^2 = 100 \) \( (14)^2 = 196 \) \( (18)^2 = 324 \) Sum of squared deviations = \(2 \times (324+196+100+36+4) = 2 \times (520+100+36+4) = 2 \times (620+36+4) = 2 \times (656+4) = 2 \times 660 = 1320\). Variance \(\sigma^2 = \frac{1320}{10} = 132\). Standard deviation \(\sigma = \sqrt{132} \approx 11.489\). Rounding to one decimal place gives 11.5. \[ \boxed{11.5} \]