Problem

A frequency table is given as follows: \[\begin{array}{|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 \\ \hline f & 4 & 2 & 3 & 1 & 5 \\ \hline \end{array}\] Where X represents the data values and f is their corresponding frequency. Find the standard deviation of the given frequency distribution.

Answer

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Answer

Step 3: Finally, the standard deviation (\(\sigma\)) is the square root of the variance. So, \(\sigma = \sqrt{0.79} = 0.89\)

Steps

Step 1 :Step 1: First, let's calculate the mean (\(\mu\)) of the distribution. Mean is calculated as \(\mu = \frac{\sum f\cdot X}{\sum f}\) where \(\sum f\cdot X\) is the sum of the product of each data value and its frequency, and \(\sum f\) is the sum of frequencies. Substituting the given values, we get \(\mu = \frac{4\cdot1 + 2\cdot2 + 3\cdot3 + 1\cdot4 + 5\cdot5}{4+2+3+1+5} = \frac{38}{15} = 2.53\)

Step 2 :Step 2: Now, let's calculate the variance (\(\sigma^2\)) of the distribution. Variance is calculated as \(\sigma^2 = \frac{\sum f\cdot(X - \mu)^2}{\sum f}\) where \(X\) are the data values, \(\mu\) is the mean, and \(f\) are the frequencies. Substituting the given values, we get \(\sigma^2 = \frac{4\cdot(1 - 2.53)^2 + 2\cdot(2 - 2.53)^2 + 3\cdot(3 - 2.53)^2 + 1\cdot(4 - 2.53)^2 + 5\cdot(5 - 2.53)^2}{15} = \frac{11.79}{15} = 0.79\)

Step 3 :Step 3: Finally, the standard deviation (\(\sigma\)) is the square root of the variance. So, \(\sigma = \sqrt{0.79} = 0.89\)

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