Answer :
We are given the following frequency distribution for the ages of employees:
[tex]\[
\begin{array}{c|ccccccc}
\textbf{Age interval (years)} & 50-55 & 45-50 & 40-45 & 35-40 & 30-35 & 25-30 & 20-25 \\[6mm]
\textbf{Frequency } (f) & 25 & 30 & 40 & 45 & 80 & 110 & 170 \\
\end{array}
\][/tex]
For easier computation, it is best to work with the age groups arranged in ascending order. Rearranging the data from lowest to highest, we have:
[tex]\[
\begin{array}{c|ccccccc}
\textbf{Age interval (years)} & 20-25 & 25-30 & 30-35 & 35-40 & 40-45 & 45-50 & 50-55 \\[6mm]
\textbf{Frequency } (f) & 170 & 110 & 80 & 45 & 40 & 30 & 25 \\
\end{array}
\][/tex]
We wish to compute the coefficient of Karl-Pearson's skewness, which is given by the formula
[tex]\[
\text{Skewness} = \frac{3 \, (\bar{x} - \text{Median})}{s}
\][/tex]
where [tex]$\bar{x}$[/tex] is the mean and [tex]$s$[/tex] is the standard deviation.
Let’s go step by step.
---------------------------------------------------------------------
Step 1. \textbf{Compute the Midpoints of Each Class}
For each class interval [tex]$[a, b]$[/tex], the midpoint is found by
[tex]\[
x_{\text{mid}} = \frac{a + b}{2}
\][/tex]
Thus, we have:
- For 20–25: midpoint [tex]$= \frac{20+25}{2} = 22.5$[/tex]
- For 25–30: midpoint [tex]$= \frac{25+30}{2} = 27.5$[/tex]
- For 30–35: midpoint [tex]$= \frac{30+35}{2} = 32.5$[/tex]
- For 35–40: midpoint [tex]$= \frac{35+40}{2} = 37.5$[/tex]
- For 40–45: midpoint [tex]$= \frac{40+45}{2} = 42.5$[/tex]
- For 45–50: midpoint [tex]$= \frac{45+50}{2} = 47.5$[/tex]
- For 50–55: midpoint [tex]$= \frac{50+55}{2} = 52.5$[/tex]
---------------------------------------------------------------------
Step 2. \textbf{Calculate the Mean}
The mean for grouped data is given by
[tex]\[
\bar{x} = \frac{\sum (f \cdot x_{\text{mid}})}{N}
\][/tex]
where [tex]$N = \sum f$[/tex] is the total frequency.
After performing the multiplication of the frequencies by their corresponding midpoints and summing, one obtains:
[tex]\[
\bar{x} \approx 31.15
\][/tex]
---------------------------------------------------------------------
Step 3. \textbf{Determine the Median}
To find the median, first compute the total frequency:
[tex]\[
N = 170 + 110 + 80 + 45 + 40 + 30 + 25 = 500
\][/tex]
Half of the total frequency is
[tex]\[
\frac{N}{2} = 250
\][/tex]
Next, form the cumulative frequency distribution until you reach or exceed 250:
[tex]\[
\begin{array}{cc}
\text{Class interval} & \text{Cumulative Frequency} \\
20-25 & 170 \\
25-30 & 170 + 110 = 280 \\
\end{array}
\][/tex]
Since 280 exceeds 250, the median lies in the 25–30 class.
Let:
- [tex]$L$[/tex] = lower boundary of the median class [tex]$= 25$[/tex]
- [tex]$f_m$[/tex] = frequency of the median class [tex]$= 110$[/tex]
- [tex]$CF_{\text{before}}$[/tex] = cumulative frequency before the median class [tex]$= 170$[/tex]
- Class width [tex]$= 30 - 25 = 5$[/tex]
The median formula for grouped data is
[tex]\[
\text{Median} = L + \left( \frac{\frac{N}{2} - CF_{\text{before}}}{f_m} \right) \times \text{class width}
\][/tex]
Substituting the values gives:
[tex]\[
\text{Median} = 25 + \left( \frac{250 - 170}{110} \right) \times 5 \approx 28.6364
\][/tex]
---------------------------------------------------------------------
Step 4. \textbf{Compute the Variance and Standard Deviation}
The variance for grouped data is calculated using
[tex]\[
\sigma^2 = \frac{\sum f \, (x_{\text{mid}} - \bar{x})^2}{N}
\][/tex]
After evaluating the sum (which involves the squared differences multiplied by the corresponding frequencies) and dividing by [tex]$N$[/tex], the standard deviation is found by taking the square root of the variance:
[tex]\[
s \approx 9.0237
\][/tex]
---------------------------------------------------------------------
Step 5. \textbf{Compute Karl-Pearson's Skewness Coefficient}
Substitute the computed values of the mean, median, and standard deviation into the formula:
[tex]\[
\text{Skewness} = \frac{3 \, (\bar{x} - \text{Median})}{s} = \frac{3 \, (31.15 - 28.6364)}{9.0237}
\][/tex]
Evaluating the above expression gives:
[tex]\[
\text{Skewness} \approx 0.8357
\][/tex]
---------------------------------------------------------------------
\textbf{Final Answer:}
The coefficient of Karl-Pearson's skewness for the given data is approximately
[tex]\[
\boxed{0.8357}
\][/tex]
[tex]\[
\begin{array}{c|ccccccc}
\textbf{Age interval (years)} & 50-55 & 45-50 & 40-45 & 35-40 & 30-35 & 25-30 & 20-25 \\[6mm]
\textbf{Frequency } (f) & 25 & 30 & 40 & 45 & 80 & 110 & 170 \\
\end{array}
\][/tex]
For easier computation, it is best to work with the age groups arranged in ascending order. Rearranging the data from lowest to highest, we have:
[tex]\[
\begin{array}{c|ccccccc}
\textbf{Age interval (years)} & 20-25 & 25-30 & 30-35 & 35-40 & 40-45 & 45-50 & 50-55 \\[6mm]
\textbf{Frequency } (f) & 170 & 110 & 80 & 45 & 40 & 30 & 25 \\
\end{array}
\][/tex]
We wish to compute the coefficient of Karl-Pearson's skewness, which is given by the formula
[tex]\[
\text{Skewness} = \frac{3 \, (\bar{x} - \text{Median})}{s}
\][/tex]
where [tex]$\bar{x}$[/tex] is the mean and [tex]$s$[/tex] is the standard deviation.
Let’s go step by step.
---------------------------------------------------------------------
Step 1. \textbf{Compute the Midpoints of Each Class}
For each class interval [tex]$[a, b]$[/tex], the midpoint is found by
[tex]\[
x_{\text{mid}} = \frac{a + b}{2}
\][/tex]
Thus, we have:
- For 20–25: midpoint [tex]$= \frac{20+25}{2} = 22.5$[/tex]
- For 25–30: midpoint [tex]$= \frac{25+30}{2} = 27.5$[/tex]
- For 30–35: midpoint [tex]$= \frac{30+35}{2} = 32.5$[/tex]
- For 35–40: midpoint [tex]$= \frac{35+40}{2} = 37.5$[/tex]
- For 40–45: midpoint [tex]$= \frac{40+45}{2} = 42.5$[/tex]
- For 45–50: midpoint [tex]$= \frac{45+50}{2} = 47.5$[/tex]
- For 50–55: midpoint [tex]$= \frac{50+55}{2} = 52.5$[/tex]
---------------------------------------------------------------------
Step 2. \textbf{Calculate the Mean}
The mean for grouped data is given by
[tex]\[
\bar{x} = \frac{\sum (f \cdot x_{\text{mid}})}{N}
\][/tex]
where [tex]$N = \sum f$[/tex] is the total frequency.
After performing the multiplication of the frequencies by their corresponding midpoints and summing, one obtains:
[tex]\[
\bar{x} \approx 31.15
\][/tex]
---------------------------------------------------------------------
Step 3. \textbf{Determine the Median}
To find the median, first compute the total frequency:
[tex]\[
N = 170 + 110 + 80 + 45 + 40 + 30 + 25 = 500
\][/tex]
Half of the total frequency is
[tex]\[
\frac{N}{2} = 250
\][/tex]
Next, form the cumulative frequency distribution until you reach or exceed 250:
[tex]\[
\begin{array}{cc}
\text{Class interval} & \text{Cumulative Frequency} \\
20-25 & 170 \\
25-30 & 170 + 110 = 280 \\
\end{array}
\][/tex]
Since 280 exceeds 250, the median lies in the 25–30 class.
Let:
- [tex]$L$[/tex] = lower boundary of the median class [tex]$= 25$[/tex]
- [tex]$f_m$[/tex] = frequency of the median class [tex]$= 110$[/tex]
- [tex]$CF_{\text{before}}$[/tex] = cumulative frequency before the median class [tex]$= 170$[/tex]
- Class width [tex]$= 30 - 25 = 5$[/tex]
The median formula for grouped data is
[tex]\[
\text{Median} = L + \left( \frac{\frac{N}{2} - CF_{\text{before}}}{f_m} \right) \times \text{class width}
\][/tex]
Substituting the values gives:
[tex]\[
\text{Median} = 25 + \left( \frac{250 - 170}{110} \right) \times 5 \approx 28.6364
\][/tex]
---------------------------------------------------------------------
Step 4. \textbf{Compute the Variance and Standard Deviation}
The variance for grouped data is calculated using
[tex]\[
\sigma^2 = \frac{\sum f \, (x_{\text{mid}} - \bar{x})^2}{N}
\][/tex]
After evaluating the sum (which involves the squared differences multiplied by the corresponding frequencies) and dividing by [tex]$N$[/tex], the standard deviation is found by taking the square root of the variance:
[tex]\[
s \approx 9.0237
\][/tex]
---------------------------------------------------------------------
Step 5. \textbf{Compute Karl-Pearson's Skewness Coefficient}
Substitute the computed values of the mean, median, and standard deviation into the formula:
[tex]\[
\text{Skewness} = \frac{3 \, (\bar{x} - \text{Median})}{s} = \frac{3 \, (31.15 - 28.6364)}{9.0237}
\][/tex]
Evaluating the above expression gives:
[tex]\[
\text{Skewness} \approx 0.8357
\][/tex]
---------------------------------------------------------------------
\textbf{Final Answer:}
The coefficient of Karl-Pearson's skewness for the given data is approximately
[tex]\[
\boxed{0.8357}
\][/tex]
