First, we need to organize the data into a frequency distribution with 5 classes. The data set is:
\[ 167, 180, 128, 132, 176, 156, 165, 224, 149, 135, 193, 205, 149, 255, 258, 242, 288, 134, 200, 177 \]
To determine the class width, we first find the range of the data:
- Minimum value: 128
- Maximum value: 288
Range = \( 288 - 128 = 160 \)
Class width = \(\frac{\text{Range}}{\text{Number of classes}} = \frac{160}{5} = 32\)
Using the class width of 32, we construct the frequency distribution:
- Class 1: 128-160
- Class 2: 161-193
- Class 3: 194-226
- Class 4: 227-259
- Class 5: 260-292
Now, count the number of data points in each class:
- 128-160: 6 (128, 132, 135, 134, 149, 156)
- 161-193: 6 (167, 180, 176, 165, 193, 177)
- 194-226: 2 (205, 200)
- 227-259: 3 (255, 258, 242)
- 260-292: 3 (288)
The frequency histogram is a graphical representation of the frequency distribution. Each class interval is represented on the x-axis, and the frequency of each class is represented on the y-axis.
Based on the frequency distribution, the histogram will have the following frequencies:
- 128-160: 6
- 161-193: 6
- 194-226: 2
- 227-259: 3
- 260-292: 3
The histogram is likely to be positively skewed because the frequencies decrease as the class intervals increase.
Frequency Distribution:
- 128-160: 6
- 161-193: 6
- 194-226: 2
- 227-259: 3
- 260-292: 3
Histogram: Constructed based on the frequency distribution.
Shape of the Histogram: Positively skewed.
\[
\boxed{
\begin{array}{c|c}
\text{Class} & \text{Frequency} \\
\hline
128-160 & 6 \\
161-193 & 6 \\
194-226 & 2 \\
227-259 & 3 \\
260-292 & 3 \\
\end{array}
}
\]