Question

1. frequency distribution and a frequency polygon. describe any patterns.

exam scores number of classes: 5
data set:
83 92 94 82 73 98 78 85 72 90 89 92
96 89 75 85 63 47 75 82

Explanation:

Step1: Determine Class Width

First, find the range of the data. The maximum value is \(98\) and the minimum value is \(47\). The range \(R = 98 - 47=51\). Since the number of classes \(k = 5\), the class width \(w=\lceil\frac{R}{k}
ceil=\lceil\frac{51}{5}
ceil = 11\) (using the ceiling function to ensure all data is included).

Step2: Define Class Intervals

Start with the minimum value \(47\). The class intervals are: \(47 - 57\), \(58 - 68\), \(69 - 79\), \(80 - 90\), \(91 - 101\).

Step3: Calculate Frequency

Count the number of data points in each class:

• For \(47 - 57\): There is \(1\) value (\(47\)).
• For \(58 - 68\): There is \(1\) value (\(63\)).
• For \(69 - 79\): The values are \(73,78,72,75,75\), so the frequency \(f = 5\).
• For \(80 - 90\): The values are \(83,82,85,90,89,85,82,89\), so the frequency \(f = 8\).
• For \(91 - 101\): The values are \(92,94,98,92,96\), so the frequency \(f = 5\).

Step4: Calculate Mid - point

The mid - point of a class interval \(a - b\) is \(\frac{a + b}{2}\).

• For \(47 - 57\): Mid - point \(m=\frac{47+57}{2}=52\).
• For \(58 - 68\): Mid - point \(m=\frac{58 + 68}{2}=63\).
• For \(69 - 79\): Mid - point \(m=\frac{69+79}{2}=74\).
• For \(80 - 90\): Mid - point \(m=\frac{80 + 90}{2}=85\).
• For \(91 - 101\): Mid - point \(m=\frac{91+101}{2}=96\).

Step5: Calculate Relative Frequency

The relative frequency \(rf=\frac{f}{\sum f}\), where \(\sum f=20\).

• For \(47 - 57\): \(rf=\frac{1}{20}=0.05\).
• For \(58 - 68\): \(rf=\frac{1}{20}=0.05\).
• For \(69 - 79\): \(rf=\frac{5}{20}=0.25\).
• For \(80 - 90\): \(rf=\frac{8}{20}=0.4\).
• For \(91 - 101\): \(rf=\frac{5}{20}=0.25\).

Step6: Calculate Cumulative Frequency

The cumulative frequency \(cf\) is calculated by adding the frequencies from the first class to the current class.

• For \(47 - 57\): \(cf = 1\).
• For \(58 - 68\): \(cf=1 + 1=2\).
• For \(69 - 79\): \(cf=2+5 = 7\).
• For \(80 - 90\): \(cf=7 + 8=15\).
• For \(91 - 101\): \(cf=15+5=20\).

Step7: Draw Frequency Polygon

Plot the mid - points on the \(x\) - axis and the frequencies on the \(y\) - axis. Connect the points \((52,1)\), \((63,1)\), \((74,5)\), \((85,8)\), \((96,5)\) with straight lines. Also, extend the polygon to the \(x\) - axis at the mid - points of the classes before the first class (\(47 - 57\)) and after the last class (\(91 - 101\)) (for example, assume a class before \(47 - 57\) with mid - point \(41\) and frequency \(0\) and a class after \(91 - 101\) with mid - point \(107\) and frequency \(0\)).

Step8: Describe Patterns

The distribution is somewhat symmetric around the \(80 - 90\) class. The highest frequency is in the \(80 - 90\) class, indicating that most of the exam scores are in this range. There are lower frequencies in the lower (\(47 - 57\) and \(58 - 68\)) and higher (\(91 - 101\)) score ranges.

Answer:

Class	Frequency	Mid - point	Relative Frequency	Cumulative Frequency
\(58 - 68\)	\(1\)	\(63\)	\(0.05\)	\(2\)
\(69 - 79\)	\(5\)	\(74\)	\(0.25\)	\(7\)
\(80 - 90\)	\(8\)	\(85\)	\(0.4\)	\(15\)
\(91 - 101\)	\(5\)	\(96\)	\(0.25\)	\(20\)

The frequency polygon has a peak at the \(80 - 90\) class (mid - point \(85\)) and is symmetric around this central class to some extent.