Question

the scores of the 48 members of a sociology lecture class on a 50 - point exam were as follows. complete parts (a) through (c) below.
click the icon to view the data.
(a) construct frequency and relative frequency distributions.
(round to the nearest integer as needed.)
full data set
41 42 35 50 44 38 22 46
42 37 50 44 45 32 46 42
37 30 47 45 36 46 42 43
34 47 45 36 28 48 43 40
47 45 41 31 48 43 44 33
49 41 39 48 43 44 33 49

Explanation:

Step1: Count the frequency for each class

• For class \(21 - 25\): Count the number of scores in this range. The score \(22\) is in this range. So \(f_{21 - 25}=1\)
• For class \(26 - 30\): Count the scores. The score \(28\) is in this range. So \(f_{26 - 30}=1\)
• For class \(31 - 35\): Count the scores \(31,33,33,35\). So \(f_{31 - 35}=4\)
• For class \(36 - 40\): Count the scores \(36,36,38,39,40\). So \(f_{36 - 40}=5\)
• For class \(41 - 45\): Count the scores \(41,41,41,42,42,43,43,43,44,44,44,45,45,45\). So \(f_{41 - 45}=14\)
• For class \(46 - 50\): Count the scores \(46,46,47,47,47,48,48,48,49,49,50,50\). So \(f_{46 - 50}=12\)

Step2: Calculate the relative frequency

The total number of data points \(n = 48\)

• For class \(21 - 25\): \(\frac{f}{n}=\frac{1}{48}\approx 2\%\)
• For class \(26 - 30\): \(\frac{f}{n}=\frac{1}{48}\approx 2\%\)
• For class \(31 - 35\): \(\frac{f}{n}=\frac{4}{48}\approx 8\%\)
• For class \(36 - 40\): \(\frac{f}{n}=\frac{5}{48}\approx 10\%\)
• For class \(41 - 45\): \(\frac{f}{n}=\frac{14}{48}\approx 29\%\)
• For class \(46 - 50\): \(\frac{f}{n}=\frac{12}{48}=25\%\)

Answer:

Class Limits	Frequency \(f\)	Relative Frequency \(\frac{f}{n}\)
\(26 - 30\)	\(1\)	\(2\%\)
\(31 - 35\)	\(4\)	\(8\%\)
\(36 - 40\)	\(5\)	\(10\%\)
\(41 - 45\)	\(14\)	\(29\%\)
\(46 - 50\)	\(12\)	\(25\%\)