Question

find the mean of the data summarized in the given frequency distribution. compare the computed mean to the actual mean of 51.8 miles per hour
speed (miles per hour)
42 - 45
46 - 49
50 - 53
54 - 57
58 - 61
frequency
29
15
6
3
2
the mean of the frequency distribution is 11 miles per hour.
(type an integer or decimal rounded to one decimal place as needed)

Explanation:

Step1: Find the mid - point of each class

For the class \(42 - 45\), the mid - point \(x_1=\frac{42 + 45}{2}=43.5\)
For the class \(46 - 49\), the mid - point \(x_2=\frac{46+49}{2}=47.5\)
For the class \(50 - 53\), the mid - point \(x_3=\frac{50 + 53}{2}=51.5\)
For the class \(54 - 57\), the mid - point \(x_4=\frac{54+57}{2}=55.5\)
For the class \(58 - 61\), the mid - point \(x_5=\frac{58 + 61}{2}=59.5\)

Step2: Calculate the sum of \(f\times x\)

The frequencies are \(f_1 = 29\), \(f_2=15\), \(f_3 = 6\), \(f_4=3\), \(f_5 = 2\)
\(\sum(f\times x)=f_1x_1+f_2x_2+f_3x_3+f_4x_4+f_5x_5\)
\(=29\times43.5+15\times47.5 + 6\times51.5+3\times55.5+2\times59.5\)
\(=29\times43.5+15\times47.5+6\times51.5 + 3\times55.5+2\times59.5\)
\(=1261.5+712.5+309+166.5+119\)
\(=1261.5+712.5=1974\), \(1974+309 = 2283\), \(2283+166.5=2449.5\), \(2449.5+119=2568.5\)

Step3: Calculate the sum of frequencies \(\sum f\)

\(\sum f=f_1+f_2+f_3+f_4+f_5=29 + 15+6+3+2=55\)

Step4: Calculate the mean \(\bar{x}\)

The formula for the mean of a frequency distribution is \(\bar{x}=\frac{\sum(f\times x)}{\sum f}\)
\(\bar{x}=\frac{2568.5}{55}\approx46.7\)

Answer:

The mean of the frequency distribution is \(46.7\) miles per hour. The computed mean (\(46.7\)) is less than the actual mean (\(51.8\))