Question

it seems these days that college graduates who are employed full - time are working more hours than ever. data are available that can help us decide if this is true. in a survey for a recent year, there were 78 respondents who were college graduates employed full - time. the reported number of hours each worked per week is summarized in the following histogram, which was displayed in a government publication. based on the histogram, estimate the mean number of hours worked per week by these respondents. carry your intermediate computations to at least four decimal places, and round your answer to at least one decimal place.

Explanation:

Step1: Find the mid - point of each class interval

Assume the class intervals:

• For the first bar (frequency = 9): If we assume the class interval is \(36 - 40\), the mid - point \(x_1=\frac{36 + 40}{2}=38\)
• For the second bar (frequency = 24): If the class interval is \(40 - 44\), the mid - point \(x_2=\frac{40+44}{2}=42\)
• For the third bar (frequency = 18): If the class interval is \(44 - 48\), the mid - point \(x_3=\frac{44 + 48}{2}=46\)
• For the fourth bar (frequency = 14): If the class interval is \(48 - 52\), the mid - point \(x_4=\frac{48+52}{2}=50\)
• For the fifth bar (frequency = 8): If the class interval is \(52 - 56\), the mid - point \(x_5=\frac{52+56}{2}=54\)
• For the sixth bar (frequency = 5): If the class interval is \(56 - 60\), the mid - point \(x_6=\frac{56+60}{2}=58\)

Step2: Calculate the sum of \(f_ix_i\)

The formula for the weighted mean \(\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}\), where \(f_i\) is the frequency and \(x_i\) is the mid - point of the class.
\(\sum_{i = 1}^{6}f_ix_i=9\times38+24\times42 + 18\times46+14\times50+8\times54+5\times58\)

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\(\sum_{i=1}^{6}f_i=9 + 24+18+14+8+5=78\)

Step3: Calculate the mean

\(\bar{x}=\frac{\sum_{i = 1}^{6}f_ix_i}{\sum_{i=1}^{6}f_i}=\frac{3600}{78}\approx46.2\)

Answer:

\(46.2\)