Question

the table shows the yearly earnings, in thousands of dollars, over a 10 - year period for college graduates. which statement is true about the distributions representing the yearly earnings? self - employed: 52, 101, 53, 96, 60, 81, 38, 51, 46, 72 wage earners: 66, 89, 64, 81, 62, 84, 44, 58, 51, 65 the mean earnings of the self - employed are higher than the mean earnings of the wage earners. the distribution of earnings for wage earners is more symmetric than the distribution of earnings for the self - employed. the iqrs of the distributions are equal. the standard deviations of the distributions are equal.

Explanation:

Step1: Calculate the mean for self - employed

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). For self - employed, \(n = 10\), \(\sum x=(52 + 101+53 + 96+60 + 81+38 + 51+46 + 72)=650\). So, \(\bar{x}_{self - employed}=\frac{650}{10}=65\)

Step2: Calculate the mean for wage earners

For wage earners, \(n = 10\), \(\sum x=(66 + 89+64 + 81+62 + 84+44 + 58+51 + 65)=664\). So, \(\bar{x}_{wage - earners}=\frac{664}{10}=66.4\). So the first option is wrong.

Step3: Check the symmetry

Sort self - employed data: \(38,46,51,52,53,60,72,81,96,101\). Sort wage earners data: \(44,51,58,62,64,65,66,81,84,89\). The wage - earners' data is more symmetric (closer to a bell - shaped curve in terms of spread around the center) compared to the self - employed's data.

Step4: Calculate the IQR for self - employed

First, find \(Q_1\) and \(Q_3\). For \(n = 10\), \(Q_1\) is the \(\frac{n + 1}{4}\)th value. \(\frac{10+1}{4}=2.75\)th value. \(Q_1=46+(51 - 46)\times0.75 = 49.75\). \(Q_3\) is the \(\frac{3(n + 1)}{4}\)th value. \(\frac{3(10 + 1)}{4}=8.25\)th value. \(Q_3=81+(96 - 81)\times0.25=84.75\). \(IQR_{self - employed}=84.75-49.75 = 35\)

Step5: Calculate the IQR for wage earners

\(\frac{10 + 1}{4}=2.75\)th value. \(Q_1=51+(58 - 51)\times0.75=56.25\). \(\frac{3(10 + 1)}{4}=8.25\)th value. \(Q_3=81+(84 - 81)\times0.25 = 81.75\). \(IQR_{wage - earners}=81.75-56.25=25.5\). So the third option is wrong.

Step6: Calculate the standard deviation (using the formula \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\))

For self - employed: \(\sum(x_{i}-\bar{x})^{2}=(52 - 65)^{2}+(101 - 65)^{2}+\cdots+(72 - 65)^{2}=4338\). \(s_{self - employed}=\sqrt{\frac{4338}{9}}\approx22\). For wage earners: \(\sum(x_{i}-\bar{x})^{2}=(66 - 66.4)^{2}+(89 - 66.4)^{2}+\cdots+(65 - 66.4)^{2}=2309.2\). \(s_{wage - earners}=\sqrt{\frac{2309.2}{9}}\approx16\). So the fourth option is wrong.

Answer:

The distribution of earnings for wage earners is more symmetric than the distribution of earnings for the self - employed.