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 mean of self - employed

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

Step2: Calculate mean of wage earners

For wage earners, $n = 10$, and $\sum_{i = 1}^{10}x_{i}=66+89+64+81+62+84+44+58+51+65 = 704$. So the mean $\bar{x}_{wage - earners}=\frac{704}{10}=70.4$. So the first option is false.

Step3: Check for symmetry

To check for symmetry, we can look at the spread. We can order the data. For self - employed: $38,46,51,52,53,60,72,81,96,101$. For wage earners: $44,51,58,62,64,65,66,81,84,89$. The wage - earners data is more symmetric as the values are more evenly distributed around the middle values.

Step4: Calculate IQR for self - employed

First, order the data: $38,46,51,52,53,60,72,81,96,101$. The median of the lower half ($Q_1$) is the median of $38,46,51,52,53$ which is $51$. The median of the upper half ($Q_3$) is the median of $60,72,81,96,101$ which is $81$. So $IQR_{self - employed}=Q_3 - Q_1=81 - 51 = 30$.

Step5: Calculate IQR for wage earners

Order the data: $44,51,58,62,64,65,66,81,84,89$. The median of the lower half ($Q_1$) is the median of $44,51,58,62,64$ which is $58$. The median of the upper half ($Q_3$) is the median of $65,66,81,84,89$ which is $81$. So $IQR_{wage - earners}=Q_3 - Q_1=81 - 58 = 23$. So the third option is false.

Step6: Calculate standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. For self - employed:
\[

$$\begin{align*} \sum_{i=1}^{10}(x_{i}-65)^{2}&=(38 - 65)^{2}+(46-65)^{2}+(51 - 65)^{2}+(52-65)^{2}+(53 - 65)^{2}+(60 - 65)^{2}+(72-65)^{2}+(81 - 65)^{2}+(96 - 65)^{2}+(101 - 65)^{2}\\ &=(- 27)^{2}+(-19)^{2}+(-14)^{2}+(-13)^{2}+(-12)^{2}+(-5)^{2}+7^{2}+16^{2}+31^{2}+36^{2}\\ &=729+361+196+169+144+25+49+256+961+1296\\ &=4186 \end{align*}$$

\]
$s_{self - employed}=\sqrt{\frac{4186}{9}}\approx21.5$.
For wage earners:
\[

$$\begin{align*} \sum_{i=1}^{10}(x_{i}-70.4)^{2}&=(44 - 70.4)^{2}+(51-70.4)^{2}+(58 - 70.4)^{2}+(62-70.4)^{2}+(64 - 70.4)^{2}+(65 - 70.4)^{2}+(66 - 70.4)^{2}+(81 - 70.4)^{2}+(84 - 70.4)^{2}+(89 - 70.4)^{2}\\ &=(-26.4)^{2}+(-19.4)^{2}+(-12.4)^{2}+(-8.4)^{2}+(-6.4)^{2}+(-5.4)^{2}+(-4.4)^{2}+(10.6)^{2}+(13.6)^{2}+(18.6)^{2}\\ &=696.96+376.36+153.76+70.56+40.96+29.16+19.36+112.36+184.96+345.96\\ &=2030.4 \end{align*}$$

\]
$s_{wage - earners}=\sqrt{\frac{2030.4}{9}}\approx15$. So the fourth option is false.

Answer:

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