Question

which data set would the mean be significantly affected by extreme values, making the median a more reliable measure of center?
money earned from babysitting by five students: $30, $32, $33, $35, $330.
tips earned by five students working as servers: $20, $22, $23, $25, $25.
earnings from a school car wash fundraiser by five students: $50, $55, $57, $60, $50.
profits from selling handmade crafts online by five students: $40, $42, $43, $45, $40.

Explanation:

Step1: Analyze the first data set

The data set for money earned from babysitting is \( \$30,\$32,\$33,\$35,\$330 \). The value \( \$330 \) is an extreme out - lier.

Step2: Recall the property of mean and median

The mean is calculated as \( \bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n} \). For \( n = 5 \), \( \bar{x}=\frac{30 + 32+33 + 35+330}{5}=\frac{460}{5}=92 \). The median for \( n = 5 \) (odd number of data points) is the third - ordered value. Ordering the data: \( \$30,\$32,\$33,\$35,\$330 \), the median is \( \$33 \). The mean is highly affected by the outlier \( \$330 \).

Step3: Analyze the second data set

The tips data set: \( \$20,\$22,\$23,\$25,\$25 \). Mean \( \bar{x}=\frac{20 + 22+23 + 25+25}{5}=\frac{115}{5}=23 \). Median (third value) is \( \$23 \). There is no extreme outlier.

Step4: Analyze the third data set

The car - wash earnings data set: \( \$50,\$55,\$57,\$60,\$50 \). Mean \( \bar{x}=\frac{50+55 + 57+60+50}{5}=\frac{272}{5}=54.4 \). Median (third value) is \( \$55 \). There is no extreme outlier.

Step5: Analyze the fourth data set

The profit data set: \( \$40,\$42,\$43,\$45,\$40 \). Mean \( \bar{x}=\frac{40+42 + 43+45+40}{5}=\frac{210}{5}=42 \). Median (third value) is \( \$42 \). There is no extreme outlier.

Answer:

The data set "Money earned from babysitting by five students: \( \$30,\$32,\$33,\$35,\$330 \)"