Question

the table shows the probabilities of certain prizes in a restaurants contest where the first 100 customers are winners. how does the $100 gift card affect the measure of center of the data? contest prizes $1 drink: 44, $5 meal: 25, $5 gift card: 15, $10 gift card: 10, $20 gift card: 5, $100 gift card: 1. it increases the mean value of the prizes. it decreases the mean value of the prizes. it increases the median value of the prizes. it decreases the median value of the prizes

Explanation:

Step1: Recall the effect of an outlier on mean and median

The mean is affected by extreme values (outliers). The median is the middle - value when data is ordered. The $100$ gift card is an outlier (a very large value compared to other prizes).

Step2: Analyze the effect on the mean

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}f_{i}}{\sum_{i = 1}^{n}f_{i}}$, where $x_{i}$ is the value of the prize and $f_{i}$ is the frequency. Since the $100$ gift card has a large value, when we calculate the mean $\frac{(1\times44)+(5\times25)+(5\times15)+(10\times10)+(20\times5)+(100\times1)}{44 + 25+15+10+5+1}=\frac{44+125 + 75+100+100+100}{100}=\frac{544}{100}=5.44$. If we remove the $100$ gift card, $\frac{(1\times44)+(5\times25)+(5\times15)+(10\times10)+(20\times5)}{99}=\frac{44+125+75+100+100}{99}=\frac{444}{99}\approx4.48$. So it increases the mean.

Step3: Analyze the effect on the median

We have $n = 100$ data points. The median is the average of the $50^{th}$ and $51^{st}$ ordered data points. When we list out the prizes in order (44 of $1$, 25 of $5$, 15 of $5$, 10 of $10$, 5 of $20$, 1 of $100$), the $50^{th}$ and $51^{st}$ values are both $5$. If we remove the $100$ gift card ($n=99$), the $50^{th}$ value is still $5$. So the median is not affected.

Answer:

It increases the mean value of the prizes.