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the table shows the probabilities of certain prizes in a restaurants co…

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? 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. contest prizes $1 drink: 44, $5 meal: 25, $5 gift card: 15, $10 gift card: 10, $20 gift card: 5, $100 gift card: 1

Explanation:

Step1: Recall the definitions of mean and median

The mean is the average value of a data set, calculated by \(\frac{\sum_{i = 1}^{n}x_{i}}{n}\). The median is the middle - value when the data is ordered. For \(n = 100\) (an even number of data points), the median is the average of the 50th and 51st ordered values.

Step2: Analyze the effect on the mean

The mean formula is \(\bar{x}=\frac{\sum_{i = 1}^{k}x_{i}f_{i}}{N}\), where \(x_{i}\) is the value of the \(i -\)th prize, \(f_{i}\) is the frequency of the \(i -\)th prize, and \(N=\sum_{i = 1}^{k}f_{i}\).
Let's assume we calculate the mean without considering the \(\$100\) gift - card first. Then, when we add the \(\$100\) gift - card (\(x = 100\), \(f=1\)), since \(100>1,5,10,20\), the sum \(\sum_{i = 1}^{k}x_{i}f_{i}\) increases. As \(N = 100\) (constant), the mean \(\bar{x}=\frac{\sum_{i = 1}^{k}x_{i}f_{i}}{N}\) increases.

Step3: Analyze the effect on the median

Order the data by prize value. The number of non - \(\$100\) prizes is \(44 + 25+15 + 10+5=99\). When we order the data (from smallest to largest prize value), the 50th and 51st values are still within the non - \(\$100\) prize group (because \(44+25 = 69>51\)). So the median (average of 50th and 51st values) does not change.

Answer:

It increases the mean value of the prizes.