Belk Department Store is having a special sale this weekend. Customers charging pur- chases of more than $50 to their Belk credit card will be given a special Belk Lottery card. The customer will scratch off the card, which will indicate the amount to be taken off the total amount of the purchase. Listed below are the amount of the prize and the per- cent of the time that amount will be deducted from the total amount of the purchase.
Prize Amountx
Probability
$ 10
.50
25
.40
50
.08
100
.02
a. What is the mean amount deducted from the total purchase amount?
b. What is the standard deviation of the amount deducted from the total purchase
The Correct Answer and Explanation is :
Given Information:
The Belk Lottery card offers four possible prize amounts, each with its own probability:
- Prize Amount of $10: Probability 0.50
- Prize Amount of $25: Probability 0.40
- Prize Amount of $50: Probability 0.08
- Prize Amount of $100: Probability 0.02
a. Mean Amount Deducted from the Total Purchase
The mean (or expected value) of a random variable is calculated by multiplying each possible outcome by its respective probability and then summing these products. The formula for the expected value E(X)E(X) is: E(X)=(x1⋅P(x1))+(x2⋅P(x2))+(x3⋅P(x3))+(x4⋅P(x4))E(X) = (x_1 \cdot P(x_1)) + (x_2 \cdot P(x_2)) + (x_3 \cdot P(x_3)) + (x_4 \cdot P(x_4))
Where:
- xix_i is the prize amount.
- P(xi)P(x_i) is the probability of receiving that prize.
Using the given data: E(X)=(10⋅0.50)+(25⋅0.40)+(50⋅0.08)+(100⋅0.02)E(X) = (10 \cdot 0.50) + (25 \cdot 0.40) + (50 \cdot 0.08) + (100 \cdot 0.02)
Let’s calculate the mean: E(X)=(10⋅0.50)+(25⋅0.40)+(50⋅0.08)+(100⋅0.02)=5+10+4+2=21E(X) = (10 \cdot 0.50) + (25 \cdot 0.40) + (50 \cdot 0.08) + (100 \cdot 0.02) = 5 + 10 + 4 + 2 = 21
So, the mean amount deducted from the total purchase is $21.
b. Standard Deviation of the Amount Deducted
The standard deviation measures the spread or dispersion of the values from the mean. It is calculated as the square root of the variance. The formula for variance Var(X)\text{Var}(X) is: Var(X)=∑(xi−E(X))2⋅P(xi)\text{Var}(X) = \sum (x_i – E(X))^2 \cdot P(x_i)
First, we compute each term (xi−E(X))2⋅P(xi)(x_i – E(X))^2 \cdot P(x_i):
- For x=10x = 10: (10−21)2⋅0.50=(−11)2⋅0.50=121⋅0.50=60.5(10 – 21)^2 \cdot 0.50 = (-11)^2 \cdot 0.50 = 121 \cdot 0.50 = 60.5
- For x=25x = 25: (25−21)2⋅0.40=42⋅0.40=16⋅0.40=6.4(25 – 21)^2 \cdot 0.40 = 4^2 \cdot 0.40 = 16 \cdot 0.40 = 6.4
- For x=50x = 50: (50−21)2⋅0.08=292⋅0.08=841⋅0.08=67.28(50 – 21)^2 \cdot 0.08 = 29^2 \cdot 0.08 = 841 \cdot 0.08 = 67.28
- For x=100x = 100: (100−21)2⋅0.02=792⋅0.02=6241⋅0.02=124.82(100 – 21)^2 \cdot 0.02 = 79^2 \cdot 0.02 = 6241 \cdot 0.02 = 124.82
Now, sum these values to find the variance: Var(X)=60.5+6.4+67.28+124.82=258\text{Var}(X) = 60.5 + 6.4 + 67.28 + 124.82 = 258
Finally, the standard deviation σ(X)\sigma(X) is the square root of the variance: σ(X)=258≈16.06\sigma(X) = \sqrt{258} \approx 16.06
So, the standard deviation of the amount deducted from the total purchase is approximately $16.06.
Conclusion:
- The mean amount deducted from the total purchase is $21.
- The standard deviation of the amount deducted from the total purchase is approximately $16.06.
Explanation:
The mean is the weighted average of the prize amounts, where each prize is weighted by its probability. In this case, the prize amounts of $10, $25, $50, and $100 have respective probabilities of 0.50, 0.40, 0.08, and 0.02. The mean represents the expected amount that will be deducted on average from a customer’s total purchase.
The standard deviation quantifies how much the prize amounts vary from the mean. It shows the typical amount of deviation from the mean in the lottery card deductions. A higher standard deviation means a wider spread of prize amounts, and a lower standard deviation would suggest that the prize amounts are more concentrated around the mean. In this case, the standard deviation of $16.06 reflects a relatively high spread in the prize amounts relative to the mean.