Question

Question 6(Mutiple Choice Worth 10 points)

(06.02 HC)

To increase sales, an online clothing store began giving a 50% off coupon to random customers. Customers didn't know whether they would receive the coupon until after the final sale. The website claimed that one in six customers received the coupon. Eight customers each made purchases from the website. Let X = the number of customers that received the 50% off coupon. What are the mean and standard deviation of X? Provide an
interpretation for each value in context.

A. μ=1.200 and o,-0.980, if the store were to sell many clothing orders, the average number of coupons is 1.200 out of 8, and they will typically give between 0.153 and 2 133 coupons out of each 8 orders.
B. -1.333 and o.-1.054, If the store were to sell many clothing orders, an average of 1.333 customers per day will receive the coupon, and they will typically give between 0.236 and 2.336 discounts per day
C. -1333 and 0,-1.054, If the store were to sell many clothing orders, the average number of coupons is 1.333 out of 8 customers, and they will typically give between 0.279 and 2.387 coupons out of each
8 orders.
D. μ-1.200 and o,-0.980; If the store were to sell many clothing orders, average of 1.200 customers per day will receive the coupon, and they will typically give between 0.153 and 2.133 discounts per day.

Answer 1

The mean of X is 1.333 and the standard deviation is 0.966. The mean represents the average number of customers out of eight that will receive the 50% off coupon, while the standard deviation indicates the typical amount of variation in the number of customers that receive the coupon out of eight orders.

option c is the correct

Step-by-step explanation:

In this scenario, X represents the number of customers that received the 50% off coupon out of the eight customers. The website claims that one in six customers received the coupon, so the probability of a customer receiving the coupon is 1/6. Since each customer's probability of receiving the coupon is independent of the others, we can use the binomial distribution to calculate the mean and standard deviation of X.

To find the mean, we multiply the number of trials by the probability of success:

μ = 8 * (1/6) = 1.333

To find the standard deviation, we use the formula:

σ = sqrt(n * p * (1-p))

σ = sqrt(8 * (1/6) * (1-(1/6))) = 0.966

The mean of X (μ=1.333) represents the average number of customers out of eight that will receive the 50% off coupon. The standard deviation (σ=0.966) indicates the typical amount of variation in the number of customers that receive the coupon out of eight orders.