Question

A large fast-food chain runs a promotion where [1]-in-[4] boxes of french fries include a coupon for a free box of french fries. Suppose that some location sells [100] of these boxes of fries per day. Let [x=] the number of coupons won per day. Find the mean and standard deviation of [x]. You may round your answers to the nearest tenth.

Answer 1

The mean number of coupons won per day is 25, and the standard deviation of the number of coupons won per day is 5.5.

Step-by-step explanation:

To find the mean and standard deviation of x, we need to make use of the given information. The promotion states that 1 in 4 boxes of french fries include a coupon for a free box of fries. This means that the probability of winning a coupon is 1/4. Since 100 boxes of fries are sold per day, the mean number of coupons won per day can be calculated as:

1. 
   
3. Mean (x) = Probability of winning a coupon * Number of boxes sold per day
4. 
   
6. Mean (x) = 1/4 * 100 = 25

Next, to find the standard deviation of x, we can use the formula for the standard deviation of a binomial distribution:

1. 
   
3. Standard Deviation (σ) = √ (Number of boxes sold per day * Probability of winning a coupon * Probability of not winning a coupon)
4. 
   
6. Standard Deviation (σ) = √ (100 * 1/4 * 3/4) = √ (75/4) = 5.5