According to a recent survey 89 of all customers will return...

According to a recent survey, 89% of all customers will return to the same grocery store. Suppose 10 customers are selected at random, what is the probability that: (Round probabilities to 4 decimal places and the expected customers to 2 decimal places) a. Exactly 5 of the customers will return? (1 point) b. All 10 will return? (1 point) c. At least 7 will return? (1 point) d. Less than 6 will return? (1 point) e. How many customers would be expected to return to the store? (1 point)

Transcript

00:01 For this exercise, we are told that 89 % of all customers will return to the same grocery store.

00:07 So that means for a randomly selected customer, the probability that that person returns to the same grocery store is 0 .89.

00:16 And we have a random selection of 10 customers.

00:20 And we are asked for some probabilities about the number of the 10 who would return to the same grocery store.

00:28 So let's let the random variable x be the number of customers.

00:34 Customers who return to the store.

00:42 Here, x is a binomial random variable.

00:54 Binomial has two parameters, the probability of success, and the number of trials.

01:04 And the probability mass function, where the binomial random variable, is given by this formula.

01:21 Now for part a, we are asked for the probability that exactly five customers will return.

01:27 So this is the probability that x is equal to five.

01:31 And using our probability mass function, this is 10 choose 5 times 0 .89 to the exponent 5, times 0 .11 to the exponent 5.

01:46 And this comes out to a probability of approximately 0 .0023.

01:56 For part b, we are asked for the probability that x equals 10.

02:00 That's all 10 return out of 10.

02:04 When x equals 10, the probability mass function simplifies to p to the exponent n, which is 10.

02:18 And this comes out to .3118.

02:25 And for c we're asked for the probability that at least 7 return...

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