According to a recent survey, 65% of all customers will return to the same grocery store. Suppose 11 customers are selected at random,what is the probability that
A) exactly 5 will return?
B) all 11 will return?
C) at least 6 will return
D) at least one will return
E) how many customers would be expected to return to the same store
To calculate the probabilities in this scenario, we can use the binomial probability formula. The formula is:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
where:
- P(X = k) represents the probability of getting exactly k successes (in this case, customers who will return).
- (n C k) denotes the number of combinations of selecting k elements from a set of n elements.
- p is the probability of a single event being a success.
- n is the number of trials (in this case, the number of customers).
- k is the number of desired successes.
Let's apply this formula to each part of the question:
A) Exactly 5 customers will return.
- p = 0.65 (probability of a customer returning)
- n = 11 (number of customers)
- k = 5 (desired number of returning customers)
Calculate:
P(X = 5) = (11 C 5) * (0.65^5) * (1 - 0.65)^(11 - 5)
B) All 11 customers will return.
- p = 0.65
- n = 11
- k = 11
Calculate:
P(X = 11) = (11 C 11) * (0.65^11) * (1 - 0.65)^(11 - 11)
C) At least 6 customers will return.
To find this probability, we need to add up the probabilities of getting 6, 7, 8, 9, 10, and 11 successes.
P(at least 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11)
D) At least one customer will return.
To find this probability, we need to subtract the probability of getting zero successes from 1.
P(at least 1) = 1 - P(X = 0)
E) The expected number of customers who will return.
The expected number can be calculated using the formula:
Expected value = n * p
where:
- n = 11 (number of customers)
- p = 0.65 (probability of a single customer returning)
Expected value = 11 * 0.65
I hope this helps!
Where is your attempt to do these problems?
They are straightforward if tedious applications of the binomial distribution.