The time required to prepare a certain specially coffee at a local coffee shop i

Question

The time required to prepare a certain specially coffee at a local coffee shop is uniformly distributed betwoen 40 and 65 seconds. Assuming a customer just ordered one of these specialty coffees, determine the probabilities described below a. What is the probability that the preparation time will be more than 43 seconds? b. What is the probability that the preparation time will be between 49 and 55 seconds? c. What percentage of these specialty coffees will be prepared within 57 seconds? d What is the standard deviation of preparation times for this specialty coffee at this shop? a. P(preparation time more than 43 seconds)- (Simplify your answer) b P(preparation time between 49 and 55 seconds)- (Simplify your answer ) % P(preparation time less than or equal to 57) (Simplify your answer) d · (Round to four decimal places as needed )

Explanation / Answer

Solution:

We are given that the random variable follows uniform distribution with a = 40 and b = 65.

Part a

We have to find P(X>43)

P(X>x) = (b – x) / (b – a)

P(X>43) = (65 – 43) / (65 – 40)

P(X>43) = 22/25 = 0.88

Required probability = 0.88

Part b

We have to find P(49<X<55)

P(49<X<55) = P(X<55) – P(X<49)

P(X<55) = (x – a) / (b – a) = (55 – 40) / (65 – 40) = 15/25 = 0.60

Required probability = 0.60

Part c

We have to find P(X57)

P(X57) = P(X<57) (X follows continuous uniform distribution)

P(X<57) = (x – a) / (b – a) = (57 – 40) / (65 – 40) = 17/25 = 0.68

Required percentage = 68%

Part d

Here, we have to find

Formula for is given as below:

= sqrt((b – a)2/12)

= sqrt((65 – 40)2/12)

= sqrt(25^2/12)

= sqrt(625/12)

= sqrt(52.08333)

= 7.216878

= 7.2169

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