Question

The coffee company American Expresso has a promotion where "if you don't get your take away coffee within 180 seconds, it's free!".

American Expresso has commissioned an independent study which found that the time to prepare a take away coffee at one of their stores is approximately normally distributed with a mean time of 151 seconds and a standard deviation of 23 seconds.

a)Calculate the percentage of take away coffees that are given to customers for free according to the promotion. Give your answer as a percentage to 1 decimal place.

Percentage of free coffees = %

b)Some shareholders are unhappy with the terms of the promotion and suggest that the time limit should be such that at most only 2.5% of take away coffees are given for free under the promotion.

Calculate the time limit such that this requirement is satisfied. Give your answer as a whole number.

Time limit =  seconds

Answer 1

Solution:

Let X be a random variable which represents the time to prepare a take away coffee at one of their stores.

Given that, X ~ N(151, 23²)

i.e. μ = 151 seconds and σ = 23 seconds.

a) It is given that, "if you don't get your take away coffee within 180 seconds, it's free!". And I have to find percentage of free coffees. It means we have to find P(X > 180).

We know that if X ~ N(μ, σ^{​​​​​​2}) then, .

η X -μ Ν(0, 1)

180 - 4 :: P(X > 180) = P(

1 180 151 :. P(X > 180) = P(Z >= 23

::P(X > 180) = P(Z > 1.2609)

Using the Excel software we get,

P(Z > 1.2609) = 0.104

::P(X > 180) = 0.104

0.104 = 10.4%

Hence, percentage of free coffees = 10.4%

b) Let the time limit be t, such that at most 2.5% of take away coffees are given for free under the promotion.

2.5% = 2.5/100 = 0.025.

Hence, P(X ≤ t) = 0.025

sc00 = >) 4

...............................(1)

=P(z="-352) – 0.025

Using the Excel software we get, P(Z ≤ -1.96) = 0.025

Comparing P(Z ≤ 1.96) = 0.025 and equation (1) we get,

t - 151 -1.96 23

=t=151 - (1.96 x 23)

t= 105.92

t = 105.92 - 106

Hence, time limit = 106 seconds.

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