Question

2) The amount of beer (on tap) that a bar sells in a day follows a Normal distribution with a mean of 15 gallons and a standard deviation of 2 gallons.

1. What is the probability that the bar sells more than 16.15 gallons of beer on a day?

2. What is the probability that the bar sells more than 1525 gallons of beer over one hundred days?

3. What is the probability (or your best approximation of this probability) that the average amount the bar sells over 15 days is less than 14.7 gallons? Do you need to use a t distribution to answer this question?

4. The bar owner is concerned whenever less than 12.5 gallons of beer are sold in a day – refer to such a day as a “Bad Demand Day.” What is the probability that over the next 100 days, the bar has more than 14 “Bad Demand Days”?

Answer 1

a) P(X > 16.15)

= P((X - )/ > (16.15 - )/)

= P(Z > (16.15 - 15)/2)

= P(Z > 0.58)

= 1 - P(Z < 0.58)

= 1 - 0.7190

b) P( > 15.25)

= P(( - )/() > (15.25 - )/())

= P(Z > (15.25 - 15)/(2/))

= P(Z > 1.25)

= 1 - P(Z < 1.25)

= 1 - 0.8944

= 0.1056

c)

P( < 14.7)

= P(( - )/() < (14.7 - )/())

= P(Z < (14.7 - 15)/(2/))

= P(Z < -0.58)

= 0.2843

There is no need to use t - distribution.

d)

P( > 14)

= P(( - )/() > (14 - )/())

= P(Z > (14 - 15)/(2/))

= P(Z > -1.94)

= 1 - P(Z < -1.94)

= 1 - 0.0262

= 0.9738