Cabin Fever Coffee is a local coffee bar located in Defiance, Ohio, and is striv
ID: 3362842 • Letter: C
Question
Cabin Fever Coffee is a local coffee bar located in Defiance, Ohio, and is striving for consistent quality in their coffee beverages served to guests. Guests often complain that their beverage is not hot enough (lukewarm) or too hot and thus leave unsatisfied. Management determines that the temperature of their brewed coffee is approximately normally distributed with a mean of 101 degrees and a standard deviation of 4 degrees. Management has also set a temperature of 95 degrees or lower to be “lukewarm” and a temperature of 106 degrees to be too hot.
Round all probability answers to four decimal places.
a) What is the probability that a customer’s beverage is lukewarm?
b) What is the probability that a customer’s beverage is too hot?
c) What is the probability that a customer leaves the store unsatisfied?
d) Suppose management randomly selects 7 customers from the day. What is the probability that at least one of them left unsatisfied?
e) From the same random sample of 7 customers in part (d), what is the probability that at most half of them left satisfied?
f) From the same random sample of 7 customers in part (d), what is the probability that the average temperature of the sample’s coffee beverages was less than 103 degrees?
g) From the same random sample of 7 customers in part (d), what is the probability that the average temperature of the sample’s coffee beverages was within management’s acceptable temperature range?
Explanation / Answer
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 101
standard Deviation ( sd )= 4
a.
P(X < 95) = (95-101)/4
= -6/4= -1.5
= P ( Z <-1.5) From Standard Normal Table
= 0.0668
b.
P(X > 106) = (106-101)/4
= 5/4 = 1.25
= P ( Z >1.25) From Standard Normal Table
= 0.1056
c.
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 95) = (95-101)/4
= -6/4= -1.5
= P ( Z <-1.5) From Standard Normal Table
= 0.0668
P(X > 106) = (106-101)/4
= 5/4 = 1.25
= P ( Z >1.25) From Standard Normal Table
= 0.1056
P( X < 95 OR X > 106) = 0.0668+0.1056 = 0.1725
d.
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
P( X < 1) = P(X=0)
= ( 7 0 ) * 0.1725^0 * ( 1- 0.1725 ) ^7
= 0.265690495
P( X > = 1 ) = 1 - P( X < 1) = 0.7343