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 102 degrees and a standard deviation of 4 degrees. Management has also set a temperature of 98 degrees or lower to be “lukewarm” and a temperature of 107 degrees to be too hot.

(Round all probability answers to four decimals)

a) What’s the probability that a customer’s beverage is lukewarm? b)

What’s the probability that a customer’s beverage is too hot?

c) What’s the probability’s the probability a customer leaves the store unsatisfied?

d) Suppose management randomly selects 10 customers from the day. What’s the probability that at least one of them left unsatisfied?

e) From the same random sample of 10 customers in part (d), what’s the probability that at most half of them left satisfied?

f) From the same random sample of 10 customers in part (d), what’s the probability that the average temperature of the sample’s coffee beverages was less than 103 degrees?

g) From the same random sample of 10 customers in part (d), what’s the probability that the average temperature of the sample’s coffee beverages was within management’s acceptable temperature range?

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	102
std deviation   =σ=	4.0000

probability that a customer’s beverage is lukewarm :

probability =

P(X<98)

=

P(Z<-1)=

0.1587

b)

probability that a customer’s beverage is too hot :

probability =

P(X>107)

=

P(Z>1.25)=

1-P(Z<1.25)=

1-0.8944=

0.1056

c)

probability’s the probability a customer leaves the store unsatisfied =P(X>107)+P(X<98)=0.1056+0.1587=0.2643

d)

probability that at least one of them left unsatisfied =1-P(all satisfied)=1-(1-0.2643)¹⁰ =0.9535

e)

probability that at most half of them left satisfied

=P(X<=5)=$\sum_{x=0}^{5}\binom{10}{x}(1-0.2643)^{x}(0.2643)^{10-x}$ =0.0960

f)

sample size       =n=	10
std error=σx̅=σ/√n=	1.2649

probability =

P(X<103)

=

P(Z<0.79)=

0.7852

g)

probability =

P(98<X<107)

=

P(-3.16<Z<3.95)=

1-0.0008=

0.9992