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 7 degrees. Management has also set a temperature of 98 degrees or lower to be "lukewarm" anda temperature of 109 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) Whats the probability's the probability a customer leaves the store unsatisfied? d) Suppose management randomly selects 8 customers from the day What's the probability that at least one of them left unsatisfied? e) From the same random sample of 8 customers in part (d), what's the probability that at most half of them left satisfied? f) From the same random sample of 8 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 8 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

Solution:-

a) The probability that a costumer's beverage is lukewarm is 0.2839.

Mean = 102, S.D = 7

x = 98

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

z = - 0.5714

P(z < -0.5714) = 0.2839

b) The probability that a costumer's beverage is too hot is 0.1587.

Mean = 102, S.D = 7

x = 109

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

z = 1.0

P(z > 1.0) = 0.1587

c) The probability that a costumer leaves the store unsatisfied is 0.4426.

x₁ = 98

x₂ = 109

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

z₁ = - 0.5714

z₂ = 1.0

P( - 0.5714 < z < 1.0) = P(z < - 0.5714) + P(z > 1.0)

P( - 0.5714 < z < 1.0) = 0.2839 + 0.1587

P( - 0.5714 < z < 1.0) = 0.4426

d) The probability that atleast one of them left unsatisfied is 0.9907.

n = 8, p = 0.4426

x = 1

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x > 1) = 0.9907

e) The probability that atmost half of them left unsatisfied is 0.7536.

n = 8, p = 0.4426

x = 4

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x < 4) = 0.7536