Question

The owner of a coffee shop, an amateur statistician, advertises that the price of coffee on any given day will be randomly picked using a normal distribution with mean $1.35 and standard deviation $0.10. If a customer buys a cup of coffee, what is the probability that he will pay at least $1.50

Answer 1

Answer:

If a customer buys a cup of coffee, 0.067 is the probability that he will pay at least $1.50.    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $1.35

Standard Deviation, σ = $0.10

We are given that the distribution of  price of coffee is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(customer will pay at least $1.50)

$P(x \geq 1.50)$

$P( x \geq 1.50) = P( z \geq \displaystyle\frac{1.50 - 1.35}{0.10}) = P(z \geq 1.5)$

$= 1 - P(z < 1.5)$

Calculation the value from standard normal z table, we have,  

$P(x \geq 1.50) = 1 - 0.933 = 0.067 = 6.7\%$

Hence, if a customer buys a cup of coffee, 0.067 is the probability that he will pay at least $1.50.