Respuesta :
Answer:
If the waiting time exceeds 22.52 minutes, the customer will receive the discount.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 15 minutes
Standard Deviation, σ = 7 minutes
We are given that the distribution of waiting time is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
We have to find the value of x such that the probability is 0.15
P(X > x) = 0.15
[tex]P(z>\displaystyle\frac{x-15}{7}) = 0.15\\\\1 - P(z<\displaystyle\frac{x-15}{7}) = 0.15\\\\P(z<\displaystyle\frac{x-15}{7}) = 0.85[/tex]
Calculating the value from standard normal table, we have,
[tex]P(z<1.036) = 0.85[/tex]
Comparing, we get,
[tex]\displaystyle\frac{x-15}{7} = 1.036\\\\x = 22.252 \approx 22.52[/tex]
If the waiting time exceeds 22.52 minutes, the customer will receive the discount.
The number of minutes they need to wait to receive the discount is approximately 22
Probability and normal distribution
The formula for calculating the z score is expressed as:
[tex]z =\frac{x-\mu}{\sigma}[/tex]
Given the following parameters;
P(X > x) = 0.15
[tex]p(z < \frac{x-15}{7})=1-0.15\\ p(z < \frac{x-15}{7})=0.85\\[/tex]
According to the standard table;
[tex]p(z < 1.036)=0.85\\[/tex]
[tex]\frac{x-15}{7} =1.036[/tex]
Solve the expression for the value of x
x - 15 = 7(1.036)
x - 15 = 7.252
x = 15 + 7.252
x = 22.352
Hence the number of minutes they need to wait to receive the discount is approximately 22
Learn more on normal distribution here: https://brainly.com/question/4079902
