Question

A soft drink machine outputs a mean of 27 ounces per cup. The machine's output is normally distributed with a standard deviation of 3 ounces. What is the probability of filling a cup between 21 and 28 ounces? Round your answer to four decimal places.

Answer 1

Answer:  0.6065

Step-by-step explanation:

Given : The machine's output is normally distributed with

$\mu=27\text{ ounces per cup}$

$\sigma=3\text{ ounces per cup}$

Let x be the random variable that represents the output of machine .

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x= 21 ounces

$z=\dfrac{21-27}{3}\approx-2$

For x= 28 ounces

$z=\dfrac{28-27}{3}\approx0.33$

Using the standard normal distribution table , we have

The p-value : $P(21$

$P(z$

Hence, the probability of filling a cup between 21 and 28 ounces=  0.6065