Question

a soft drink machine outputs a mean of 25 ounces per cup. the machines output is normally distributed with a standard deviation of 3 ounces. what is the probability of filling a cup between 23 and 31 ounces? round your answer to four decimal places.

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 25\) (mean), \(\sigma=3\) (standard deviation).
For \(x = 23\):
\(z_1=\frac{23 - 25}{3}=\frac{-2}{3}\approx - 0.67\)
For \(x = 31\):
\(z_2=\frac{31 - 25}{3}=\frac{6}{3}=2\)

Step2: Use the standard normal distribution table

We know that \(P(23<X<31)=P(-0.67<Z<2)\)
Since \(P(-0.67<Z<2)=P(Z < 2)-P(Z<-0.67)\)
From the standard normal distribution table, \(P(Z < 2)=0.9772\) and \(P(Z<-0.67) = 0.2514\)

Step3: Calculate the probability

\(P(-0.67<Z<2)=0.9772-0.2514 = 0.7258\)

Answer:

\(0.7258\)