A soft drink machine outputs a mean of 24 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 18 and 30 ounces

The probability is 0.9545

Step-by-step explanation:

We start by using z-scores here

Mathematically;

z-score = (x-mean)/SD

mean = 24 ounces

SD = 3 ounces

For 18, we have

z-score = (18-24)/3 = -6/3 = -2

For 30, we have

z-score = (30-24)/3 = 6/3 = 2

So the probability we want to calculate is;

P( -2 < x < 2)

We can use the standard normal distribution table for this

From the standard normal distribution table;

P(-2 < x < 2) = 0.9545

idk

step-by-step explanation: