Question

The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 ounces and a standard deviation of 0.30 ounce. Each can holds a maximum of 12.75 ounces of soda. Every can that has more than 12.75 ounces of soda poured into it causes a spill and the can must go through a special cleaning process before it can be sold. What is the probability that a randomly selected can will need to go through this process?

Answer 1

Answer:

0.1587

Step-by-step explanation:

According to the situation, the solution and the data provided is as follows

mean = 12.45 ounces

Standard deviation = 0.30 ounces

maximum = 12.75 ounces

More than ounces of soda = 12.75

Based on the above information, the probability is

$Z=\frac{X-\mu }{\sigma } \\\\Z=\frac{12.75-12.45 }{0.30 } \\\\\Z=\frac{0.30 }{0.30 } \\\\Z= 1 \\\\P(X> 12.75)=1-P(X< 12.75) \\\\\P(X> 12.75)=1-P(Z< 1)$

As we know that

P(Z<1) = 0.8413

So,

P (X > 12.75) = 1 - 0.8413

= 0.1587