Question

The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce. Each can holds a maximum of 12.50 ounces of soda. Every can that has more than 12.50 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? A) .1587 B) .6587 C) .8413 D) .3413

Answer 1

Answer: A) .1587

Explanation:

Given : The amount of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.30 ounces and a standard deviation of 0.20 ounce.

i.e.
`\mu=12.30` and
`\sigma=0.20`

Let x denotes the amount of soda in any can.

Every can that has more than 12.50 ounces of soda poured into it must go through a special cleaning process before it can be sold.

Then, the probability that a randomly selected can will need to go through the mentioned process = probability that a randomly selected can has more than 12.50 ounces of soda poured into it =

`P(x>12.50)=1-P(x\leq12.50)\\\\=1-P((x-\mu)/(\sigma)\leq(12.50-12.30)/(0.20))\\\\=1-P(z\leq1)\ \ [\because z=(x-\mu)/(\sigma)]\\\\=1-0.8413\ \ \ [\text{By z-table}]\\\\=0.1587`

Hence, the required probability= A) 0.1587