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

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. [tex]\mu=12.30[/tex] and [tex]\sigma=0.20[/tex]

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 =

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

Hence, the required probability= A) 0.1587