Q.55 - The Practice of Statistics for AP

Question

Bottling cola A hattling compamy uses a fillimg maichine to fill plastic botles with cola. The bottles are supposed to contain $300$ milliliters (ml) . In fact, the contents vary according to a Normal distribution with mean  $\mu =298$ml and standard deviation $\sigma =3 ml$

(a) What is the probability that in individual bottle contains less than $295 ml$ ? Show you work.

(b) What is the probability that the mean contents of six randomly selected bottles is less than  $295 ml$? Show your work.

Step-by-Step Solution

Answer

(a) The probability is $0.1587$

(b) The probability is $0.0071$

Part (a) Step-1 Given Information

 Given in the question that,

population mean $\left(\mu \right)=298$$\left(\mu \right)=298$

Population standard deviation $\left(\sigma \right)=3$

we have to find that  the probability that in individual bottle contains less than  $295 ml$. 

Part (a) Step-2 Explanation

The formula to compute the Z- score is:

$z=\frac{x-\mu }{\sigma }$

$x$ is raw score

$\mu$ is population mean

$s$ is population standard deviation

Consider, $X$ be the random variable that shows the amount of cola in plastic bottles follows the normal distribution with mean $=298 ml$ and standard deviation $=3 ml$ .

The probability that an individual bottle would contain less than$295 ml$ cola can be computed as:

$P( X<295)=P\left(\frac{x-\mu }{\sigma }<\frac{195-\mu }{\sigma }\right)$

                     $=P\left(Z<\frac{295-298}{3}\right)$

            $=P(Z<-1) (From s\tan dard normal table )$

                      $=0.1587$

Thus, the required probability is $0.1587$.

Part (b) Step-1 Given Information

Given in the question that sample size $\left(n\right)=6$ we have to find that the probability that the mean contents of six randomly selected bottles is less than  $295 ml$.

Part (b) Step-2: Explanation

The probability that mean content in randomly chosen $6$ bottles is less than $295 ml$ is calculated as follows:

$P(\overline{X}<295)=P\left(\frac{{\displaystyle \frac{\overline{x-\mu }}{\sigma }}}{\sqrt{n}}<\frac{{\displaystyle \frac{295-\mu }{\sigma }}}{\sqrt{n}}\right)$

                   $=$ $P\left(Z<\frac{{\displaystyle \frac{295-298}{3}}}{\sqrt{6}}\right)$

      $P(Z<-2.45) (From s\tan dard normal table)$$=P(Z<-2.45) (From s\tan dard normal table)$

                   $=0.0071$

Thus the require probability is $0.0071$