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 $\sigma =3ml$

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

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

Short answer

(a) The probability is $0.1587$

(b) The probability is$0.0071$

Step-by-step solution

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 $295ml$.

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 $=298ml$and standard deviation $=3ml$.

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

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

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

$=P(Z<-1)(Froms\tan dardnormaltable)$

$=0.1587$

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 $295ml$.

Part (b) Step-2:  Explanation 

The probability that mean content in randomly chosen $6$bottles is less than $295ml$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)(Froms\tan dardnormaltable)$$=P(Z<-2.45)(Froms\tan dardnormaltable)$

$=0.0071$

Thus the require probability is$0.0071$