The amount of cola in a 355 ml bottle from a certain company is a random variable with a mean of 355 ml and a standard deviation of 2 ml. For a sample of size 32, perform the following calculations. a. Find an approximate probability that the sample mean is less than 354.8 ml. b. Suppose the amount of cola is distributed as N(355, 4). Find an approximate probability that 10 of the bottles in the sample contain less than 354.8 ml of cola.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the amount of cola in a 355 ml bottle from a certain company, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu=355,\sigma=2)[/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

On this case [tex]\bar X \sim N(355,\frac{2}{\sqrt{32}})[/tex]

Part a

We want this probability:

[tex]P(\bar X<354.8)=P(\bar X<354.8)[/tex]

The best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

If we apply this formula to our probability we got this:

[tex]P(\bar X <354.8)=P(Z<\frac{354.8-355}{\frac{2}{\sqrt{32}}})=P(Z<-0.57)[/tex]

[tex]P(\bar X<354.8)=0.284[/tex]

Part b

Let X the random variable that represent the amount of cola in a 355 ml bottle from a certain company, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu=355,\sigma=4)[/tex]