Question

The average weight of a chicken egg is 2.25 ounces with a standard deviation of 0.2 ounces. You take a random sample of a dozen eggs. What is the probability that the mean weight of the eggs in the sample will be less than 2.2 ounces?

Answer 1

`P(\bar X <2.2)=P(Z<(2.2-2.25)/((0.2)/(√(12))))=P(Z<-0.866)`

`P(\bar X<2.2)= P(Z<-0.866)=0.193`

Explanation:

Previous concepts

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 interest on this case, and for this case we know the distribution for X is given by:

`X \sim N(\mu,\sigma)`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

On this case
`\bar X \sim N(2.25,(0.2)/(√(12)))`

Solution to the problem

We want this probability:

`P(\bar X<2.2)`

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

`z=(x-\mu)/((\sigma)/(√(n)))`

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

`P(\bar X <2.2)=P(Z<(2.2-2.25)/((0.2)/(√(12))))=P(Z<-0.866)`

`P(\bar X<2.2)= P(Z<-0.866)=0.193`