Question

The population of scores on a nationally standardized test forms a normal distribution with μ = 300 and σ = 50. If you take a random sample of n = 25 students, what is the probability that the sample mean will be less than M = 280?

Answer 1

`P(\bar X <280)=P(Z<(280-300)/((50)/(√(25)))=-2)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<-2)=0.0228`

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".

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

`X \sim N(300,50)`

Where
`\mu=300` and
`\sigma=50`

Since the distribution for X is normal then we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

We can find the individual probability like this:

`P(\bar X <280)=P(Z<(280-300)/((50)/(√(25)))=-2)`

And using a calculator, excel or the normal standard table we have that:

`P(Z<-2)=0.0228`