Question

The sat scores have an average of 1200 with a standard deviation of 120. a sample of 36 scores is selected. what is the probability that the sample mean will be larger than 1224? round your answer to three decimal places.

Answer 1

We first standardise
`\bar{x}`:
P(1224 >
`\bar{x}`)=P(
`(1224-1200)/((120)/( √(36) ))`>
`\frac{\bar{x}-1200}{(120)/(√(36))}`)
which reduces to finding P(Z > 1.2) = 0.115

Answer 2

There is a 11.51% probability that the sample mean will be larger than 1224.

Explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation
`(\sigma)/(√(n))`

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The sat scores have an average of 1200 with a standard deviation of 120. This means that
`\mu = 1200, \sigma = 120`.

A sample of 36 scores is selected. what is the probability that the sample mean will be larger than 1224?

This probability is 1 subtracted by the pvalue of Z when
`X = 1224`.

By the central limit theorem, we have that:

`s = (120)/(√(36)) = 20`

So

`Z = (X - \mu)/(\sigma)`

`Z = (1224 - 1200)/(20)`

`Z = 1.2`

`Z = 1.2` has a pvalue of 0.8849.

This means that there is a 1-0.8849 = 0.1151 = 11.51% probability that the sample mean will be larger than 1224.