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($\frac{1224-1200}{\frac{120}{ \sqrt{36} }}$>$\frac{\bar{x}-1200}{\frac{120}{\sqrt{36}}}$)
which reduces to finding P(Z > 1.2) = 0.115

Answer 2

Answer:

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

Step-by-step 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 $\frac{\sigma}{\sqrt{n}}$

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:

$Z = \frac{X - \mu}{\sigma}$

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:

$s = \frac{120}{\sqrt{36}} = 20$

So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1224 - 1200}{20}$

$Z = 1.2$

$Z = 1.2$ 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.