Question

A sample of n = 9 scores is selected from a normal population with a mean of µ = 80 and a standard deviation of s = 12. The probability that the sample mean will be greater than M = 86 is equal to the probability of obtaining a z-score greater than z = 1.50.True / False.

Answer 1

Answer:

$P(\bar X >86)=P(Z>\frac{86-80}{\frac{12}{\sqrt{9}}}=1.5)$

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

$P(Z>1.5)=0.0668$

TRUE.

Step-by-step 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 variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(80,12)$  

Where $\mu=80$ and $\sigma=12$

Since S follows a normal distribution then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We can find the probability of interest with the following expression:

$P(\bar X >86)=P(Z>\frac{86-80}{\frac{12}{\sqrt{9}}}=1.5)$

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

$P(Z>1.5)=0.0668$

TRUE.