A set of exam scores is normally distributed and has a mean of 74.4 and a standard deviation of 8.5. What is the probability that a randomly selected score will be between 69 and 77?

Answer = (round to four decimal places)

Note: Be careful...only use the Z Table here...do not use technology or the 68-95-99.7 Rule.

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 who represent the exam scores for the population, and for this case we know the distribution for X is given by:

[tex]X \sim N(74.4,8.5)[/tex]

Where [tex]\mu=74.4[/tex] and [tex]\sigma=8.5[/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

What is the probability that a randomly selected score will be between 69 and 77?

For this case we can use the z score formula given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

[tex]P(69<x<77)=P(\frac{69-74.4}{8.5}<Z<\frac{77-74.4}{8.5})=P(-0.635<Z<0.306)=P(Z<0.306)-P(Z<-0.635)=0.6202-0.2627=0.3575[/tex]

And that correspond with the 35.75% of the data.