A stockbroker has kept a daily record of the value of a particular stock over the years and finds that prices of the stock form a normal distribution with a mean of $8.52 with a standard deviation of $2.38. The percentile rank of a price of $13.87 is

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 prices of the stock of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(8.52,2.38)[/tex]

Where [tex]\mu=8.52[/tex] and [tex]\sigma=2.38[/tex]

If we want to find the percentile for 13.87. We are interested on this probability

[tex]P(X<13.87)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:[tex]P(X<13.87)=P(\frac{X-\mu}{\sigma}<\frac{13.87-\mu}{\sigma})=P(Z<\frac{13.87-8.52}{2.38})=P(z<2.25)[/tex]And we can find this probability using the normal standard table or excel and we got:

[tex] P(X<13.87) = P(Z<2.25) =0.9877[/tex]

So then we can conclude that the value of $13.87 would be approximately the 98th percentile