Finding the z-Value that Corresponds to a Given Area

Question

In one region of the Caribbean Sea, daily water temperatures are normally distributed with a mean of 77.9 Fahrenheit and a standard deviation of 2.4 Fahrenheit. What is the third quartile for water temperatures in this region?

Answer 1

The third quartile is the the value that would divide the bottom 75% of the data from the top 25%, so we want P(X< Q3 = .75. Since the data follows a normal distribution with mean 77.9 and standard deviation of 2.4, that means we want Q3 to equal a Z-Score of 0.675.

P(Z<0.675) = .75,

Z = (x-77.9)/2.4,

0.675=(x-77.9)/2.4 => (2.4*0.675)+77.9 = x = 79.52

The third quartile is 79.52.

Answer 2

Generally speaking the Z score for the 3rd quartile is .67, 1st quartile is -.67 since they are symmetric opposites, and the median (50th percentile) is 0 since its at the center of the model.

to find the random variable X that represents the 3rd quartile of this distribution (temperatures) you:

.67 = (x - 77.9)/2.4

.67(2.4) = x - 77.9

.67(2.4) + 77.9 = x

x = 79.51 degrees Fahrenheit is the 3rd quartile of temperatures for the model or distribution