A set of average city temperatures in August are normally distributed with a mean of 21.25∘C21.25 ^\circ \text{C}21.25∘C21, point, 25, degrees, start text, C, end text and a standard deviation of 2∘C2 ^\circ \text{C}2∘C2, degrees, start text, C, end text.

What proportion of temperatures are between 23.71∘C23.71^\circ \text{C}23.71∘C23, point, 71, degrees, start text, C, end text and 26.17∘C26.17^\circ \text{C}26.17∘C26, point, 17, degrees, start text, C, end text?

You may round your answer to four decimal places.

And we can find this probability with this difference:

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  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 temperaturs in August, and for this case we know the distribution for X is given by:

Where and

We are interested on this probability

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

If we apply this formula to our probability we got this:

And we can find this probability with this difference:

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

Using the normal distribution, it is found that 0.1024 = 10.24% of temperatures are between 23.71ºC and 26.17ºC.

--------------------------

Normal Probability Distribution

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. Looking at the z-score table and find the p-value associated with this z-score, which is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

--------------------------

X = 26.17

has a p-value of 0.9931.

X = 23.71

has a p-value of 0.8907.

0.9931 - 0.8907 = 0.1024

0.1024 = 10.24% of temperatures are between 23.71ºC and 26.17ºC.

A similar problem is given at link

step-by-step explanation: