Question

a set of average city temperatures in august are normally distributed with a mean of 21.25°c and a standard deviation of 2°c. what proportion of temperatures are between 19.63°c and 20.53°c? you may round your answer to four decimal places.

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set.
For $x = 19.63$, $z_1=\frac{19.63 - 21.25}{2}=\frac{- 1.62}{2}=-0.81$.
For $x = 20.53$, $z_2=\frac{20.53 - 21.25}{2}=\frac{-0.72}{2}=-0.36$.

Step2: Use the standard normal table

We want to find $P(-0.81We know that $P(-0.81From the standard - normal table, $P(Z < - 0.36)=0.3594$ and $P(Z < - 0.81)=0.2090$.
So $P(-0.81

Answer:

$0.1504$