a set of average city temperatures in october are normally distributed with a mean of 20.6°c and a standard deviation of 2°c. what proportion of temperatures are between 18°c and 20.6°c? you may round your answer to four decimal places.

Question

Accepted Answer

# Explanation:
## Step1: Calculate the z - score for 18°C
The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean and $\sigma$ is the standard deviation. Given $\mu = 20.6$, $\sigma=2$ and $x = 18$. Then $z=\frac{18 - 20.6}{2}=\frac{- 2.6}{2}=-1.3$.
## Step2: Find the proportion using the standard normal distribution
We want to find $P(-1.3<Z<0)$. Since the standard normal distribution is symmetric about $Z = 0$, and $P(Z < 0)=0.5$, and we know that we can use the standard - normal table to find $P(Z<-1.3)$. From the standard - normal table, $P(Z < - 1.3)=0.0968$. Then $P(-1.3<Z<0)=P(Z < 0)-P(Z < - 1.3)=0.5 - 0.0968 = 0.4032$.

# Answer:
0.4032

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the z - score for 18°C

Step2: Find the proportion using the standard normal distribution

Answer: