Question

Temperatures in Decatur in December follow a normal model with a mean of 32.5 degrees and a standard deviation of 10.1 degrees. What percent of the time will the temperatures December be less than 23.2 degrees?

Answer 1

Answer:

17.88% of the time the temperatures December be less than 23.2 degrees.

Step-by-step explanation:

We are given that temperatures in Decatur in December follow a normal model with a mean of 32.5 degrees and a standard deviation of 10.1 degrees.

Let X = temperatures in Decatur in December

So, X ~ N($\mu=32.5,\sigma^{2}=10.1^{2}$)

Now, the z score probability distribution is given by;

          Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean temperatures in Decatur in December = 32.5 degrees

            $\sigma$ = standard deviation = 10.1 degrees

So, probability that temperatures in December will be less than 23.2 degrees is given by = P(X < 23.2 degrees)

   P(X < 23.2) = P( $\frac{X-\mu}{\sigma}$ < $\frac{23.2-32.5}{10.1}$ ) = P(Z < -0.92) = 1 - P(Z $\leq$ 0.92)

                                                        = 1 - 0.8212 = 0.1788 or 17.88%

Therefore, 17.88% of the time the temperatures in December will be less than 23.2 degrees.

Answer 2

Answer:

17.9%

Step-by-step explanation:

We have been given that temperatures in Decatur in December follow a normal model with a mean of 32.5 degrees and a standard deviation of 10.1 degrees. We are asked to find the percentage of the time when temperatures in December be less than 23.2 degrees.

First of all, we will find z-score corresponding to 23.2 degrees as:

$z=\frac{x-\mu}{\sigma}$

$z=\frac{23.2-32.5}{10.1}$

$z=\frac{-9.3}{10.1}$

$z=-0.920$

Now, we will use normal distribution table to find area under a z-score of $-0.920$ as:

$P(z<-0.920)=0.17879$

Let us convert $0.17879$ into percentage as:

$0.17879\times 100\%=17.879\%\approx 17.9\%$

Therefore, approximately 17.9% of the time the temperatures in December will be less than 23.2 degrees.