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.
<em>Let X = temperatures in Decatur in December</em>
So, X ~ N()
Now, the z score probability distribution is given by;
Z = ~ N(0,1)
where, = mean temperatures in Decatur in December = 32.5 degrees
= 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( < ) = P(Z < -0.92) = 1 - P(Z 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.