Answer:
The probability is P=0.9545.
The empirical rule is consistent, as it would have estimate a probability of 95%.
Step-by-step explanation:
We have a random variable with mean 31.8 in. and standard deviation of 2.4 in.
We have to find the probability that a randomly selected part from this supplier will have a value between 27.0 and 36.6 in.
If we assumed a normal-like distribution, we can calculate the z-score for both values:
[tex]z_1=\dfrac{X_1-\mu}{\sigma}=\dfrac{27-31.8}{2.4}=\dfrac{-4.8}{2.4}=-2\\\\\\z_2=\dfrac{X_2-\mu}{\sigma}=\dfrac{36.6-31.8}{2.4}=\dfrac{4.8}{2.4}=2[/tex]
That can be interpreted as an interval that is 2 standard deviations wide, centered on the mean.
This interval has a probability of:
[tex]P(27.0<x<36.6)=P(-2<z<2)=0.9545[/tex]
If we apply the empirical rule, with 2 standard deviations from the mean, we would have estimate a probability of 95%, which is accurate.
