Question

A manufacturer of isopropyl alcohol claims that their product has a mean content of 480 mL. and a standard 

deviation of 21.5 mL. Assume that the variable is normally distributed.

a. If a sample is selected, what is the probability that the content will be less than 505 mL?

b. If a sample of 6 isopropyl alcohol is selected, what is the probability that the mean of the sample will be 

less than 505?

Answer 1

a.$P(X<505)=(Z<\frac{505-480}{21.5})=P(z<1.16)=0.8770$

b. $P(X<505)=(Z<\frac{505-480}{21.5/\sqrt{6}})=P(z<2.85)=0.9978$