Question #336487

2. 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?


1
Expert's answer
2022-05-03T09:25:08-0400

a.

P(X<505)=(Z<50548021.5)P(X<505)=(Z<\frac{505-480}{21.5})




=P(Z<1.1628)=0.8775=P(Z<1.1628)=0.8775

b. 

P(X<505)=(Z<50548021.5/6)P(X<505)=(Z<\frac{505-480}{21.5/\sqrt{6}})




=P(Z<2.8482)=0.9978=P(Z<2.8482)=0.9978


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