Antonio measures the alcohol content of whiskey for his Chem

Antonio measures the alcohol content of whiskey for his Chemistry 101 lab. He actually measures the mass of 6.7 milliliters of whiskey -- a chemical calculation then finds the percent alcohol from the mass. The standard deviation of students\' measurements of mass is = 11.4 milligrams (mg). Antonio repeats the measurement 6 times and records the mean x of his 6 measurements.

Averages of several measurements are less variable than individual measurements. The true mass of the whiskey sample is 4.7 grams, or 4700 milligrams (mg). Antonio\'s measurements have the normal distribution with mean 4700 mg and standard deviation 11.4 mg. In this case, the mean of his 6measurements also has a normal distribution.

I mainly need help on C/D. Thanks!

Solution

Answers for C and D

c) given mean = 4700mg and standard deviaiton = 11.4 mg

Antonio misses the true mass by more than 7.752 mg in either direction means

if he get less than 4700-7.752=4692.248 or greater than 4700+ 7.752=4707.752

this means X ( mass of the wihiskey) is X<4692.248 or X>4707.752

if he makes one measurement then probability that Antonio misses the true mass by more than 7.752 mg in either direction = P(X<4692.248) + P(X>4707.752)

= P[z< (4692.248-4700)/11.4) + P[z> (4707.752-4700)/11.4]= P(z<-0.68)+ P(Z>0.68)=2*P(z<-0.68) ( since normal distribution is symmetric)

=0.4965

d) if there are 6 measurements then SD of the mean = 11.4/sqrt(6) = 4.654

probability that the mean of 6 independent measurements misses the true mass by more than 7.752 mg is

P(X<4692.248) + P(X>4707.752) == P[z< (4692.248-4700)/4.654) + P[z> (4707.752-4700)/4.654]= P(z<-1.666)+ P(Z>1.666)=2*P(z<-1.666) ( since normal distribution is symmetric)

=0.0958