SOLUTION: The process for manufacturing ball bearing results in weights that have an approximately normal distribution with mean of 0.15g and standard deviation 0.003g. A. If you select on

Question 1106314: The process for manufacturing ball bearing results in weights that have an approximately normal distribution with mean of 0.15g and standard deviation 0.003g.
A. If you select one ball bearing at random, what is the probability that it weights less than 0.148g?
B. If you select four ball bearings at random, what is the probability that their mean weight is less than 0.148g?
C. If you select ten ball bearings at random, what is the probability that their mean weight is less than 0.148g?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
z=(x-mean)/sd or (xbar-mean)/sigma/sqrt(n)
z=(0.148-0.150)/0.003=-2/3. probability z< (-2/3) is 0.2525
z=-0.002*sqrt(4)/0.003 or -4/3. Probability z < -(4/3) is 0.0912
z=-0.002*sqrt(10)/0.003=0.0174. Notice the inverting of the denominator fraction so that sqrt(n) is in the numerator.
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