Answer:
Step-by-step explanation:
Given that the Acme Company manufactures widgets, which have a mean of 60 ounces and a standard deviation of 7 ounces
We know that 95% of the area lie between -2 and 2 std deviations from the mean.
i.e. Probability for lying in the middle of 95%
Z score
Between 46 and 74 oz.
b) Between 12 and 57
convert into Z score
P(-6.86<z<-0.43)
=0.5-0.1664=0.3336
c) X<30 gives Z<-4.83
i.e. P(X<30) =0.00