Question

The weight of snickers produced by a certain machine follows a normal distribution with an average weight of 6 oz and a standard dev of 0.06. We need to find the sample size . IF we want the probability of getting a sample mean under 5.98 to be .03 what sample size should be used.

Answer 1

Ans:

P(Z<z)=0.03

z=normsinv(0.03)=-1.88

z=(5.98-6)/(0.06/sqrt(n))

-1.88=-0.02*sqrt(n)/0.06

sqrt(n)=1.88*0.06/0.02

sqrt(n)=5.64

n=31.81 $\approx 32$

sample size of 32 should be used for the the probability of getting a sample mean under 5.98 to be .03