Question

If the mean of breaking strength of a copper wire is 575 pounds with a standard deviation of 8.3 pounds how large a sample must be used in order that there be one chance in 100 that the mean breaking strength of the sample is less than 572 pounds?

Answer 1

z=(x-mean)/$\sigma/\sqrt{n}=(572-575)/8.3/\sqrt{n}$ or |z|=$3\sqrt{n}/8.3$

we need to find that value of x for which

0.01=area to the right at the variate z

that is area to the left=1-0.01=0.99=0.5+0.49

From the areas under the standard normal curve the corresponding value of z is 2.33. Hence

|z|=$3\sqrt{n}/8.3$ gives 2.33=3$\sqrt{n}$ /8.3

3$\sqrt{n}=2.33*8.3=19.339,\sqrt{n}=6.446,n=6.446^2=41.55=42$