Answer to Question #124759 in Statistics and Probability for Fikiri

Question

Ball bearings of a given brand weigh 0.5 oz with a standard deviation of 0.02 oz. What is the probability that two lots of 1,000 ball bearings each, will differ in weight by more than 2 oz?

Answer 1

Let X₁ and X₂ denote the mean weights of ball bearings in the two lots. Then μ_X1-X2 =μ_X1-μ_X2 =0.50-0.50=0

σ_X1-X2=$\sqrt{\sigma^2_1/n_1+\sigma^2_2/n_2}$ = 0.000895

The standardized variable for the difference in means is Z=((X₁-X₂)-0)/0.000895 and is very nearly normally distributed.

A difference of 2 oz in the lots is equivalent to a difference of 2/1000=0.002 oz in the means. This can occur either if X_1-X₂ ≥0.002 or X₁-X₂ ≤-0.002 ,i.e.

Z≥(0.002-0)/0.000895=2.23 or Z≤(-0.002-0)/0.000895=-2.23

Then

P(Z≥2.23 or Z≤-2.23)=P(Z≥2.23) +P( Z≤-2.23)=2(0.5-0.4871)=0.0258