Question

A manufacturer of ball bearings claims that this product has a mean weight of 5.02 g and a standard deviation of 0.30 g. What is the probability that a random sample of 100 ball bearings will have combined weight:

a. between 496 g and 500 g?

b. more than 510 g?

Answer 1

X ~ N(5.02, 0.30^2)

n = 100

a.

$P(4.96< \bar{X}<5.00) = P(\frac{4.96-5.02}{\frac{0.3}{\sqrt{100}}} <Z< \frac{5.00-5.02}{\frac{0.3}{\sqrt{100}}}) \\ = P(\frac{4.96-5.02}{\frac{0.3}{\sqrt{100}}} <Z< \frac{5.00-5.02}{\frac{0.3}{\sqrt{100}}}) \\ = P(-2.00 < Z < -0.666) \\ = P(Z< -0.666) -P(Z< -2) \\ = 0.2527 -0.0227 \\ = 0.2300 \\ = 23.00 \; \%$

b.

$P(\bar{X}>5.10) = 1 -P(\bar{X}<5.1) \\ = 1 -P(Z< \frac{x-μ}{\frac{σ}{\sqrt{n}}}) \\ = 1 - P(Z< \frac{5.10-5.02}{\frac{0.3}{\sqrt{100}}}) \\ = 1 -P(Z<2.666) \\ = 1 -0.9961 \\ = 0.0039 \\ = 0.39 \; \%$