Question

A manufacturer receives a shipment of 500 spare parts from a supplier who claims that the lengths of the spare parts are approximately normally distributed having a mean of 2.5cm and a standard deviation of 0.04 cm. If the manufacturer takes a 10% random sample from the shipment, what is the probability that he gets a mean length of:

A. More than 2.22 cm?

B. More than 2.40 cm?

Answer 1

Mean lentgh of 500 * 10% = 50 random variables is distributed as normal random variable with mean m = 2.5cm and standard deviation $= 0.04 / \sqrt 50 = 0.0057cm$ .

A. So, according to distribution function of a normal random variable X with mean = 2.5, and standard deviation = 0.0057:

$Pr(X > X_1) = 1 - Pr(X < X_1) = 1 - \Phi(\frac{X_1 - 2.5}{0.0057})$ , where $\Phi(x)$ is distribution function of a normal random variable with mean = 0 and standard deviation = 1

For X1 = 2.22cm: Pr(X > X1) = 1

B. By analogy:

For X1 = 2.4cm: Pr(X >X1) = 1