Question

The breaking strength X of a certain rivet used in a machine engine is normally distributed with mean 5000 psi and standard deviation 400 psi. Find the probability that the difference (in absolute value) between a randomly chosen rivet and the mean is within 250 psi.

Answer 1

Solution:

Given in the question

X is a normal random variable

Mean = 5000

Standard deviation = 400

We need to calculate

P(4750<Xbar<5250) = P(Xbar<5250) -P(Xbar<4750)

Z =(4750-5000)/400 = -0.625

Z = (5250-5000)/400 = 0.625

From Z table we found p- value

P(4750<Xbar<5250) = P(Xbar<5250) -P(Xbar<4750) = 0.7340 -0.2660 = 0.468

So there is 46.8% chances that a randomly chosen river and the mean is within 250psi.