Question

The tensile strength of a metal part is normally distributed with mean 35 pounds and standard deviation 5 pounds. If 10,000 parts are produced, How many would have a tensile strength in excess of 43 pounds?

Answer 1

This is a normal distribution question with

P(x > 43.0)=?
The z-score at x = 43.0 is,

z = 1.6
This implies that
P(x > 43.0) = P(z > 1.6) = 1 - 0.9452
P(x > 43.0) = 0.0548

Since there are 10000 parts produced, 0.0548*10000 = 548 parts have tensile strength in excess of 43 pounds
PS: you have to refer z score table to find the final probabilities.
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