Question

I am not able to tackle this question. Somebody please help me out

The minimum load at which a certain kind of iron wire breaks can be supposed to define a random variable normally distributed with expected value 80N and standard deviation 3N. Find the probability that such a wire will break when the load is (a) 74N
(b)89N

Answer 1

The wire breaks if the load
`X` exceeds some amount. So what you're asked to do is find
`P(X\ge74)` and
`P(X\ge89)`.

In either case, transform
`X` to the random variable
`Z` that's normally distributed with expected value 0 and standard deviation 1.

`P(X\ge74) = P\left(\frac{X-80}3 \ge \frac{74-80}3\right) \\ = P(Z \ge -2) \\ = 1-P(Z < -2) \approx 0.9773`

`P(X\ge89) = P\left(\frac{X-80}3 \ge \frac{89-80}3\right) \\= P(Z \ge 3) \\= 1 - P(Z < 3) \approx 0.0013`