Answered

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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

Sagot :

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

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

[tex]P(X\ge74) = P\left(\dfrac{X-80}3 \ge \dfrac{74-80}3\right) \\ = P(Z \ge -2) \\ = 1-P(Z < -2) \approx 0.9773[/tex]

[tex]P(X\ge89) = P\left(\dfrac{X-80}3 \ge \dfrac{89-80}3\right) \\= P(Z \ge 3) \\= 1 - P(Z < 3) \approx 0.0013[/tex]

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