The yield force of a steel-reinforcing bar of a certain type is found to be norm

Question

The yield force of a steel-reinforcing bar of a certain type is found to be normally distributed with the mean of 8600 pounds and standard deviation of 95 pounds. Round all probabilities to 3 decimal places, as needed. A. If one such bar is to be used on a certain project, find the probability that it has yield force in excess of 8650 pounds. .299 Correct: Your answer is correct. B. If five such bars are to be used, what is the probability that they all have yield forces in excess of 8650 pounds? 0 Incorrect: Your answer is incorrect. C. If five such bars are to be used, what is the probability that at least one will have yield force in excess of 8650 pounds? D. What yield force should a bar have so that only 10% of such bars exceed it? (round to a whole number).

Explanation / Answer

Here mean yield force = 8600 pounds

Standard deviation = 95 pounds

A. Part A is ccorrectly solved by you.

B. If five such bars are to be used, what is the probability that they all have yield forces in excess of 8650 pounds?

Answer :

If five such bars are used then what is the probability tht they all have yield forces in excess of 8650 pounds. Now, this is a question of binomial experiment where n = 5 and p = 0.299

so Pr(X = 5) = BIN (X = 5 ; n = 5 ; p = 0.299) = 0.2995 = 0.0024

(b) Here now we have to find that if five such bars are to be used, what is the probability tht at least one would have yield force in excess of 8650 pounds?

Pr(at least one would have yield force in excess of 8650 pounds) = 1 - Pr(no bar would have yield force in excess of 8650 pounds)

Again this question become binomail question. where n = 5 and p = 1 - 0.299 = 0.701

Pr(no bar would have yield force in excess of 8650 pounds) = BIN (X < 8650; 5; 0.701) = 0.7015 = 0.168871

Pr(at least one would have yield force in excess of 8650 pounds) = 1 - 0.168871 = 0.831

(D) Here let sy that yield force is X0

so,

Pr(X > X0 ) = 0.10

Pr(X = < X0 ) = 1 - 0.10 = 0.90

so finding the Z - value for the given probbility from Z - table is

Z = 1.282

Z = (X0 - 8600)/ 95

1.282 = (X0 - 8600)/ 95

X0 = 1.282 * 95 + 8600 = 8722

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