Question

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with μ = 44 and σ = 5.0.

(a) What is the probability that yield strength is at most 39? Greater than 62? (Round your answers to four decimal places.) at most 39

greater than 62

(b) What yield strength value separates the strongest 75% from the others? (Round your answer to three decimal places.) ksi

Answer 1

P(X < A) = P(Z < (A - )/)

= 44

= 5

a) P(at most 39) = P(X < 39)

= P(Z < (39 - 44)/5)

= P(Z < -1)

= 0.1587

P(greater than 62) = P(X > 62)

= 1 - P(X < 62)

= 1 - P(Z < (62 - 44)/5)

= 1 - P(Z < 3.6)

= 1 - 0.9998

= 0.0002

b) Le the yield value that separates the strongest 75% from the others be S

P(X > S) = 0.75

P(X < S) = 1 - 0.75 = 0.25

P(Z < (S - 44)/5) = 0.25

From standard normal distribution table, take z value corresponding to 0.25

(S - 44)/5 = -0.67

S = 40.650 ksi

Note: If z value is taken as -0.674, S = 40.630 ksi