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 38? Greater than 64? (Round your answers to four decimal places.)

at most 38
greater than 64

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

Answer 1

µ = 44, σ = 5

a) probability that yield strength is at most 38, P(X <= 38) =

= P( (X-µ)/σ <= (38-44)/5 )

= P(z <= -1.2)

Using excel function:

= NORM.S.DIST(-1.2, 1)

= 0.1151

probability that yield strength is greater than 64, P(X > 64) =

= P( (X-µ)/σ > (64-44)/5)

= P(z > 4)

= 1 - P(z < 4)

Using excel function:

= 1 - NORM.S.DIST(4, 1)

= 0.00

b)

P(x > a) = 0.75

= 1 - P(x < a) = 0.75

= P(x < a) = 0.25

Z score at p = 0.25 using excel = NORM.S.INV(0.25) = -0.6745

Value of X = µ + z*σ = 44 + (-0.6745)*5 = 40.628 ksi