Question

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with μ = 44 and σ = 4.5.

(a) What is the probability that yield strength is at most 38? Greater than 60? (Round your answers to four decimal places.)

at most 38
greater than 60

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

Answer 1

Solution :

Given that,

mean = = 44

standard deviation = =4.5

A ) P( x 38 )

P ( x -  / ) ( 38- 44 / 4.5)

P ( z -6 / 4.5 )

P ( z -1.33 )

Using z table

= 0.0918

Probability = 0.0918

B ) P ( x > 60 )

= 1 - P (x < 60 )

= 1 - P ( x -  / ) < ( 60 - 44 / 4.5 )

= 1 - P ( z < 16 / 4.5 )

= 1 - P ( z < 3.56)

Using z table

= 1 -0.9998

= 0.0002

Probability = 0.0002

Using standard normal table,

P(Z > z) = 75%

1 - P(Z < z) = 0.75

P(Z < z) = 1 - 0.75 = 0.25

P(Z < -0.6745) = 0.99

z = -0.6745

Using z-score formula,

x = z * +

x = - 0.67 * 4.5 + 44

= 40.985

X = 40.985