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 40? Greater than 62? (Round your answers to four decimal places.) at most 40 greater than 62 (b) What yield strength value separates the strongest 75% from the others? (Round your answer to three decimal places.)

Step-by-step explanation:

Using normal distribution,

z = (x - μ) / σ

μ = mean = 44 and

σ = standard deviation = 5.0

a) The probability that yield strength is at most 40=

P (x lesser than or equal to 40)

z = (40-44) / 5 = - 0.8

Looking at the normal distribution table,

P (x lesser than or equal to 40) = 0.2119

b) P (x greater than 62) = 1 - P (x lesser than or equal to 62)

z = (62-44) / 5 = 3.6

Looking at the normal distribution table,

P (x greater than 62) = 1 - 0.99984

= 0.00016

c) P (42 lesser than or equal to x lesser than or equal to 62)

= P (x lesser than or equal to 62) - P (x lesser than or equal to 40)

= 0.99987-0.2119 = 0.78797

d) What yield strength value separates the strongest 75% from the others.

To get x for strongest 75, we get the z value corresponding to 0.75 from the table

z = 0.675 = (x-44) / 5

x = 3.375+44 = 47.375

The rest is 25% = 0.25

we get the z value corresponding to 0.25 from the table)

z = - 0.67 = (x-44) / 5

-3.35 = x - 44

x = - 44+3.35 = 40.65

yield strength value that separates the strongest 75% from the others

=47.375-40.65 = 6.725