An article suggested that yield strength (ksi) for A36 grade steel is normally d

Question

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with = 42 and = 5.0 (a) What is the probability that yield strength is at most 40? Greater than 60? (Round your answers to four decimal places.) at most 40 greater than 60 (b) what yield strength value separates the strongest 75% from the others? (Round your answer to three decimal places.) ksi You may need to use the appropriate table in the Appendix of Tables to answer this question Need Help?Read It Talk to a Tutor

Explanation / Answer

X: Yield strength

Mean = 46

standard deviation = 5

a)

P( Yield is at most 40 ) = P ( X <= 40 )

We use z score formula

Z = (x - mean ) / standard deviation = (40 - 42) / 5 = -2/5 = - 0.4

P ( Z <= - 0.4 ) = 0.3446

SO

at most 40 = 0.3446

For P ( yield strength Greater than 60) = P ( X > 60 )

for finding P ( X > 60 )

First find P ( X <= 60 )

z = ( 60 - 42 ) / 5 = 18/5 = 3.6

P ( X <= 60 ) = P ( Z <= 3.6 ) = 0.9998

P ( X <=60 ) = P( Z < = 3.6) = 0.9998

P ( X > 60 ) = 1 - P (Z<=3.6) = 1 - 0.9998 = 0.0002

Greater than 60 = 0.0002

b )

Here in this question assume x yield strength value separates the strongest 75% from the others. So we need area below x as 25% and above x as 75%

P ( Z < z ) = 0.25

by using the table in a reverse way we get z = - 0.67

z = ( x - mean ) / standard deviation

- 0.67 = ( x - 42 ) / 5

-0.67 * 5 = x - 42

-3.35 = x- 42

x = 42 - 3.35

x = 38.35

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