Question

Help with b

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with u = 45 and g = 5.5. in USE SALT (a) What is the probability that yield strength is at most 38? Greater than 65? (Round your answers to four decimal places.) at most 38 0.1020 greater than 65 0.0001 (b) What yield strength value separates the strongest 75% from the others? (Round your answer to four decimal places.) 48.685 X ksi

Answer 1

Let X be the random variable denoting the yeild strength of the steel.

We know that X ~ N(45 , 5.5²)

b) Here we need the minimum strength which is in the strongest 75% of the steel =

Let the minimum strength be x.

So, P(X > x) = 0.75

P(X < x) = 0.25

P(Z < (x-45)/5.5) = 0.25

Using the standard normal table, we get,

(x-45)/5.5 = -0.674

x = (-0.674*5.5) + 45

x = 41.293

So, the minimum strength which separates the strongest 75% yeild strength is 41.2930 ksi.

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