7. +-13 points DevoreStatg 4.E.040 My Notes ÷ Ask Your Teacher An article sugges
ID: 3133125 • Letter: 7
Question
7. +-13 points DevoreStatg 4.E.040 My Notes ÷ Ask Your Teacher An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with = 43 and = 4.5 (a) What is the probability that yield strength is at most 38? Greater than 61? (Round your answers to four decimal places.) at most 38 at mOS greater than 61 (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 E Submit AnswerSave F rogressExplanation / Answer
a)
i)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 38
u = mean = 43
s = standard deviation = 4.5
Thus,
z = (x - u) / s = -1.11
Thus, using a table/technology, the left tailed area of this is
P(z < -1.11 ) = 0.1335 [ANSWER]
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ii)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 61
u = mean = 43
s = standard deviation = 4.5
Thus,
z = (x - u) / s = 4
Thus, using a table/technology, the right tailed area of this is
P(z > 4 ) = 0 [ANSWER]
*******************
b)
First, we get the z score from the given left tailed area. As
Left tailed area = 1 - 0.75 = 0.25
Then, using table or technology,
z = -0.67
As x = u + z * s,
where
u = mean = 43
z = the critical z score = -0.67
s = standard deviation = 4.5
Then
x = critical value = 39.985 [ANSWER]
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