A student has submitted a concrete cylinder to the \"strength contest\" in Engin
ID: 3132923 • Letter: A
Question
A student has submitted a concrete cylinder to the "strength contest" in Engineering Open House, and is scheduled to be the last contestant for load testing. The process producing her concrete cylinder yields strength that is normally distributed with mean = 80 kips and standard deviation = 20 kips (a kip is a non-SI unit of force). Immediately prior to her cylinder being tested, the two highest strengths in the contest thus far are 100 kips and 70 kips.
A. What is the probability that this student will be the second place winner?
ANS: 0.5328
B. Now the cylinder (from A) is being tested and it has not shown any kind of distress at a load of 90 kips. What is the probability that this student will win first place?
I posted this earlier but part B was incorrect. Please show all steps with a correct answer for a rating. Thanks in advance. :)
Explanation / Answer
We need it to be greater than 100 kips, given that it is greater than 90 kips.
Hence, here, we need
P(x>100|x>90) = P(x>100)/P(x>90)
FOR P(X>100):
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 100
u = mean = 80
s = standard deviation = 20
Thus,
z = (x - u) / s = 1
Thus, using a table/technology, the right tailed area of this is
P(z > 1 ) = 0.158655254
FOR P(X>90):
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 90
u = mean = 80
s = standard deviation = 20
Thus,
z = (x - u) / s = 0.5
Thus, using a table/technology, the right tailed area of this is
P(z > 0.5 ) = 0.308537539
Hence,
P(x>100|x>90) = 0.158655254/0.308537539 = 0.51421702 [ANSWER]
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