7) The time needed to complete a test in a particular college course is normally
ID: 3271629 • Letter: 7
Question
7) The time needed to complete a test in a particular college course is normally distributed with a mean of 90 minutes and a standard deviation of 15 minutes. Answer the following questions.
Please include what to type into Excel to get the answers.
a) What is the probability of completing the test in one hour or less?
b) What is the probability that a student will complete it in more than 60 minutes but less than 105 minutes?
c) Assume that the class has 60 students and that the testing period is 120 minutes in length. How many students do you expect will be unable to complete the test in the allotted time?
Explanation / Answer
a) Solution
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 60
u = mean = 90
s = standard deviation = 15
Thus,
z = (x - u) / s = -2
Thus, using a table/technology, the left tailed area of this is
P(z < -2 ) = 0.0228 [ANSWER]
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b) -Solution
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 60
x2 = upper bound = 105
u = mean = 90
s = standard deviation = 15
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2
z2 = upper z score = (x2 - u) / s = 1
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.0228
P(z < z2) = 0.1587
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.1359 [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 120
u = mean = 90
s = standard deviation = 15
Thus,
z = (x - u) / s = 2
Thus, using a table/technology, the right tailed area of this is
P(z > 2 ) = 0.0228
Hence, we expect 60*0.0228 = 1.368 students to not complete the exam. [ANSWER: 1.368 ]
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