Homework SourseQUESTION The time needed to typically complete a final exam
QUESTION:
The time needed to typically complete a final exam in a standard college course is normally distributed with a mean of 90 minutes, and with a standard deviation of 15 minutes. Please answer the following...
1.) What would be the probability of completing the exam in one hour or less?
2.) What would be the probability that a student will complete the exam in more than 60 mins. but less than 105 mins.?
3.) Let\'s go ahead and assume that the class in question has 60 students, and that examination period is 120 mins. long in length. How many students do you expect will be unable to complete the exam in the time that is allowed?
Solution
1.
Note that 1 hour = 60 min.
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 ) = 0.022750132 [ANSWER]
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2.
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) = 0.022750132
P(z < z2) = 0.841344746
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.818594614 [ANSWER]
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3.
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 ) = 0.022750132
Hence, we expect 0.022750132*60 = 1.36500792 or approximately just 1 student [ANSWER]
