A nurse supervisor has found that staff nurses, on average, complete a certain t
ID: 3132218 • Letter: A
Question
A nurse supervisor has found that staff nurses, on average, complete a certain task in 10 minutes. If the time required to complete the task is approximately normally distributed with a standard deviation of 3 minutes.
A. What is the probability that a randomly selected nurse will be able to complete the task in less than 12 minutes?
B. What time limit should the nurse supervisor set so that only 2.5% or less nurses will not be able to complete the task in this time limit?
C. The nurse supervisor has also found that nurse assistants, on average, complete this task in 15 minutes. The time required for the nurse assistants to complete this task is approximately normally distributed with a standard deviation of 5 minutes. If the supervisor set the time limit such that 97.5% or more nurses will complete this task, what proportion of nurse assistants will be able to complete this task within this time limit?
D. What is the probability that 2 of the 9 nurses selected complete the task between 10 and 12 minutes?
E. If repeated samples are selected from the nurse assistants, how large must the samples be for 95% of the sample means to lie within 1 minute of the population mean?
F. Assume the time required to complete this task for nurses and nurse assistants are independent of each other. If we randomly select a nurse and a nurse assistant, what is the probability that both of them will be able to complete the task within 15 minutes?
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 12
u = mean = 10
s = standard deviation = 3
Thus,
z = (x - u) / s = 0.666666667
Thus, using a table/technology, the left tailed area of this is
P(z < 0.666666667 ) = 0.747507462 [ANSWER]
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b)
First, we get the z score from the given left tailed area. As
Left tailed area = 1 - 0.025 = 0.975
Then, using table or technology,
z = 1.959963985
As x = u + z * s,
where
u = mean = 10
z = the critical z score = 1.959963985
s = standard deviation = 3
Then
x = critical value = 15.87989195 [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 = 15.87989195
u = mean = 15
s = standard deviation = 5
Thus,
z = (x - u) / s = 0.17597839
Thus, using a table/technology, the left tailed area of this is
P(z < 0.17597839 ) = 0.56984454 [ANSWER]
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d)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 10
x2 = upper bound = 12
u = mean = 10
s = standard deviation = 3
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 0
z2 = upper z score = (x2 - u) / s = 0.666666667
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.5
P(z < z2) = 0.747507462
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.247507462
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 9
p = the probability of a success = 0.247507462
x = the number of successes = 2
Thus, the probability is
P ( 2 ) = 0.301296771 [ANSWER]
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