2. Assume that heart rate (in beats per minute, or bpm) for STA 100 students is
ID: 3021907 • Letter: 2
Question
2. Assume that heart rate (in beats per minute, or bpm) for STA 100 students is distributed normal, with a mean of 110 bpm and a standard deviation of 20.2 bpm. Assume all students in the following problem are selected from this population.
(a) Find the probability that a randomly selected students heart rate is above 125.
(b) What percentage of randomly selected students could we expect to have a heart rate between 90 and 130?
(c) What is the rst and third quartile of heartrates for randomly selected students?
(d) What is the 8th percentile for heart rates among randomly selected students?
Explanation / Answer
A)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 125
u = mean = 110
s = standard deviation = 20.2
Thus,
z = (x - u) / s = 0.742574257
Thus, using a table/technology, the right tailed area of this is
P(z > 0.742574257 ) = 0.228869739 [answer]
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B)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 90
x2 = upper bound = 130
u = mean = 110
s = standard deviation = 20.2
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.99009901
z2 = upper z score = (x2 - u) / s = 0.99009901
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.161062864
P(z < z2) = 0.838937136
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.677874273 = 67.79% [ANSWER]
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c)
FIRST QUARTILE:
First, we get the z score from the given left tailed area. As
Left tailed area = 0.25
Then, using table or technology,
z = -0.67448975
As x = u + z * s,
where
u = mean = 110
z = the critical z score = -0.67448975
s = standard deviation = 20.2
Then
x = critical value = 96.37530705 [ANSWER]
****
THIRD QUARTILE:
First, we get the z score from the given left tailed area. As
Left tailed area = 0.75
Then, using table or technology,
z = 0.67448975
As x = u + z * s,
where
u = mean = 110
z = the critical z score = 0.67448975
s = standard deviation = 20.2
Then
x = critical value = 123.624693 [ANSWER]
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d)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.08
Then, using table or technology,
z = -1.40507156
As x = u + z * s,
where
u = mean = 110
z = the critical z score = -1.40507156
s = standard deviation = 20.2
Then
x = critical value = 81.61755448 [ANSWER]
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