Assume that heart rate (in beats per minute, or bpm) before an exam for STA 100
ID: 2929718 • Letter: A
Question
Assume that heart rate (in beats per minute, or bpm) before an exam for STA 100 students is distributed normal, with a mean of 95 bpm and a standard devation of 18.5 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 110.
b) What percentage of randomly selected students could we expect to have a heart rate between 90 and 120?
c) What is the first quartile of heartrates for randomly selected students?
d) What is the third quartile of heartrates for randomly selected students?
e) What is the 8th percentile for heart rates among randomly selected students?
f) If we know a randomly selected students heart rate is over 100 (it is given), what is the probability that it is under 125?
If you could show your work so I can work to understand how these problems are done that would be great, thanks!
Explanation / Answer
Solution:
a) We first get the z score for the critical value. As z = (x - )/, then as
x = critical value = 110
= mean = 95
= standard deviation = 18.5
Thus,
z = (x - ) / = (110-95)/18.5 = 0.8108
Thus, using a table/technology, the right tailed area of this is
P(z > 0.8108) = 0.209 [answer]
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b) We first get the z score for the two values. As z = (x - ) / , then as
x1 = lower bound = 90
x2 = upper bound = 120
= mean = 105
= standard deviation = 21.21
Thus, the two z scores are
z1 = lower z score = (x1 - )/ = (90-105)/21.21 = -0.7072
z2 = upper z score = (x2 - ) / = (120-105)/21.21 = 0.7072
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.2389
P(z < z2) = 0.7611
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.5222= 52.22%
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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 = -0.67449
As x = + z * ,
where
= mean = 95
z = the critical z score = -0.67449
= standard deviation = 18.5
Then
x = critical value = 82.521935 [ANSWER]
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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 = + z * ,
where
= mean = 95
z = the critical z score = 0.67449
= standard deviation = 18.5
Then
x = critical value = 107.478065 [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 = -1.40507
As x = + z * ,
where
= mean = 95
z = the critical z score = -1.40507
= standard deviation = 18.5
Then
x = critical value = 69.006205
e) For 8th percentile z = -1.4051
hence corresponding heartrate = + z * = 69.01
f) P(X<125|X>100) =P(100<X<125)/P(X>100)
= P(0.2703<Z<1.6216)/(1-P(Z<0.2703))
=(0.9476-0.6065)/(1-0.6065) = 0.562