Assume that heart rate (in beats per minute, or bpm) before an .... for STA 100
ID: 2929626 • Letter: A
Question
Assume that heart rate (in beats per minute, or bpm) before an .... for STA 100 students is distributed normal, with a mean of 95 bpm and a standard deviation 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 student?
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?
Explanation / Answer
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 95
standard Deviation ( sd )= 18.5
a.
P(X > 110) = (110-95)/18.5
= 15/18.5 = 0.8108
= P ( Z >0.8108) From Standard Normal Table
= 0.2087
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 90) = (90-95)/18.5
= -5/18.5 = -0.2703
= P ( Z <-0.2703) From Standard Normal Table
= 0.3935
P(X < 120) = (120-95)/18.5
= 25/18.5 = 1.3514
= P ( Z <1.3514) From Standard Normal Table
= 0.9117
P(90 < X < 120) = 0.9117-0.3935 = 0.5182
c.
P ( Z < x ) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is -0.6745
P( x-u/s.d < x - 95/18.5 ) = 0.25
That is, ( x - 95/18.5 ) = -0.6745
--> x = -0.6745 * 18.5 + 95 = 82.5219
d.
P ( Z < x ) = 0.75
Value of z to the cumulative probability of 0.75 from normal table is 0.6745
P( x-u/s.d < x - 95/18.5 ) = 0.75
That is, ( x - 95/18.5 ) = 0.6745
--> x = 0.6745 * 18.5 + 95 = 107.4781
e.
P ( Z < x ) = 0.08
Value of z to the cumulative probability of 0.08 from normal table is -1.4051
P( x-u/s.d < x - 95/18.5 ) = 0.08
That is, ( x - 95/18.5 ) = -1.4051
--> x = -1.4051 * 18.5 + 95 = 69.0062