SOLUTION: Assuming the normal heart rate (H.R) in normal healthy individuals is normally distributed with Mean = 70 and Standard Deviation =10 beats/min. a) What is the probability that th

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Question 1190385: Assuming the normal heart rate (H.R) in normal healthy individuals is normally distributed with
Mean = 70 and Standard Deviation =10 beats/min.
a) What is the probability that the heart rate is above 80 beats/min?
b) What is the probability that the heart rate is above 90 beats/min?
c) What is the probability that the heart rate is between 50-90 beats/min?
d) What is the probability that the heart rate is above 100 beats/min?
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!

Part (a)

mu = 70 = population mean
sigma = 10 = population standard deviation

Convert x = 80 to its corresponding z score
z = (x-mu)/sigma
z = (80-70)/10
z = 10/10
z = 1

The question of asking P(X > 80) is identical to P(Z > 1) when we have the parameters of mu = 70 and sigma = 10.

Use a Z table in the back of your textbook to find that
P(Z < 1) = 0.84134

Here's a free Z table if you don't have your stats textbook with you
https://www.ztable.net/

From that we could say:
P(Z > 1) = 1 - P(Z < 1)
P(Z > 1) = 1 - 0.84134
P(Z > 1) = 0.15866

Answer: Approximately 0.15866

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Part (b)

You'll follow the same steps as the previous part.
x = 90 converts to z = 2. I'm skipping steps a bit.
P(X > 90) is equivalent to P(Z > 2)

Use the table to find that
P(Z < 2) = 0.97725
So,
P(Z > 2) = 1 - P(Z < 2)
P(Z > 2) = 1 - 0.97725
P(Z > 2) = 0.02275

Answer: Approximately 0.02275

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Part (c)

x = 50 converts to z = -2
x = 90 converts to z = 2

Computing P(50 < X < 90) is equivalent to P(-2 < Z < 2)

The table shows that
P(Z < -2) = 0.02275
P(Z < 2) = 0.97725

This leads to:
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-2 < Z < 2) = P(Z < 2) - P(Z < -2)
P(-2 < Z < 2) = 0.97725 - 0.02275
P(-2 < Z < 2) = 0.9545
This fits with the Empirical Rule which says roughly 95% of the normal distribution is within 2 standard deviations of the mean.

Answer: Approximately 0.9545

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Part (d)

x = 100 converts to z = 3
P(Z < 3) = 0.99865
P(Z > 3) = 1 - P(Z < 3)
P(Z > 3) = 1 - 0.99865
P(Z > 3) = 0.00135

Answer: Approximately 0.00135

You can use a normalCDF calculator to compute the more accurate versions of each approximate answer above.

Answer by ikleyn(52847)   (Show Source):
You can put this solution on YOUR website!
.

Go to online (free of charge) normal distribution probability calculator
https://onlinestatbook.com/2/calculators/normal_dist.html

Input the given parameters of each part into the appropriate window of the calculator and get the answers
to your questions.

The calculator has perfect description and design, so EVERY person, even beginner, may work with it on his or her own,
even having minimum knowledge on the subject.

Happy calculations ( ! )

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