Question

Assume that systolic blood pressure (SBP) in a population is normally distributed with a mean of 120 mmHg and a standard deviation of 10 mmHg

- What percent of the population has an SBP > 130 mmHg?

- What percent of the population has an SBP in the range of 110 to 140 mmHg inclusive?

- What percent of the population has an SBP > 150 or < 110 mmHg?

Fill in blanks of the following statements.

Bottom 10% of this population would have SBP of less than _______ mmHg

Top 5 % of this population would have SBP of at least _______ mmHg

Answer 1

Let
a)
Percent of the population that has an SBP > 130 mmHg
P(X > 130) :
Converting to standard normal:

P(Z>1) = 1 - P(Z<1)
P(Z<1) = 0.8413
P(Z>1) = 1-0.8413 = 0.1587

b)
Percent of the population that has an SBP in the range 110 to 140 mmHg
P(110 < X < 140), Converting to standard normal

P(-1 < Z < 2) = P(Z < 2) - P(Z < -1)
P(Z < 2) = 0.9772
P(Z < -1) = 0.1587
P(-1 < Z < 2) = 0.9772 - 0.1587 = 0.8185

c)
Percent of population that has an SBP of > 150 or < 110:
P(X > 150) :

P(Z>3) = 1 - P(Z<3)
P(Z<3) = 0.9987
P(Z>3) = 1-0.9987 = 0.0013
P(X<110):

P(Z < -1) = 0.1587
Percent of population that has an SBP of > 150 or < 110: = P(X > 150) + P(X < 110)= 0.0013 + 0.1587 = 0.16

4)
For bottom 10% z-score:
We look for the value closes to 0.10 in standard normal table i.e. 0.1003
z-score corresponding to 0.1003 = -1.28

x = (-1.28 x 10) + 120 = 107.2
Bottom 10% of this population would have SBP of less than 107.2 mmHg


5)
For top 5% z-score:
We look for the value closes to 1-0.05 = 0.95 in standard normal table
z-score corresponding to 0.95 = 1.645

x = (1.645 x 10) + 120 = 136.45
Top 5% of this population would have SBP of atleast 136.45 mmHg
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