Question

Let us assume that it is known that the systolic blood pressure (SBP) for a particular population of individuals is approximately normally distributed with a mean of 120 mmHg with a known standard deviation of 25 mmHg. We will assume the previous to be true. Use this information to answer the following Part (a) (WebWorkiR ) An individual with a SBP of 140 or more is classified as hypertensive. What percentage of individuals in this population are hypertensive? Please submit your answer as a proportion rounded to 3 decimal places Part (b) (WebWorkiR ) A SBP of 160 or more is classified as Stage II Hypertension, while a SBP of 140-159.9 is classified as Stage I Hypertension. Of all individuals who are classified as hypertensive (Stage I or I), what percentage are Stage ll Hypertensive? Please report your answer as a proportion rounded to 3 decimal places Part (c) (WebWorkiR) What is the median SBP for this population? Part (d) (WebWorkiR) What is the 90h percentle of SBP for this population. Please report your answer rounded to one decimal place. Part (e) ( WebWorkiR) Suppose an independent random sample of 43 individuals is chosen from this population. What is the probability that the sample mean SBP for these 43 individuals is less than or equal to 117mmHg? Please report your answer as a proportion rounded to 3 decimal places.

Answer 1

Given:

SBP follows normal distribution with mean 120 and standard deviation 25.

$Mean,\;\mu=120$

$Standard\;Deviation,\;\sigma=25$

$Z=\frac{x-\mu}{\sigma}$

a) Proportion of hypertensive:

Hypertensive if SBP is more than or equal to 140.

Z-score for 140 is given as

$Z=\frac{140-120}{25}=\frac{20}{25}=0.80$

P(x > 140) = P(Z > 0.80) = 1 - P(Z < 0.80)

From standard normal table or using Excel, the P(Z < 0.80) is calculated as 0.788 (function used in Excel to get the probability value is NORMSDIST and input for the function is the z-value which is 0.80. So, type =NORMSDIST(0.80) in any of the cells in Excel to get the probability value of 0.788)

Therefore,

P(x > 140) = P(Z > 0.80) = 1 - P(Z < 0.80) = 1 - 0.788 = 0.212

b) Proportion of Stage II hypertensive in total hypertensive subjects:

Stage II hypertensive is defined as SBP greater than or equal to 160. Therefore, in general the probability of SBP greater than or equal to 160 is given as

Z-score for 160 is given as

$Z=\frac{160-120}{25}=\frac{40}{25}=1.6$

P(x > 160) = P(Z > 1.60) = 1 - P(Z < 1.60)

From standard normal table or using Excel, the P(Z < 1.60) is calculated as 0.945 (function used in Excel to get the probability value is NORMSDIST and input for the function is the z-value which is 1.60. So, type =NORMSDIST(1.60) in any of the cells in Excel to get the probability value of 0.945)

Therefore,

P(x > 160) = P(Z > 1.60) = 1 - P(Z < 1.60) = 1 - 0.945 = 0.055

The overall stage II hypertensive percentage is given as 0.055*100 = 5.50%

Therefore, the percentage of stage II hypertensive in overall hypertensive subjects is given as 0.055/0.212 = 0.259

c) Median

Since, SBP follows normal distribution the median will be same as mean which is 120.

d) 90th Percentile

P(X < x) = 0.90

From the standard normal table or Excel we can calculate the z-score for 0.90 which is 1.2816. (The function used in Excel to calculate the z-score is NORMSINV and the input for this is the probability value of 0.90. Therefore in excel in any of the cell type =NORMSINV(0.9) which will result in 1.2816)

Now we have z-score for a value x which is 1.2816. Therefore, X-value that corresponds to 1.2816 is calculated as below

$Z=\frac{x-120}{25}=1.2816$

$x-120=1.2816\times 25$

$x-120=32.04$

$x=32.04&plus;120=152.04\sim 152.0$$x=32.04&plus;120=152.04$

90th percentile of SBP for this population is given as 152.0

e) Mean less than or equal to 117

Now the probability has to be calculated from the sample, z is given as below

$Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

n = 43

$\bar{x}$ is the sample mean

Therefore, Z is given as below

$Z=\frac{117-120}{\frac{25}{\sqrt{43}}}$

$Z=\frac{-3}{\frac{25}{6.5574}}=\frac{-3}{3.8125}=-0.7869$

$P(\bar{x}\le 117)=P(Z\le -0.7869)$

From standard normal table or using Excel, the P(Z < -0.7869) is calculated as 0.216 (function used in Excel to get the probability value is NORMSDIST and input for the function is the z-value which is -0.7869. So, type =NORMSDIST(-0.7869) in any of the cells in Excel to get the probability value of 0.216)

Therefore,

$P(\bar{x}\le 117)=P(Z\le -0.7869)=0.216$

Final Answers:

a) 0.212

b) 0.259

c) 120

d) 152.0

e) 0.216