Answer to Question #311004 in Statistics and Probability for Eve

Question

Ophthalmology Glaucoma is an eye disease that is manifested by high intraocular pressure (IOP). The distribution of IOP in the general population is approximately normal with mean =16mmHg and standard deviation 3 mmHg. a. If the normal range for IOP is considered to be between 12 and 20 mmHg, then what percentage of the general population would fall within this range? b. If the top 5% of IOP is considered to be in danger range, then what minimum value of danger range? c. What is the 85% percentile of IOP?

Answer 1

solution

Population mean $\mu =16$

Standard deviation $\sigma=3$

(a) Normal range (between 12 and 20)

We calculate Z values for values below 12 and above 20

$Z=\dfrac{X-\mu}{\sigma}$

$Z_{12}=\dfrac{12-16}{3}=-1.333$

$Z_{20}=\dfrac{20-16}{3}=1.333$

From the normal distribution tables

$P(Z_{12})=0.09176$

$P(Z_{20})=0.90824$

Since we are calculating the probability for values above $Z_{20}$ we subtract the value from 1.

$=1-0.90824=0.09176$

Percentage in the normal range $(12 \to 20)$

$[1-(0.09176+0.09176)]\times100\%$

$=81.648\%$

(b)Top $5\%$ of the population

$P=0.95$

From the normal distribution tables

$Z=1.65$

$Z=\dfrac{X-\mu}{\sigma}$

$X=Z\sigma+\mu$

$X=1.65\times3+16$

$X=20.95$

(c) 85^th percentile of the population

$P=0.85$

From normal distribution tables

$Z=1.04$

$Z=\dfrac{X-\mu}{\sigma}$

$X=Z\sigma+\mu$

$X=1.04\times3+16$

$X=19.12$