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

$Z=1.04$

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

$X=Z\sigma+\mu$

$X=1.04\times3+16$

$X=19.12$