Question

Glaucoma is a disease of the eye that is manifested by high intraocular pressure. The distribution of intraocular pressure in the general population is approximately normal with mean 16 mm Hg and standard deviation 3 mm Hg.
(a) What percentage of people have an intraocular pressure lower than 12 mm Hg?
(b) Fill in the blank. Approximately 80% of adults in the general population have an intraocular pressure that is greater than ________ (how many?) mm Hg.

Answer 1

Answer:

(a) 9.18% of people have an intraocular pressure lower than 12 mm Hg.

(b) 80% of adults in the general population have an intraocular pressure that is greater than 13.47 mm Hg.

Step-by-step explanation:

We are given that the distribution of intraocular pressure in the general population is approximately normal with mean 16 mm Hg and standard deviation 3 mm Hg.

Let X = intraocular pressure in the general population

So, X ~ Normal($\mu=16,\sigma^{2} = 3^{2}$)

The z score probability distribution for normal distribution is given by;

                             Z  =  $\frac{ X-\mu}{\sigma } }$  ~ N(0,1)

where, $\mu$ = population mean = 16 mm Hg

           $\sigma$ = standard deviation = 3 mm Hg

(a) Percentage of people have an intraocular pressure lower than 12 mm Hg is given by = P(X < 12 mm Hg)

       P(X < 12) = P( $\frac{ X-\mu}{\sigma } }$ < $\frac{ 12-16}{3 } }$ ) = P(Z < -1.33) = 1 - P(Z $\leq$ 1.33)

                                                   = 1 - 0.9082 = 0.0918 or 9.18%

The above probability is calculated by looking at the value of x = 1.33 in the z table which has an area of 0.9082.

(b) We have to find that 80% of adults in the general population have an intraocular pressure that is greater than how many mm Hg, that means;

        P(X > x) = 0.80      {where x is the required intraocular pressure}

        P( $\frac{ X-\mu}{\sigma } }$ > $\frac{ x-16}{3 } }$ ) = 0.80

        P(Z > $\frac{ x-16}{3 } }$ ) = 0.80

Now, in the z table the critical value of z which represents the top 80% of the area is given as -0.842, that is;

                           $\frac{ x-16}{3 } } = -0.842$

                           $x -16 = -0.842 \times 3$

                           x = 16 - 2.53 = 13.47 mm Hg

Therefore, 80% of adults in the general population have an intraocular pressure that is greater than 13.47 mm Hg.