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 = <u><em>intraocular pressure in the general population</em></u>
So, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean = 16 mm Hg
= 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( <
) = P(Z < -1.33) = 1 - P(Z
1.33)
= 1 - 0.9082 = <u>0.0918</u> 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( >
) = 0.80
P(Z > ) = 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;
x = 16 - 2.53 = <u>13.47</u> mm Hg
Therefore, 80% of adults in the general population have an intraocular pressure that is greater than 13.47 mm Hg.
