Answer:
P (X \leq 11) = 0.0478
) for z = 80% , X = 13.88 mm Hg
Step-by-step explanation:
given data:
mean = 16 mm Hg
standard deviation 3 mm Hg
let x be the percentage of individual which have intraocular pressure
a) P (X \leq 11)
[tex]P(\frac{x -\mu}{\sigma} < \frac{11 - 16}{3})[/tex]
[tex]P(z \leq - 1.66) = 0.0478[/tex]
b) for z = 80%
P(Z>z ) = 0.8
[tex]P(\frac{x -\mu}{\sigma} < \frac{x - 16}{3}) = 0.80[/tex]
[tex]z = \frac{x - 16}{3})[/tex]
[tex]0.80 = \frac{x - 16}{3})[/tex]
x = 13.8 mm Hg