Answer to Question #244348 in Statistics and Probability for boo

Question

The Prior to medical visits, patients are asked to complete a self-administered medical history
form. The time it takes each patient to complete the form is approximately normally distributed,
with a mean of 12.4 minutes and a standard deviation of 2.1 minutes.
a) What proportion of patients completes the medical history form in 10 minutes or less?
b) What proportion of patients takes 20 minutes or more to complete the medical history form?
c) What is the 90th percentile of the time to complete the medical history form?

Answer 1

$\mu=12.4 \\ \sigma= 2.1$

a)

$P(X<10) = P(Z< \frac{10-12.4}{2.1}) \\ = P(Z< -1.14) \\ = 0.1271$

b)

$P(X>20) = 1 -P(X<20) \\ = 1 -P(Z< \frac{20-12.4}{2.1}) \\ = 1 -P(Z< 3.62) \\ = 1 -0.9999 \\ = 0.0001$

c)

$P(X<x) =0.9 \\ P(Z< \frac{x-12.4}{2.1}) = 0.9 \\ \frac{x-12.4}{2.1} = 1.282 \\ x = 12.4 + 2.1 \times 1.282 \\ x = 15.09$