Question

Problem 1. Hospital records show that a certain surgical procedure takes on an average of 120 minutes with
a standard deviation of 3 minutes. Between how many minutes must be the lengths of at least
82% of these surgical procedures?

Answer 1

Given that,

mean = $\mu$ = 120

standard deviation = $\sigma$ =3

Using standard normal table,

P(Z > z) = 82%

= 1 - P(Z < z) = 0.82

= P(Z < z ) = 1 - 0.82

= P(Z < z ) = 0.18

z = -0.92 (using standard normal (Z) table )

Using z-score formula  

x = z * $\sigma$ + $\mu$

x= -0.92*3 + 120

x= 117.24