Question

using historical data plan a planner is sequencing surgeries. Average time for one procedure is 120min with sd of 20min.

a. what % will be more than 120min

b. what % will be less than 150min to complete

c. what % of procedures will be between 120min to 160min to complete

Answer 1

Solution :

Given that ,

mean = = 120

standard deviation = = 20

a.

P(x > 120) = 1 - P(x < 120)

= 1 - P((x - ) / < (120 - 120) / 20)

= 1 - P(z < 0)

= 1 - 0.5

= 0.5

percent = 0.005%

b.

P(x < 150) = P((x - ) / < (150 - 120) / 20)

= P(z < 1.5)

= 0.9332

percent = 93.32%

c.

P(120 < x < 160) = P((120 - 120)/ 20) < (x - ) / < (160 - 120) / 20) )

= P(0 < z < 2)

= P(z < 2) - P(z < 0)

= 0.9772 - 0.5

= 0.4772

percent = 47.72%