Answer to Question #118077 in Statistics and Probability for Bridgit Almos

Question

Access time to a supercomputer mainframe is normally distributed with an average (mean) of 92 minutes a day with a standard deviation of 11 minutes. Find the probability that on access time is

a).less than 60 minutes.



b).more than 100 minutes





c).between 70 and 90 minutes.



d).between 1 and 2 hours.



e). On 15% of the days, the access time is more than how many minutes?

Answer 1

a) P(x<60)

z=(x-μ)/σ= (60-92)/11=-2.9

using table P(x<60)=0.00187

b) P(x>100)=1-P(x<100)

z=(100-92)/11=0.73

P(x<100)=0.76424

P(x>100)=1-0.76424=0.23576

c)P(70<x<90)=P(x<90)-P(x<70)

P(x<90), z=(90-92)/11=0.18, P(x<90)=0.57142

P(x<70), z=(70-92)/11=-2, P(x<70)=0.02275

P(70<x<90)=0.57142-0.02275=0.54867

d)P(60<x<120)=P(x<120)-P(x<60)

P(x<120), z=(120-92)/11=2.54, P(x<120)=0.99446

P(x<60), z=(60-92)/11=-2.9, P(x<60)=0.00187

P(60<x<120)=0.99446-0.00187=0.99259

e) P(x>X)=0.15

1-P(x<X)=0.15; P(x<X)=0.85

using table z=1.04

(x-92)/11=1.04

x=(11*1.04)+92=103.44