Question

1. Suppose the population of commute times for junior managers of corporations in a large city is normal with mean 87 minutes and standard deviation 22 minutes.
   1. What is the chance a junior manager has commute time of more than one hour?

2. What is the chance a junior manager has commute time of less than half an hour?

3. What is the chance a junior manager has commute time between 1 and 2 hours?

4. What is the 90^thpercentile of commute times?

5. What is the 25^thpercentile of commute times?

Answer 1

Solution :

Given that ,

mean = = 87 minutes

standard deviation = = 22 minutes

a) 1 hour = 60 minutes

P(x > 60) = 1 - P(x < 60 )

= 1 - P[(x - ) / < (60-87) /22 ]

= 1 - P(z <-1.23 )

=1 - 0.1093 =0.8907

Probability = 0.8907

b ) half an hour = 30 minutes

P(x < 30 ) = P[(x - ) / < (30-87) /22 ]

= P(z < -2.59 )

= 0.0048

Probability = 0.0048

c ) 1 and 2 hours = 60 and 120 minutes

P(60 < x <120 ) = P[(60-87)/22 ) < (x - ) /  < (120-87) / 22) ]

= P(-1.23 < z <1.5 )

= P(z < 1.5 ) - P(z < -1.23 )

= 0.9332 - 0.1093 =0.8239

Probability = 0.8239

d ) 90% is

P(Z < z ) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x =1.28 * 22+87

x = 115.16

e) 25 % is

P(Z < z ) = 0.25

z = - 0.6745

Using z-score formula,

x = z * +

x = (-0.6745) * 22+87

x = 72.161