Question

0. 6.8 FlyHigh Airlines determined that the distance

   traveled per aircraft per year is normally distributed,

with a mean of 60 thousand miles and a standard deviation of 10 thousand miles.

1. What proportion of aircrafts can be expected to travel between

   44 and 54 thousand miles in a year?

2. What percentage of aircrafts can be expected to travel either

   less than 25 or more than 70 thousand miles in a year?

3. How many miles will be traveled by at least 70 percent of the

   aircrafts?

4. What are your answers to (a) through (c) if the standard devia-

   tion is 12 thousand miles?

Answer 1

Answer:

Given,

Mean = 60

Standard deviation = 10

a)

P(44 < X < 54) = P((44 - 60)/10 < (x-u)/s < (54 - 60)/10)

= P(-1.6 < z < -0.6)

= P(z < -0.6) - P(z < -1.6)

= 0.2742531 - 0.0547993 [since from z table]

= 0.2195

b)

P(X < 25) + P(X > 70) = 1 - P(25 < X < 70)

= 1 - [P(25 < X < 70)]

= 1 - P((25 - 60)/10 < z < (70 - 60)/10)

= 1 - [P(-3.5 < z < 1)]

= 1 - [P(z < 1) - P(z < -3.5)]

= 1 - [0.8413447 - 0.0002326] [since from z table]

= 1 - 0.8411

= 0.1589

c)

Here for top 70%, critical z value is z = -0.52

Consider,

x = mean + z*sd

substitute values

= 60 - 0.52*10

= 60 - 5.2

= 54.8