Question

The ages of commercial aircraft are normally distributed with a mean of 12.0 years and a standard deviation of 7.8895 years. What percentage of individual aircraft have ages between 9 years and 15 ​years? Assume that a random sample of 81 aircraft is selected and the mean age of the sample is computed. What percentage of sample means have ages between 9 years and 15 years?

The percentage of individual aircraft that have ages between 9 years and 15 years is?

Answer 1

Solution :

Given that,

mean = $\mu$ = 12.0

standard deviation = $\sigma$ = 7.8895

a) P( 9 < x < 15 ) = P[(9 - 12.0)/ 7.8895) < (x - $\mu$) /$\sigma$  < (15 - 12.0) / 7.8895) ]

= P(-0.38 < z < 0.38)

= P(z < 0.38) - P(z < -0.38)

Using z table,

= 0.6480 - 0.3520

= 0.2960

percentage = 29.60%

b) n = 81

$\mu$_{$\bar x$} = $\mu$ = 12.0

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 7.8895 / $\sqrt$ 81 = 0.877

P(9 < < 15)  

T

= P[(9 - 12.0) / 0.877 < (
T
- $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (15 - 12.0) / 0.877)]

= P( -3.42 < Z < 3.42 )

= P(Z < 3.42 ) - P(Z < -3.42)

Using z table,  

= 0.9997 - 0.0003

= 0.9994

percentage = 99.94%