Question

The ages of commercial aircraft are normally distributed with a mean of 13.013.0 years and a standard deviation of 8.11428.1142 years. What percentage of individual aircraft have ages between 1010 years and 1616 ​years? Assume that a random sample of 8181 aircraft is selected and the mean age of the sample is computed. What percentage of sample means have ages between 1010 years and 1616 ​years?

Answer 1

Let , X be the age of commercial aircraft

X follows normal distribution with mean = 13 years and standard deviation = 8.1142 years.

a)

We have to find percentage of individual aircraft have ages ( x ) between 10 years and 16 ​years

i.e we have to find P( 10 < x < 16 )

P( 10 < x < 16 ) = P( x < 16 ) - P( x < 10 )

Using Excel function , =NORMDIST( x ,, , 1 )

P( x < 16 ) = NORMDIST( 16 , 13 , 8.1142 , 1 ) = 0.644205

P( x < 10 ) = NORMDIST( 10 , 13 , 8.1142 , 1 ) = 0.355795

So, P( 10 < x < 16 ) = 0.644205 - 0.355795 = 0.2884 = 28.84%

The percentage of individual aircraft have ages between 10 years and 16 ​years is 28.84%

b)

We have to find percentage of sample means ( ) have ages between 10 years and 16 ​years.

i.e we have to find P( 10 < < 16 )

{ suppose sample of size ' n ' drawn from normal distribution with mean and standard deviation = then

distribution of sample mean () is approximately normal with mean and standard deviation  

}

For this example , n = 81 , = 13 , = 8.1142

Mean of = = 13

Standard deviation of =

P( 10 < < 16 ) = P( < 16 ) - P( < 10 )

Using Excel function , =NORMDIST( x ,, , 1 )

P( < 16 ) = NORMDIST( 16 , 13 ,0.90158 , 1 ) = 0.999562

P( < 10 ) = NORMDIST( 10 , 13 , 0.90158, 1 ) = 0.000438

So, P( 10 < < 16 ) = 0.999562 - 0.000438 = 0.9991 = 99.91%

The percentage of sample means have ages between 10 years and 16​ years 99.91%