The ages of commercial aircraft are normally distributed with a mean of 13.5 years and a standard deviation of 7.6 years. What percentage of individual aircraft have ages greate…

Question

asked May 13, 2021 16.5k views

1 Answer

Tilleryj · Answer 1 · 2021-05-18T01:13:43+0000

Answer:

(a) 42.1% of individual aircraft have ages greater than 15 years.

(b) 5.71% of sample means have ages greater than 15 years.

Explanation:

We are given that the ages of commercial aircraft are normally distributed with a mean of 13.5 years and a standard deviation of 7.6 years.

Let X = ages of commercial aircraft

SO, X ~ Normal(
$\mu=13.5 ,\sigma^(2) =7.6^(2)$ )

The z score probability distribution for normal distribution is given by;

Z =
$(X-\mu)/(\sigma)$ ~ N(0,1)

where,
$\mu$ = population mean = 13.5 years

$\sigma$ = standard deviation = 7.6 years

(a) Percentage of individual aircraft having ages greater than 15 years is given by = P(X > 15 years)

P(X > 15 years) = P(
$(X-\mu)/(\sigma)$ >
(15-13.5)/(7.6) ) = P(Z > 0.20) = 1 - P(Z
$\leq$ 0.20)

= 1 - 0.57926 = 0.42074 or 42.1%

The above probability is calculated by looking at the value of x = 0.20 in the z table which has an area of 0.57926.

(b) Assume that a random sample of 64 aircraft is selected and the mean age of the sample is computed.

Now, The z score probability distribution for sample mean is given by;

Z =
$(\bar X-\mu)/((\sigma)/(√(n) ) )$ ~ N(0,1)

where,
$\mu$ = population mean = 13.5 years

$\sigma$ = standard deviation = 7.6 years

n = sample of aircraft = 64

Let
$\bar X$ = sample mean

So, Percentage of individual aircraft that have ages greater than 15 years is given by = P(
$\bar X$ > 15 years)

P(
$\bar X$ > 15 years) = P(
$(\bar X-\mu)/((\sigma)/(√(n) ) )$ >
(15-13.5)/((7.6)/(√(64) ) ) ) = P(Z > 1.58) = 1 - P(Z
$\leq$ 1.58)

= 1 - 0.94295 = 0.05705 or 5.71%

The above probability is calculated by looking at the value of x = 1.58 in the z table which has an area of 0.94295.