Question

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 greater than 15 ​years? Assume that a random sample of 64 aircraft is selected and the mean age of the sample is computed. What percentage of sample means have ages greater than 15 ​years? The percentage of individual aircraft that have ages greater than 15 years is nothing​%.

Answer 1

(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.