Respuesta :
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.
Step-by-step 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([tex]\mu=13.5 ,\sigma^{2} =7.6^{2}[/tex])
The z score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 13.5 years
[tex]\sigma[/tex] = 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( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{15-13.5}{7.6}[/tex] ) = P(Z > 0.20) = 1 - P(Z [tex]\leq[/tex] 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 = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 13.5 years
[tex]\sigma[/tex] = standard deviation = 7.6 years
n = sample of aircraft = 64
Let [tex]\bar X[/tex] = sample mean
So, Percentage of individual aircraft that have ages greater than 15 years is given by = P([tex]\bar X[/tex] > 15 years)
P([tex]\bar X[/tex] > 15 years) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{15-13.5}{\frac{7.6}{\sqrt{64} } }[/tex] ) = P(Z > 1.58) = 1 - P(Z [tex]\leq[/tex] 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.
