Mathematics, 06.05.2020 03:10 denareena97

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%.

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Answer from: qholmes02

(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( $\mu=13.5 ,\sigma^{2} =7.6^{2}$ )

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

Z = $\frac{X-\mu}{\sigma}$ ~ 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( $\frac{X-\mu}{\sigma}$ > $\frac{15-13.5}{7.6}$ ) = 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 = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ 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( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{15-13.5}{\frac{7.6}{\sqrt{64} } }$ ) = 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.

Answer from: Quest

where is the rest of the

step-by-step explanation:

Answer from: Quest

dont think so

step-by-step explanation:

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The ages of commercial aircraft are normally distributed with a mean of 13.5 years and a standard de...