The ages of commercial aircraft are normally distributed with a mean of 13.0 years and a standard deviation of 7.5 years what percentage of individual aircraft have ages greater than 15 years assume that a random sample of 49 aircraft is selected and the mean age of the sample is computed. What percentage of sample means have ages greater than 15 years
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To find the percentage of individual aircraft that have ages greater than 15 years, we can calculate the z-score and use the standard normal distribution table.
First, we calculate the z-score using the formula:
z = (X - μ) / σ
where X is the value (15 years), μ is the mean (13.0 years), and σ is the standard deviation (7.5 years).
z = (15 - 13.0) / 7.5
z = 0.2667
Next, we look up the z-score in the standard normal distribution table. The table provides the percentage of values below a given z-score. To find the percentage of values above the z-score, we subtract the table value from 1.
Using the table, we find that the percentage of values below 0.2667 is approximately 0.6049.
Therefore, the percentage of individual aircraft with ages greater than 15 years is approximately (1 - 0.6049) = 0.3951, or 39.51%.
Now let's move on to the percentage of sample means that have ages greater than 15 years.
The distribution of sample means is also normally distributed. The mean of the sample means is equal to the population mean, which is 13.0 years. The standard deviation of the sample means, denoted as σ/√n, where σ is the population standard deviation and n is the sample size.
In this case, the sample size is 49. Therefore, the standard deviation of the sample means is 7.5 / √49 = 7.5 / 7 = 1.07 years.
We repeat the previous steps for the z-score calculation using the X value of 15 years, the new mean of 13.0 years, and the new standard deviation of 1.07 years:
z = (15 - 13.0) / 1.07
z = 1.87
Next, we look up the z-score of 1.87 in the standard normal distribution table. The table value is approximately 0.9699, which represents the percentage of sample means below the given z-score.
To find the percentage of sample means above the z-score, we subtract the table value from 1:
1 - 0.9699 = 0.0301
Therefore, the percentage of sample means with ages greater than 15 years is approximately 0.0301, or 3.01%.
To find the percentage of individual aircraft with ages greater than 15 years, we need to calculate the probability using the normal distribution.
First, let's calculate the z-score for the age of 15 years using the formula: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
z = (15 - 13) / 7.5
z = 2 / 7.5
z ≈ 0.267
Next, we'll find the corresponding percentage using a standard normal distribution table or a calculator. The z-score of 0.267 corresponds to a percentage of 0.3933 or 39.33%.
So, approximately 39.33% of individual aircraft have ages greater than 15 years.
To find the percentage of sample means having ages greater than 15 years, we need to use the sampling distribution of the sample means. According to the Central Limit Theorem, as the sample size increases, the sampling distribution of the sample means approaches a normal distribution.
Since the population follows a normal distribution, the sample means will also follow a normal distribution. The mean of the sample means will be equal to the population mean (13.0 years), and the standard deviation of the sample means (also known as the standard error) will be equal to the population standard deviation divided by the square root of the sample size.
Standard error (σx̄) = σ / √n
where σ is the population standard deviation and n is the sample size.
In this case, the sample size is 49, so the standard error is:
σx̄ = 7.5 / √49
σx̄ = 7.5 / 7
σx̄ ≈ 1.071
Now, we can use the z-score formula to find the percentage of sample means with ages greater than 15 years, using the z-score calculated earlier (z ≈ 0.267) and the new standard error (σx̄ ≈ 1.071).
z = (x - μ) / σx̄
z = (15 - 13) / 1.071
z ≈ 1.87
Using the standard normal distribution table or a calculator, the z-score of 1.87 corresponds to a percentage of 96.75%.
Therefore, approximately 96.75% of sample means have ages greater than 15 years.
For the individual aircraft,
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score. Multiply by 100.
For the mean,
Z = (score-mean)/SEm
SEm = SD/√n
Use the same table.