The ages of commercial aircraft are normally distributed with a mean of 13.5 years and a standard deviation of 7.3 years. What percentage of individual aircraft have ages greater than 15 years?
15 is 1.5 above the mean
... this is 1.5/7.3 standard deviations ... .2055
use a z-score table to determine the fraction of the population this represents
15 yr or less ... 58%
Question was 15 years or more.. so the answer is 41.9% or 42%
To find the percentage of individual aircraft that have ages greater than 15 years, we need to use the properties of the normal distribution.
Step 1: Standardize the value 15 using the formula:
z = (x - μ) / σ
where
x = 15 (the value we want to standardize),
μ = 13.5 (the mean of the distribution), and
σ = 7.3 (the standard deviation of the distribution).
Plugging in the values, we have:
z = (15 - 13.5) / 7.3 = 1.5 / 7.3 = 0.2055 (rounded to four decimal places).
Step 2: Look up the area to the right of the standardized value 0.2055 in the standard normal distribution table. This represents the percentage of individual aircraft with ages greater than 15 years.
Using the standard normal distribution table, we find that the area to the left of 0.2055 is 0.5811.
Step 3: Subtract the obtained area from 1 to get the area to the right of the standardized value:
1 - 0.5811 = 0.4189 (rounded to four decimal places).
Step 4: Convert the obtained area to a percentage by multiplying by 100:
0.4189 * 100 = 41.89%.
So, approximately 41.89% of individual aircraft have ages greater than 15 years.
To find the percentage of individual aircraft that have ages greater than 15 years, we can use the properties of the normal distribution.
Step 1: Convert the problem into the standard normal distribution.
The standard normal distribution has a mean of 0 and a standard deviation of 1. To convert a value from the original normal distribution to the standard normal distribution, we use the formula:
z = (x - μ) / σ
where:
- z is the z-score in the standard normal distribution,
- x is the value from the original normal distribution,
- μ is the mean of the original normal distribution, and
- σ is the standard deviation of the original normal distribution.
In our case, we have:
- x = 15 (years)
- μ = 13.5 (mean of the original normal distribution)
- σ = 7.3 (standard deviation of the original normal distribution)
Substituting these values into the formula, we get:
z = (15 - 13.5) / 7.3
z ≈ 0.2055
So, 15 years in the original normal distribution corresponds to approximately 0.2055 in the standard normal distribution.
Step 2: Find the percentage using the standard normal distribution table.
Using the standard normal distribution table (also known as the z-table), we can find the percentage of values greater than the z-score we calculated earlier.
In the standard normal distribution table, find the closest value to 0.2055 (in the positive side of the table) and look up the corresponding percentage. Let's assume we find that the percentage is P.
Step 3: Calculate the final answer.
The percentage of individual aircraft with ages greater than 15 years is approximately P%.
Note: You can also use statistical software or calculators with built-in functions to directly find the percentage without looking up the z-table.