Answer :
The proportion of Boeing 787-8 Dreamliners that are between 185 feet and 187 feet is approximately 78.74%.
Step 1
To find the proportion of Boeing 787-8 Dreamliners that are between 185 feet and 187 feet, we can use the properties of the normal distribution. Specifically, we can calculate the z-scores for 185 feet and 187 feet, and then use these z-scores to find the corresponding probabilities.
The z-score formula is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\( X \)[/tex] is the value, [tex]\( \mu \)[/tex] is the mean, and [tex]\( \sigma \)[/tex] is the standard deviation.
Given:
- Mean [tex](\( \mu \))[/tex]= 185.4 feet
- Standard deviation [tex](\( \sigma \))[/tex] = 0.5 feet
First, we calculate the z-score for 185 feet:
[tex]\[ z_{185} = \frac{185 - 185.4}{0.5} = \frac{-0.4}{0.5} = -0.8 \][/tex]
Next, we calculate the z-score for 187 feet:
[tex]\[ z_{187} = \frac{187 - 185.4}{0.5} = \frac{1.6}{0.5} = 3.2 \][/tex]
Step 2
Now, we need to find the probabilities corresponding to these z-scores using the standard normal distribution table (or a calculator):
- The probability for [tex]\( z = -0.8 \)[/tex] is approximately 0.2119.
- The probability for [tex]\( z = 3.2 \)[/tex] is approximately 0.9993.
To find the proportion of Dreamliners between 185 feet and 187 feet, we subtract the smaller probability from the larger probability:
[tex]\[ P(185 \leq X \leq 187) = P(z_{187}) - P(z_{185}) = 0.9993 - 0.2119 = 0.7874 \][/tex]
Therefore, the proportion of Boeing 787-8 Dreamliners that are between 185 feet and 187 feet is approximately 0.7874, or 78.74%.
Answer:
78.74% of the 787-8 airplanes are between 185' and 187'.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 185.4, \sigma = 0.5[/tex]
What proportion of the 787-8 airplanes are between 185' and 187'?
This is the pvalue of Z when X = 187 subtracted by the pvalue of Z when X = 185. So
X = 187
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{187 - 185.4}{0.5}[/tex]
[tex]Z = 3.2[/tex]
[tex]Z = 3.2[/tex] has a pvalue of 0.9993.
X = 185
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{185 - 185.4}{0.5}[/tex]
[tex]Z = -0.8[/tex]
[tex]Z = -0.8[/tex] has a pvalue of 0.2119.
0.9993 - 0.2119 = 0.7874
78.74% of the 787-8 airplanes are between 185' and 187'.
