Respuesta :
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'.