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Using the normal distribution, it is found that the operating cost for the lowest 3% of the airplanes is of at most $1,630.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
[tex]\mu = 2100, \sigma = 250[/tex].
The operating cost for the lowest 3% of the airplanes is of at most the 3rd percentile, hence at most X when Z = -1.88.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.88 = \frac{X - 2100}{250}[/tex]
X - 2100 = -1.88(250)
X = 1630.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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