Using the normal distribution, it is found that:
- The percent of area associated with $54.20 is of: 27.43%.
- The percent of area associated with $86.60 is of: 88.49%.
- Adding the two together, the percent of his expenses between $54.20 and $86.60 is of: 61.06%.
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 = 65, \sigma = 18[/tex]
We find the percentage for each measure(the p-value of Z when X = measure), then subtract them.
X = 86.6:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{86.6 - 65}{18}[/tex]
Z = 1.2
Z = 1.2 has a p-value of 0.8849 = 88.49%.
X = 54.2:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{54.2 - 65}{18}[/tex]
Z = -0.6
Z = -0.6 has a p-value of 0.2743 = 27.43%.
88.49 - 27.43 = 61.06%.
More can be learned about the normal distribution at https://brainly.com/question/24537145
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