Respuesta :
Answer:
$425.6 should be budgeted for weekly repairs and maintenance.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean $400 and standard deviation $20.
This means that [tex]\mu = 400, \sigma = 20[/tex]
How much should be budgeted for weekly repairs and maintenance to provide that the probability the budgeted amount will be exceeded in a given week is only 0.1?
This is the 100 - 10 = 90th percentile, which is X when Z has a p-value of 0.9, so X when Z = 1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 400}{20}[/tex]
[tex]X - 400 = 20*1.28[/tex]
[tex]X = 425.6[/tex]
$425.6 should be budgeted for weekly repairs and maintenance.
