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The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $400 and standard deviation $20. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05

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raphealnwobi

Answer:

$432.9

Step-by-step explanation:

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given as:

[tex]z=\frac{x-\mu}{\sigma}\\\\where\ x=raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]

Given that μ = $400, σ = $20

P(z > z*) = 0.05

P(z < z*) = 1 - 0.05 = 0.95

z* = 1.645

[tex]z=\frac{x-\mu}{\sigma} \\\\1.645=\frac{x-400}{20}\\\\x-400=32.9\\\\x=432.9[/tex]

Therefore $432.9 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05.

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