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 to provide that the probability the budgeted amount will be exceeded in a given week is only 0.1?

Answers

Answer 1

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.


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state wether the product of √3 and √9 is rational or irrational. explain your answer​

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Step-by-step explanation:

Please help me!! Perimeter: about ______ ft squared
Area: about ________ ft squared

Answers

Answer:

55.1 ft ; 236.7 ft^2

Step-by-step explanation:

radius = 7.5 ft

Perimeter of the semicircles = 2 * radius * pi = 2 * 7.5 * pi = 47,12389 ft

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The sum of the first eight terms in a Geometric Series is 19680 and the sum of the first four terms is 240. A) Find the first term. B) Find the common ratio. C) Justify your answers by showing steps that demonstrates your answers generate S8=19680 and S4=240.

Answers

Answer:

First Term = 6

Common Ratio = 3

Step-by-step explanation:

According to the Question,

Given, The sum of the first eight terms in a Geometric Series is 19680 and the sum of the first four terms is 240 .

Thus, [tex]S_{8} = 19680[/tex] & [tex]S_{4} = 240[/tex] .

The Sum of n-term of Geometric Mean is [tex]S_{n} = \frac{a(r^{n-1)} }{r-1}[/tex] Where, r>1 , a=First term of G.P & r=common Ratio .

Now, on solving  [tex]\frac{S_{8} }{S_{4} }[/tex]  we get,

[tex]\frac{19680}{240} = \frac{\frac{a(r^{8-1)} }{r-1}}{\frac{a(r^{4-1)} }{r-1}}[/tex]  

[tex]82 = \frac{r^{8}-1 }{r^{4}-1 }[/tex]

[tex]82r^{4}-82 = r^{8}-1\\r^{8}-82r^{4}+81 = 0\\r^{8}-81r^{4}-r^{4}+81 = 0\\(r^{4}-81)( r^{4}-1) =0[/tex](r=1 is not possible so neglect [tex]( r^{4}-1) =0[/tex] )

So, r=3 Now Put this value in [tex]S_{4} = {\frac{a(r^{4-1)} }{r-1}}[/tex] We get a=6 .

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