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suppose that the time required to complete a 1040R tax form is normally distributed with a mean of 90 minutes and a standard deviation of 10 minutes. What proportion of 1040R tax forms will be completed in more than 77 minutes?

Sagot :

Finding the proportion is the same as finding the following probability:

[tex]P(X\ge77)[/tex]

in a normal distribution with mean 90 and standard deviation 10. To find this probability we use the z-score, defined by:

[tex]Z=\frac{x-\mu}{\sigma}[/tex]

where x is the value we need (in this case x=77), hence the probability we are looking for can be calculated as:

[tex]P(X\ge77)=P(Z\ge\frac{77-90}{10})=P(Z\ge-1.3)[/tex]

Using the tables of a normal distribution we have that:

[tex]P(Z\ge-1.3)=0.9032[/tex]

Hence:

[tex]P(X\ge77)=0.9032[/tex]

Therefore, 90.32% of tax forms will be completed in more than 77 minutes.

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