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Ahmad Mobin Haqyar Ahmad Mobin Haqyar
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4 years ago
According to an IRS study, it takes a mean of 340 minutes for taxpayers to prepare, copy, and electronically file a 1050 tax form. This distribution of times follows the normal distribution and the standard deviation is 70 minutes. A consumer watch-dog agency selects a random sample of 35 taxpayers.
A. What is the standard error of the mean in this example?
B. What is the likelihood the sample mean is greater than 310 minutes?
C. What is the likelihood the sample mean is between 300 and 400 minutes?
D. What is the likelihood the sample mean is greater than 450 minutes?
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Educator
4 years ago
A. What is the standard error of the mean in this example?

Use: \(SE_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\)

\(SE_{\overline{x}}=\frac{70}{\sqrt{35}}=11.83\)

Quote
B. What is the likelihood the sample mean is greater than 310 minutes?

\(P\left(\overline{x}>320\right)=P\left(\overline{x}-\mu >320-\mu \right)\)

Where: \(\mu =340\)

\(P=\left(z>\frac{320-340}{\frac{70}{\sqrt{35}}}\right)\)

\(P=\left(z>-1.69\right)\)

\(P=\left(z<1.69\right)\)

\(P=0.5+P\left(0<x<1.69\right)\)

Find in chart 0.4545 is for 1.69



** Subtract that number for 1.69 by 0.50 to 0.4545

0.9545
wrote...
Educator
4 years ago
Does this help so far? Let me know if you have any questions before I continue with C and D...
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