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According to an IRS study, it takes a mean of 400 minutes for taxpayers to prepare,...

According to an IRS study, it takes a mean of 400 minutes for taxpayers to prepare, copy and file a tax return. The standard deviation is 100 minutes. A consumer watchdog study select a random sample of 81 sample. What is the likelihood the sample mean is greater than 425 minutes?

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SOLUTION:

From given data,

According to an IRS study, it takes a mean of 400 minutes for taxpayers to prepare, copy and file a tax return. The standard deviation is 100 minutes. A consumer watchdog study select a random sample of 81 sample.

The distribution of time follows the normal distribution with mean 400 and standard deviation 100.Random sample of 81 sample.

mean = = 400

standard deviation = = 100

sample size = n = 81

The standard error of the mean is,

= / sqrt(n)

= 100 / sqrt(81)

= 11.1111

= = 400

z = - / = - 400 / 11.1111

The standard error of the mean is 11.1111.

What is the likelihood the sample mean is greater than 425 minutes?

The likelihood the sample mean is greater than 425 minutes is,

P( > 425) = P((- ) / ​​​​​​​ > (425- 400) / 11.1111​​​​​​​)

P( > 425) = P(z > 2.25)

P( > 425) = 1 - P(z < 2.25)

P( > 425) = 1 - 0.98778 (FROM STANDARD NORMAL DISTRIBUTION TABLE)

P( > 425) = 0.01222

The likelihood the sample mean is greater than 425 minutes is 0.01222

orchestra answered 1 year ago
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