Question 926925: According to an IRS study, it takes a mean of 330 minutes for taxpayers to prepare, copy, and electronically file a 1040 tax form. This distribution of times follows the normal distribution and the standard deviation is 80 minutes. A consumer watchdog agency selects a random sample of 40 taxpayers.
(a) What is the standard error of the mean in this example? (Round your answer to 3 decimal places.)
Standard error of the mean
(b)
What is the likelihood the sample mean is greater than 320 minutes? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability
(c)
What is the likelihood the sample mean is between 320 and 350 minutes? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability
(d)
What is the likelihood the sample mean is greater than 350 minutes? (Round z value to 2 decimal places and final answer to 4 decimal places.)
Probability
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! mean = 320mi, standard deviation is 80
sample: n = 40
a) SE = 80/sqrgt(40) = 6.325 rounded
z = (x - 320)/6.325
b) P(x > 320) = P(z > 0) = .50 0r 50%
c) P(320< x < 350) = P( 0 < z < 30/6.325) = P(z < .4216) - P(z < 0) = .6633-.5 = .1633
0r Using a TI calculator 0r similarly a Casio fx-115 ES plus
P( 0 < z < 30/6.325) = normalcdf(0,.4216)= .1633 0r 16.33%
d )p(x > 350) = P(z > .4216) = normalcdf(.4216, 100)= .3367 0r 33.67%
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