SOLUTION: The New York Times reported (Laurie J. Flynn, �Tax Surfing,� The New York Times, March 25, 2002, C10) that the mean time to download the homepage from the internal revenue service

Question 203268: The New York Times reported (Laurie J. Flynn, �Tax Surfing,� The New York Times, March 25, 2002, C10) that the mean time to download the homepage from the internal revenue service web site www.irs.gov was 0.8 second. Suppose that the download time was normally distributed with a standard deviation of 0.2 second. If you selected at random sample of 30 download times,
a) What is the probability that the sample mean is less than 0.75 second?
b) What is the probability that the sample mean is between than 0.70 and 0.90 second?
c) The probability is 80% that the sample mean is between what two values symmetrically distributed around the population mean?
d) The probability is 80% that the sample mean is less than what value?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The New York Times, March 25, 2002, C10) that the mean time to download the homepage from the internal revenue service web site www.irs.gov was 0.8 second. Suppose that the download time was normally distributed with a standard deviation of 0.2 second. If you selected at random sample of 30 download times,
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a) What is the probability that the sample mean is less than 0.75 second?
Find the z-value of 0.75.
z(0.75) = (0.75-0.8)/0.2 = -0.05/0.2 = -0.05/0.2 = -.25
P(x<0.75) = P(z<-0.25) = normalcdf(-100,-0.25) = 0.4013
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b) What is the probability that the sample mean is between than 0.70 and 0.90 second?
Find the z-scores of 0.7 and of 0.9
P(0.7 ------------------------------------------
c) The probability is 80% that the sample mean is between what two values symmetrically distributed around the population mean?
Find the z-values allociated with 10%ile and 90%ile
z for 10%ile is InvNorm(0.1) = -1.2816
z for 90%ile is InvNorm(0.9) = 1.2816
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Find the corresponding x-values using x = zs + u:
x(-1.2816) = -1.2816*0.2 + 0.8 = 0.5437
x(1.2816) = 1.05632
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d) The probability is 80% that the sample mean is less than what value?
Draw a normal curve and mark the mean position.
Put a point at the 80%ile position.
Find the z-value of that 80%ile position = InvNorm 0.8416
Find the corresponding x-value: x = 0.8416*0.2 + 0.8 = 0.9683
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Cheers,
Stan H.

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