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hellokym · Dec 12, 2007, 02:54 PM

The mean time to download the home page of a company was 0.8 second. Suppose the
download time was normally distributed, with a standard deviation of 0.2 second. If you select a
random sample of 30 download times,
a. What is the probability that the sample mean is less than 0.75 seconds?
b. What is the probability that the sample mean is between 0.70 and 0.90 seconds?

Ketan3 · Oct 24, 2009, 11:14 AM

(.8-.75)/.2=.25

+((.7-.75)/.2)=.25

((-.7-.75)/.2)=-7.25

galactus · Oct 24, 2009, 12:35 PM

The mean time to download the home page of a company was 0.8 second. Suppose the
download time was normally distributed, with a standard deviation of 0.2 second. If you select a
random sample of 30 download times,

a. What is the probability that the sample mean is less than 0.75 seconds?

$z=\frac{(x-{\mu})\sqrt{n}}{\sigma}=\frac{(.75-.8)\sqrt{30}}{.2}=-1.37$

Look it up in the z table.

b. What is the probability that the sample mean is between 0.70 and 0.90 seconds?

$z=\frac{(.7-.8)\sqrt{30}}{.2}=-2.74; \;\ z=\frac{(.9-.8)\sqrt{30}}{.2}=2.74$

Look them up in the z table and subtract.

Ositadima · Jul 9, 2010, 04:33 PM

Galactus your solution has really helped me to regain my balance on this problem. I somehow got lost and your solution rebounced me back. Thank you for taking your time to contribute mathematically. It surely does not come so easy. Keep on with the good work and always make sure that you answer correctly.

Thanks again,

Ositadima Chukwuemeka Alozie