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Probability
Message posted by Cathy (via 64.12.104.48) on June 4, 2001 at 5:48 PM (ET)
Please help with this...I am lost.
Carl is doing a project for his ecology class. He read that the average person uses about 100 gallons of water per day. He estimates the standard deviation to be about 35 gallons per day. Assuming that Carl's figures are correct, find the probability that a random sample of 49 people will have a "sample mean" water use of 90 gallons per day or less.
READERS RESPOND:
(In chronological order. Most recent at the bottom.)
Re: Probability
Message posted by JG (via 128.8.22.18) on June 5, 2001 at 2:27 AM (ET)
.0228
Re: Probability
Message posted by nancy diehl (via 129.176.151.121) on June 5, 2001 at 12:30 PM (ET)
First of all, you should draw yourself a normal curve, and drop a mean line through that normal curve and mark that with a value of 100. Drop another line through that normal curve at about where you think 90 would fall. You are looking for the probablity for the average occurrence of a value of less than 90, so shade in the area under the normal curve to the left of 90. This is the probability you are looking for.So what you want to do is translate the value of 90 to what it represents on a Normal Z scale given a mean of 100, sd of 35 and n=49. Take this Z value and find the area it represents under the normal curve by using a Z table.
To find the Z that corresponds to 90: = (90-100)/(35/SQRT(49)) = -10*7/35 = -2.0. Most likely the Z table you are using won't list negative Zs, but the areas are symmetrical, so -2.0 would equal the area for Z=2.0. Again, depending upon the Z-table you are using, it will list the area as either .4772 or .0228. Look at how the table is defined. Are they giving you the tail area (.0228) or the area between the Z-value and the mean (.4772). For this problem you want the tail area which equals .0228.
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