Question 997753: The weekly amounts that a family of four spends on groceries are normally distributed. The mean amount it $150. 7% of the families spend more than $185. Find the standard deviation.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! it's a normal distribution.
the mean is 150.
7% spend more than 185.
look up in the z-score table for a percentage of .93 to the left of the z-score.
that z-score will be 1.48.
that means the z-score is 1.48 standard deviations above the mean.
if 185 is the raw score that generates a z-score of 1.48, then 185 - 150 is equal to 1.48 standard deviations which means the standard deviation is 35/1.48 = 23.65
the z-score table is not as accurate as a z-score calculator, but it gets you close.
the table shows .9292 and .9306 as area under the normal distrubiton curve to the left of the z-score.
that means that .9292 will be slightly above 7% to the right of it and .9306 will be slightly less than .07 to the right of it.
since the table shows you the area under the curve to the left of the z-score, if you want to find the area under the curve to the right of the z-score, you just subtract the area to the left of the z-score from 1 and you get the area to the right of the z-score.
1 is the maximum area under the distribution curve.
1 is equivalent to 100%.
you could interpolate and you would probably get a z-score around 1.476 which might make your standard deviation a little different.
35/1.476 give you a standard deviation of 23.71 versus 23.65 with a z-score of 1.48.
that's not a very big difference and probably inconsequential.
bottom line is either one of those answers would be sufficient for practical purposes.
the z-score formula is z = (x - m) / s
z is the z-score (1.48)
x is the raw score (185)
m is the mean (150)
s is the standard deviation
the formula becomes 1.48 = (185 - 150) / s
solve for s to get s = (185 - 150) / 1.48 = 35 / 1.48 = 23.65
here's a link to the z-score table that i used.
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
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