Question 1068826: The London Enquirer, in its Sunday business supplement, reported that the mean number of hours worked per week by those employed full time is 43.9. The article further indicated that about one third of those employed full time work less than 40 hours per week. (a) Given this information and assuming that the number of hours worked follows the normal distribution, what is the standard deviation of the number of hours worked? (b) The article also indicated that 20 percent of those working full time work more than 49 hours per week. Determine the standard deviation with this information. Are the two estimates of the standard deviation similar? What would you conclude?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean = 43.9
standard deviation = s
probability that a score is less than 40 is 1/3.
use the z-score table to find the z-score that has an area to the left of it as 1/3 = approximately .3333
you will find that the z-score is equal to somewhere between -.43 and -.44.
since it's closer to -.43, we'll use that.
formula for z-score is z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation
using what we know, the formula becomes:
-.43 = (40-43.9)/s
solve for s to get s = -3.9/-.43 = 9.069767 rounded to 6 decimal places.
to find the probability that a z-score is greater than 80% of the population, look for the area to the left of the z-core of .80.
you will find that the z-core is between .84 and .85.
since it is close to .84, we'll use that.
same z-score formula of z = (x-m)/s becomes:
.84 = (49-43.9)/s.
solve for s to get s = 5.1/.84 = 6.071429 rounded to 6 decimal laces.
since the two standard deviations are different, you can conclude that each scenario used a different population.
if they were the same population, the standard deviation should have been the same.
the z-score table that i used can be found at the following link:
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
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