Question 138607: To Stanbon,
Using the first problem you solved on z-score please check this problem I tried to solve:
The mean weight of 545 women is 99lbs. and s is 5, Assuming that the wts. of women are normally distributed, how many would you expect to weigh less than 105?
Z= 105-99/5
z=1.2
from table: Z score = .3848 or 38.48% will weigh more than 105 that is 210 women
computing for women who will weigh less than 105:
100-38.48 = 61.52%
545x.6152 = 335 women who will weigh less 105 lbs.
Thanks again.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The mean weight of 545 women is 99lbs. and s is 5, Assuming that the wts. of women are normally distributed, how many would you expect to weigh less than 105?
Z= (105-99)/5
z=1.2
That is correct.
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P(0 < x < 105) = P(-19.8 < z < 1.2) = 0.8848 or 88.48%
from table: Z score = .3848 or 38.48% will weigh more than 105
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No: 38.48% will weigh more than the MEAN score of 99 lbs.
You have to include the 50% who weigh less than 99 lbs.
What that means is you need a z-score for "0" lbs.
z(0) = (0-99)/5 = -19.8
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Then 88.48% of 545 is 482 ladies who will weigh less than 105 lbs.
So 545-482 = 63 will weight more than 105 lbs.
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Recommendation: I think if you draw the normal curve and
put the problem values on the horizontal axis and shade the
area under the curve that corresponds to the probability
you are looking for, you will avoid some errors you might
otherwise make when you only manipulate the numbers.
You will be able to see how your answers make sense,or don't.
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Cheers,
Stan H.
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