klewis
02-14-2006, 02:01 PM
Regarding Question #2
[U][Second Question/U]
In grading apples into "A","B" and "C" a large orchard uses weights to distinguish apples. Any apple weighing more than 2 ounces is classified as Grade "A" while apple weighing less than 0.75 ounces is classified as Grade "C". If the days pick shows 16.6% are Grade "A" and 6.68% are Grade C, determine the mean and standard deviation. Assume weights are normally distributed.
Here is a HINT that I dont get - First calculate the appropriate z values for the probabilities given, then set up your equations to solve for the mean and standard deviation
=====> I understand that x=2.0 and x=.75 .....I have been taught to draw a symmetric curve to solve for x and z values, but never taught how to solve for u (mean) and s (standard deviation)
the formula I have been taught to use is z= x - u / s ===> or x=u-(s*z)
I will admit my algebra is not good, been a while
Also, that half of the curve is .5 and other half is .5
P(o<z) = P(z<o) and usually add or subtract .5 accordingly
thus z value for 2.0 is 100-16.6 = .834 - .5 is .334 (z value on chart is 0.97)
and z value for 0.75 is 100-6.68 =.9332 - .5 is .4332 (z value on chart is 1.5)
is there an easier way than this to find z chart value??
I used the formulas that you suggested for Shannon
u = x - (z*s)
but I don't understand how she arrived at s = 1.78
and how to calculate for u after getting value for s
[U][Second Question/U]
In grading apples into "A","B" and "C" a large orchard uses weights to distinguish apples. Any apple weighing more than 2 ounces is classified as Grade "A" while apple weighing less than 0.75 ounces is classified as Grade "C". If the days pick shows 16.6% are Grade "A" and 6.68% are Grade C, determine the mean and standard deviation. Assume weights are normally distributed.
Here is a HINT that I dont get - First calculate the appropriate z values for the probabilities given, then set up your equations to solve for the mean and standard deviation
=====> I understand that x=2.0 and x=.75 .....I have been taught to draw a symmetric curve to solve for x and z values, but never taught how to solve for u (mean) and s (standard deviation)
the formula I have been taught to use is z= x - u / s ===> or x=u-(s*z)
I will admit my algebra is not good, been a while
Also, that half of the curve is .5 and other half is .5
P(o<z) = P(z<o) and usually add or subtract .5 accordingly
thus z value for 2.0 is 100-16.6 = .834 - .5 is .334 (z value on chart is 0.97)
and z value for 0.75 is 100-6.68 =.9332 - .5 is .4332 (z value on chart is 1.5)
is there an easier way than this to find z chart value??
I used the formulas that you suggested for Shannon
u = x - (z*s)
but I don't understand how she arrived at s = 1.78
and how to calculate for u after getting value for s