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ShannonL
11-05-2005, 10:25 PM
Hi...have completed my assignment (and have come a long way with some helpful tips on this forum) but am stuck on one problem:

In grading apples into "A," "B", and "C", an orchard uses weights to distinguish apples. Any apple weighing more than 2 ounces is classified as Grade "A", while an apple weighing less than 0.75 ounces is classified as Grade "C". If the day's pick shows 16.6% are Grade "A" and 6.68% are Grade "C", determine the mean and the standard deviation (assume weights are normally distributed).

We are to calculate the z values for the probabilities given and then set up the equations to solve for the mean and the standard deviation but I'm lost...I have been calculating z values so far using the mean and the standard deviation and just cannot solve this problem.

Not wanting answers...just hoping someone will be able to make this a little clearer for me. Thanks very much...Shannon

JohnM
11-05-2005, 11:26 PM
Shannon,

In this problem, for the equation z = (x - u)/s, you are given x=2.0 and x=0.75.

So, z = (2.0 - u)/s and z = (0.75 - u)/s

You are also told that 2.0 or above contains 16.6% of the normal curve, and 0.75 or below contains 6.68% of the normal curve. Look up the z values for each of those proportions (hint - 16.6% in the upper tail, and 6.68% in the lower tail) and plug them into the equations above.

Now you'll have two equations with two unknowns (u and s). Using a little algebra to solve two equations with two unknowns will get you to the answers for u and s.

Hint: for the two equations, set them up like this-

z = (x - u)/s ---> u = x - (z*s)

Let us know if you get stuck again.

John

ShannonL
11-05-2005, 11:51 PM
Hi John...thanks very much for your reply!

I'm fairly early on in my intro stats course so please bear with me...I have been using the formula for z values but am unclear on how to look up the z values for the proportions as you mention (not familiar with the terms upper/lower tail). If I can figure this out, I think I would be able to complete the equations.

Been working on this course for the better part of the last few days and chances are good I may be overlooking the obvious...thanks for any help you can offer, John...

Shannon

JohnM
11-06-2005, 12:06 AM
Shannon,

Here's a link to a normal distribution table:
http://www.math.unb.ca/~knight/utility/NormTble.htm (http://www.math.unb.ca/%7Eknight/utility/NormTble.htm)

Since we are given that x=2.0 divides the upper 16.6% of the normal curve from the lower 83.4%, we need to find the z value that corresponds to 0.834. Look for 0.834 in this table - it is in the row where the far-left column is 0.9, and it is in the column where the top row is 0.07. Therefore the z value is 0.97.

I'm sure your textbook has a section on how to use the normal distribution table to look up z values??

ShannonL
11-06-2005, 02:08 PM
Hi John...thanks very much for the link and the information.

I've calculated the following:

z values are .97 and 1.5

so

.97s = 2 - u
1.5s = .75 - u

s = 1.78

u = 2 - (.97 x 1.78)
u = .75 - (1.5 x 1.78)

u = .82

Hope this is correct...

Again, many thanks for your help...have a great day! Shannon

JohnM
11-06-2005, 02:39 PM
Shannon,

Awesome job - however the z values are 0.97 and -1.50

Make the change, and recalculate, and you should get the correct answers.

John

ShannonL
11-06-2005, 07:03 PM
Hi John...

Thanks for clarifying that...makes perfect sense, of course.

Based on these z-values, I now have s = .51 and u = 1.51.

You've been a huge help, John...thanks very much for your patience.

Regards,
Shannon

bquigz
12-24-2005, 12:54 PM
Hi,
I am working on the same problem. I have been working on it for a few days now and while I have been able to figure out the z-values, and have somehow got to the point where I figured the mean = 1.51959 and s = 0.495268041, which worked with z=0.97 and x=2
but when I tried to use the mean and s for z=-1.5 and x=0.75 it did not work out.
I have then tried shifting around numbers of the mean and s to try and find a standard deviation that fits with both z and x values but have had no luck.
Perhaps I used the wrong method to calculate the mean. I didn't know how to get the mean from the z-value and as it is a normal dist., I thought I would figure out the median. To get this I tried finding the high and low values by multiplying 2 by 0.166 (the amount given above 2) and then adding it to 2, and multiplying 0.75 by the amount below it:0.0668, and adding that to 0.75.
So then I used those as the ranges and added them (2.332 +0.6999) and divided by 2 to get the median=1.51595.
What am I doing wrong?

Thank you, Bq.

JohnM
12-26-2005, 08:36 AM
Please state the exact problem you're working on and then we'll be able to give you some guidance.

Thanks,
John

bquigz
12-31-2005, 01:10 PM
Using 1.51595 as a mean
Z1= x1 – µ / Ơ
0.97 = 2ounces – 1.51595/Ơ
Ơ (0.97) = 2ounces – 1.51595 = 0.48405
Ơ (0.97)/ 0.97= 0.48405/0.97
Ơ= 0.499020618
Using this Ơ= 0.499020618, and µ = 1.51595 with x2 and z2 we get:
Z2= -1.53490652
The problem I am having is that the z values and x values do not align up exactly using the mean and standard deviation I have found. Is this the wrong way to come up with the mean and standard deviation or the numbers given in the text question (A=more than 2 ounces=16.6%, C=less than 0.75 ounces=6.68%) just approximate values?
Thank you, Bq.

JohnM
12-31-2005, 02:06 PM
Your concept is fine - you probably just have some slight rounding error. From the original discussion, you should have gotten 0.51 for s (0.5061 to be exact), not 0.499, and that may account for the small offset.

z = (x - u)/s

z = (2.0 - u)/s
z = (0.75 - u)/s

0.97 = (2.0 - u)/s
-1.5 = (0.75 - u)/s

0.97s = (2.0 - u)
-1.5s = (0.75 - u)

u = 2.0 - 0.97s
u = 0.75 + 1.5s

2.0 - 0.97s = 0.75 + 1.5s
1.25 = 2.47s
1.25/2.47 = s

s = 0.5061

then u = 1.5091

bquigz
01-07-2006, 09:22 PM
Thank you John,
I have been over it a few times now and have figured out how to do it a few ways and get the correct answer. You are right, it was mostly rounding error, once I did it with only rounding at the final stage it came out perfectly. Happy new years. B. Quigley.