Distribution

J

JeckShirauw

New member
Joined
May 28, 2020
Messages
16
Two people produce apples. The first one with
a diameter normally distributed with an average of 3 cm and standard deviation
of 0.1 cm.
The second one with a diameter which is normally distributed with an average of 3.04 cm
and standard deviation 0.02 cm.
We only need a diameter between 2.9 cm and 3.1 cm. We randomly take an apple.
Now it appears that this apple is too big and is therefore not accepted. What is the chance that the apple
was made by the first device?
My answer: 99 %

Is my answer correct?
 
1) Device? You mean TREE?
2) Too big? So what is p(apple > 3.1 cm) for each distribution?
3) How did you use this information to arrive at your conclusion?
4) Are the deliveries from each source the same size by count? This may change your answer.
 
tkhunny said:
1) Device? You mean TREE?
2) Too big? So what is p(apple > 3.1 cm) for each distribution?
3) How did you use this information to arrive at your conclusion?
4) Are the deliveries from each source the same size by count? This may change your answer.
Click to expand...
Yes I meant tree.
I came to my answer in this way:
P(Z>1)= 0.15866
P(Z>3)=0.00135
0.15866/(0.15866+0.00135)= 99%
 
That's pretty good.

You didn't answer my point #4. How do you know they are the same size populations? Are you told in some part of the problem statement? Does it make any difference if one or the other is twice the number of apples?
 
Thank you and unfortunately, that information is not provided in the exercise.

Was my answer really correct or was it sarcasm? :)
 
That's fine, so long as you state any required assumption. As the problem statement does not tell you the two types have the same number, it would be wise to get that out there. This shows 1) You know the problem was flawed, and 2) You are paying attention. Both good things to show.
 
tkhunny said:
That's fine, so long as you state any required assumption. As the problem statement does not tell you the two types have the same number, it would be wise to get that out there. This shows 1) You know the problem was flawed, and 2) You are paying attention. Both good things to show.
Click to expand...
Thank you! But I do not understand what the difference would be if they were not equal?
 
P(Z>1)= 0.15866
P(Z>3)=0.00135
0.15866/(0.15866+0.00135)= 99%

Ship 10,000 Type 1, that is expected to include 1,587 too big.
Ship 20,000 Type II, that is expected to include 27 too big.

1587/(1587+27) = 0.9833
 
Thank you! So should I include the amount “n” somewhere in my formulas?
So like n*0.15866/n*(0.15866+0.00135)= 99%
where I then scrape the “n”?
 
Well, if [math]n_{1} = n_{2}[/math], it is fo no consequence.

Personally, I would just say your solution depends on equal numbers of apples in the two shipments.
 
You must log in or register to reply here.
Top