normal distribution

1a2b3c2212 · 2010-03-09 02:28:55

A machine grades apples according to their masses. Apples with a mass exceeding 125g are rejected as too large and apples with a mass less than 75g are rejected as too small. A large batch of apples is graded and it is found that 10% are rejected as too large and 13% are rejected as too small. Assuming a normal distribution, find the mean mass of a randomly chosen apple from the batch.

Plz help coz I don't even understand normal distribution..

White_Owl · 2010-03-09 02:55:29

Place all your apples in one long line. Starting with the smallest one, ending with the largest one.
Normal distribution means that difference between sizes of two adjusted apples would be same as the difference between any other pair of adjusted apples.

So in your case you have 10% apples which weigh more than 125g and 13% which are less than 75g. That leaves us with 100-10-13=77% apples in the range of 75-125g. Let's assume that in the original batch we had 100 apples (1 apple = 1%).
Equal distribution of 77 good apples through the 100g will give us 1.3g difference in weight between two adjusted apples.
So, to find the smallest apple we can do: 75-13*1.3 = 58.1. The largest apple: 125+10*1.3 = 138.
Once we know the smallest and largest apple in the batch we find the mean: (58.1+138)/2 = 98.05

taymaz · 2010-03-09 20:24:35

I want to know thise question's answer.
let x be normal (a,b) what is E(1/x) & E(xlogx)?

bobbym · 2010-03-10 06:09:21

Hi taymaz;

The expected value of a function is the sum or the integral of that function times by the PDF.

qcao · 2010-03-16 04:57:37

REPLY TO POST #1:

Let X be the mass of an apple. Assume that X follow a normal distribution with mean m (for mu) and standard deviation s (for sigma). You always wand to transform any normal distribution to a standard normal distribution (with mean 0 and standard deviation 1) by first subtracting from m and then dividing by s.

P[Z > (125-m)/s] = 0.1 and P[Z < (75-m)/s] = 0.13, where Z is a standard normal random variable.

You can find the values of z to satisfy the above probability statements from a standard normal probability table, or by using NORMSINV function from Excel:

=NORMSINV(1-0.1) results in 1.281552 and =NORMSINV(0.13) results in -1.12639.

Let's solve a system of two equations and two unknowns:

(125-m)/s = 1.281552 and (75-m)/s = -1.12639.

The answer is m = 98.3891 and s = 20.7646.

Note that the answer from quote #2 (98.05) is a pretty good approximation, but (a) it does not use the normal probability statement which is the main point of the exercise, and (b) it assumes equal distribution of the 77 good apples, meaning the distribution is uniform but not normal. This approximation gets worse if the two given probabilities are not balanced, e.g. 0.05 and 0.4.

REPLY TO POSTS #3 and #4:

The normal distribution is anything but normal. Its cdf is not in closed form. I do not know if the integrals of f(x)/x and xlogx*f(x) are in closed form for the normal distribution.

bobbym · 2010-03-16 17:44:36

H qcao;

An equation is said to be a closed-form solution if it solves a given problem in terms of functions and mathematical operations from a given generally accepted set. For example, an infinite sum would generally not be considered closed-form. However, the choice of what to call closed-form and what not is rather arbitrary since a new "closed-form" function could simply be defined in terms of the infinite sum.

As the above quote states it is a little muddy because the CDF for normal distribution is

Whether you would consider that closed or not depends on whether you include the error function in you set.

Math Is Fun Forum

#1 2010-03-09 02:28:55

normal distribution

#2 2010-03-09 02:55:29

Re: normal distribution

#3 2010-03-09 20:24:35

Re: normal distribution

#4 2010-03-10 06:09:21

Re: normal distribution

#5 2010-03-16 04:57:37

Re: normal distribution

#6 2010-03-16 17:44:36

Re: normal distribution

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