# Thread: Binomial Distribution (LEVEL 2 CALCULUS)

1. ## (SOLVED) Binomial Distribution (LEVEL 2 CALCULUS)

Question:

A hundred samples of 5 resistors were taken from a large batch. 59 samples had no defective resistors, 33 had 1, 7 had 2, 1 had 3, and no samples had 4 or 5 defective resistors. Show that the distribution is approximately binomial and estimate the overall percentage of defective components.

I need some help in determining the variables p,q,n to solve this. Please correct me if I'm wrong:

q = possibility of no defect in a sample = 1/5
p = possibility of having defects in a sample = 4/5
n (times) = 100

2. n (times) = 100
This is wrong. Here n=number of defects in each sample.
Show that the distribution is approximately binomial...
I think this means you have to FIT the given distribution with a binomial distribution having suitable parameters. To do this, first compute the average number of defectives from the given samples (the formula is sum of (value*frequency) / sum of frequencies which will serve as the "possibility of no defect" that is p. So now you have a binomial distribution of parameters (n=100,p). From this, calculate the probabilities of X=0,1,2,3,4,5

3. So now you have a binomial distribution of parameters (n=100,p)
I'm a bit unclear regarding this. Earlier you said it's wrong? Below is some of the workings with your help. There was a typo in my earliest post: p = probability of having defect and q = probability of having no defect.
Do take a look and comment on the binomial equation I came up with, please.

4. I apologize for two reasons:
I'm a bit unclear regarding this. Earlier you said it's wrong?
I meant to say you have a binomial distribution of parameters (n=5,p)
And the second one is : the formula for getting p is { sum of (value*frequency) / sum of frequencies } / n so it will be 0.1 (divide your result by n that is by 5).

Apart from this, your method is correct. Sorry or the confusion; I am getting mad at my browser which is making a lot of trouble now-a-days.

5. I got the answer! Thank you very much for your help