Thread: If the probability of a defective bolt is

1. If the probability of a defective bolt is

Question : If the probability of a defective bolt is 0.1, find (i) the mean and (ii) the standard deviation for the distribution of defective bolts in a total of 400.
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Pr(X = def) = 0.1
Pr(Y = not def)= 1-0.1 = 0.9

$\displaystyle \mu = \sum x. Pr(X)$................I dont know what should i put in the value of x

2. I don't know man - after awhile some folks are going to stop responding to these questions because you are really asking the same question over and over again, and seemingly NOT retaining any of the material from question to question (and most definitely not from class session to class session). This isn't just an easy question, it is a basic question that should be covered in the first couple of pages introducing you to discrete random variables and their distributions.

Each individual bolt is a Bernoulli trial, with two outcomes, defective (0) or non-defective (1), and has Bernoulli distribution of:

X P(X)
0 0.1
1 0.9

Therefore 400 Bernoulli Bolts compromising your sample constitute a Binomial distribution. The mean of the binomial distribution is n*p, or the sum of of the mean of 400 independent Bernoulli Bolt trials. As the mean of a Bernoulli trial is simply P(X=1), you are simply multiplying 400 by P(X=1).

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find out the standard deviation from the number of defective screws if the probability for a screw to be defective is 0.1

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