Probability of a defective sample.

[WendysRules]

Homework Statement

In a manufacturing plant, a sample of a 100 items is taken from an assembly line. For each item in the sample, the probability of being defective is .06.

What is the probability that there are 3 or more defective units in the sample?

Homework Equations

z = (x - mean)/standard deviation
mean = n*p
standard deviation = sqrt(n*p*(1-p)

The Attempt at a Solution

Well, mean is just $$ 100*.06 = 6$$. The standard deviation is just $$ \sqrt{100*.06*.94} = 2.375 $$ so the Z = (3-6)/(2.375) = -3/2.375.

So, the prob(x>3) is just $$ 1- \frac{1}{\sqrt{2\pi}} \int^{-3/2.375}_{-\infty} e^{-.5x^2} dx = .8967 $$ Which isn't an answer choice I'm given. So, obviously, I did something wrong! Any help is appreciated.

Answer choices given: .7040, .8784, .9306, .8498, .8843.

 

[Ray Vickson]

Homework Statement

In a manufacturing plant, a sample of a 100 items is taken from an assembly line. For each item in the sample, the probability of being defective is .06.

What is the probability that there are 3 or more defective units in the sample?

Homework Equations

z = (x - mean)/standard deviation
mean = n*p
standard deviation = sqrt(n*p*(1-p)

The Attempt at a Solution

Well, mean is just $$ 100*.06 = 6$$. The standard deviation is just $$ \sqrt{100*.06*.94} = 2.375 $$ so the Z = (3-6)/(2.375) = -3/2.375.

So, the prob(x>3) is just $$ 1- \frac{1}{\sqrt{2\pi}} \int^{-3/2.375}_{-\infty} e^{-.5x^2} dx = .8967 $$ Which isn't an answer choice I'm given. So, obviously, I did something wrong! Any help is appreciated.

Answer choices given: .7040, .8784, .9306, .8498, .8843.

One of your choices is fairly close to the true value, and is also not far from the normal approximation, when you do it properly. Remember the "1/2" correction: If the true random variable $N$ is binomial and the approximate random variable $X$ is normal (with the same mean and variance as $N$), then $P(N \leq k) \doteq P(X \leq k + 1/2)$ for integer values of $k$.

There are several ways to see this, but one way is to note that for a discrete random variable $N$ on $\{ 0,1,2,\ldots, n \}$ we have $P(N \leq k ) = P(N \leq k + 0.5)$ exactly (because $N$ takes integer values). Use the normal approximation on $k + .5$ instead of $k$. For moderate values of $n$ and if $np$ is not really small or really large, the 1/2-correction often gives you one or two more decimal places of accuracy.

However, in this problem you can just as easily get an exact answer by performing the relatively easy calculation $P(N \leq 2)$ for the exact binomial.