Question

A factory makes silicon chips for use in computers. It is known that about 90% of the chips meet specifications. Every hour a sample of 18 chips is selected at random for testing and the number of chips that meet specifications is recorded. What is the approximate mean and standard deviation of the number of chips meeting specifications?
a. m = 1.625; s = 1.265
b. m = 162; s = 1.414
c. m = 16.2; s = 1.62
d. m = 16.2; s = 4.025
e. m = 16.2; s = 1.27

Answer 1

Answer:

e. m = 16.2; s = 1.27

Step-by-step explanation:

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we can assume that the distribution for X is given by:

$X \sim Binom(n=18, p=0.9)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Normal approximation

We need to check the condition in order to use the normal approximation.

$np=18*0.9=16.2 \geq 10$

So we see that we satisfy the condition and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=18*0.9=16.2$

$\sigma=\sqrt{np(1-p)}=\sqrt{18*0.9(1-0.9)}=1.273$

The best answer would be:

e. m = 16.2; s = 1.27