Question

Consider the probability that exactly 89 out of 158 CDs will not be defective. Assume the probability that a given CD will not be defective is 66%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Answer:

Using the exact distribution we got:

$P(X=89) =0.0027$

For this case we can use the following Excel code: "=BINOM.DIST(89,158,0.66,FALSE)"

We need to check if we can use the normal approximation , the conditions are:

$np=158*0.66=104.28>10$ and $n(1-p)=158*(1-0.66)=53.72>10$

Since we satisfy both conditions the normal approximation makes sense

The expected value is given by this formula:

$E(X) = np=158*0.66=104.28$

And the standard deviation for the random variable is given by:

$sd(X)=\sqrt{np(1-p)}=\sqrt{158*0.66*(1-0.66)}=5.954$

$X\sim N (\mu =104.28, \sigma =5.954)$

But for this case we can find the probability $P(X=89)$ since is a continuos approximation and the probability below the curve is 0

Step-by-step explanation:

Previous concepts  

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".  

Solution to the problem

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=158, p=0.66)$  

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

We want this probability:

$P(X=89)$

Using the exact distribution we got:

$P(X=89) =0.0027$

For this case we can use the following Excel code: "=BINOM.DIST(89,158,0.66,FALSE)"

Normal approximation

We need to check if we can use the normal approximation , the conditions are:

$np=158*0.66=104.28>10$ and $n(1-p)=158*(1-0.66)=53.72>10$

Since we satisfy both conditions the normal approximation makes sense

The expected value is given by this formula:

$E(X) = np=158*0.66=104.28$

And the standard deviation for the random variable is given by:

$sd(X)=\sqrt{np(1-p)}=\sqrt{158*0.66*(1-0.66)}=5.954$

$X\sim N (\mu =104.28, \sigma =5.954)$

But for this case we can find the probability $P(X=89)$ since is a continuos approximation and the probability below the curve is 0