Question

uppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is at most 30?

Answer 1

Answer:

Using the binomial distribution we can find this probability with the following excel code:

"=BINOM.DIST(30,200,0.1,TRUE)", and we got:

$P(X \leq 30)=0.9905$

Using the normal approximation

We can use the continuity factor correction:

$P(X \leq 30) = P(X< 30+0.5) = P(X<30.5)$

And we can use the z score formula given by:

$z = \frac{x-\mu}{\sigma}$

And we got:

$P(X<30.5) = P(Z< \frac{30.5-20}{4.243}) = 2.475)= 0.9933$

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:  

$X \sim Binom(n=200, p=0.1)$  

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 to find this probability;

$P(X \leq 30)$

Using the binomial distribution we can find this probability with the following excel code:

"=BINOM.DIST(30,200,0.1,TRUE)", and we got:

$P(X \leq 30)=0.9905$

The other way is using the normal approximation for the binomial distribution

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

$np=200*0.1=20 \geq 10$  

$n(1-p)=20*(1-0.1)=18 \geq 10$  

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

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

$E(X)=np=200*0.1=20$  

$\sigma=\sqrt{np(1-p)}=\sqrt{200*0.1(1-0.1)}=4.243$  

And we can use the continuity factor correction:

$P(X \leq 30) = P(X< 30+0.5) = P(X<30.5)$

And we can use the z score formula given by:

$z = \frac{x-\mu}{\sigma}$

And we got:

$P(X<30.5) = P(Z< \frac{30.5-20}{4.243}) = 2.475)= 0.9933$