Question

A quality control inspector has drawn a sample of 19 light bulbs from a recent production lot. Suppose 20% of the bulbs in the lot are defective. What is the probability that less than 12 but more than 9 bulbs from the sample are defective? Round your answer to four decimal places.

Answer 1

Answer:

$P(9 < x < 12)=P(X=10)+P(X=11)$

$P(X=10)=(19C10)(0.2)^{10} (1-0.2)^{19-10}=0.00127$

$P(X=11)=(19C11)(0.2)^{11} (1-0.2)^{19-11}=0.00026$

$P(9 < X < 12)=0.00127 +0.00026=0.0015$

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:

$X \sim Binom(n=19, p=0.2)$

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:

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

And we want to find this probability:

$P(9 < x < 12)=P(X=10)+P(X=11)$

$P(X=10)=(19C10)(0.2)^{10} (1-0.2)^{19-10}=0.00127$

$P(X=11)=(19C11)(0.2)^{11} (1-0.2)^{19-11}=0.00026$

$P(9 < X < 12)=0.00127 +0.00026=0.0015$