Question

The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of twelve syringes taken from the batch. Suppose the batch contains 5% defective syringes.(a) Make a histogram showing the probabilities of r = 0, 1, 2, 3, ..., 11 and 12 defective syringes in a random sample of twelve syringes.(b) Find μ. (Enter your answer to two decimal places.)μ = syringesWhat is the expected number of defective syringes the inspector will find? (Enter your answer to two decimal places.)‪ syringes(c) What is the probability that the batch will be accepted? (Round your answer to three decimal places.)(d) Find σ. (Round your answer to three decimal places.)σ = syringes

Answer 1

Answer:

a) We can use the following R code:

> x <- seq(0,12,by = 1)

> y <- dbinom(x,12,0.05)

> plot(x,y,type="h",main="Histogram for n=12, p=0.05")

And we got the figure attached.

b) $\mu = E(X) = n*p = 12*0.05=0.60$

So the expected number is 0.6 syringes

c) $P(X<2) = P(X=0)+P(X=1)$

$P(X=0)=(12C0)(0.05)^0 (1-0.05)^{12-0}=0.5404$

$P(X=1)=(12C1)(0.05)^1 (1-0.05)^{12-1}=0.34128$

$P(X<2) = 0.5404+0.34128=0.882$

d) For this case we need to find first the variance given by:

$\sigma^2= Var(X) = n*p*(1-p) =12*0.05*(1-0.05)=0.57$

And the deviation would be the square root of the variance

$\sigma =Sd(X) = \sqrt{0.57}=0.755$

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:

$X \sim Binom(n=12, p=0.05)$

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:

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

Part a

We can use the following R code:

> x <- seq(0,12,by = 1)

> y <- dbinom(x,12,0.05)

> plot(x,y,type="h",main="Histogram for n=12, p=0.05")

And we got the figure attached.

Part b

For this case the mean is given by the following formula:

$\mu = E(X) = n*p = 12*0.05=0.60$

So the expected number is 0.6 syringes

Part c

The batch is rejected if we have two or more defective syringes. So then the probability that would be accepted is:

$P(X<2) = P(X=0)+P(X=1)$

$P(X=0)=(12C0)(0.05)^0 (1-0.05)^{12-0}=0.5404$

$P(X=1)=(12C1)(0.05)^1 (1-0.05)^{12-1}=0.34128$

$P(X<2) = 0.5404+0.34128=0.882$

Part d

For this case we need to find first the variance given by:

$\sigma^2= Var(X) = n*p*(1-p) =12*0.05*(1-0.05)=0.57$

And the deviation would be the square root of the variance

$\sigma =Sd(X) = \sqrt{0.57}=0.755$