Question

18. It seems that 96% of people bitten by vampires become vampires themselves. This Halloween 35 people will be bitten by vampires. a) What is the probability no more than33 of those bitten will become vampires?b) Justify an appropriatesimplified approximatemethod for answering the question in a), carry it out, and compare the two answers. (Hint: Not becoming a vampire has lowprobability).

Answer 1

a)
`P(X \leq 33)=1-P(X>33)= 1- P(X\geq 34) = 1-[P(X=34)+P(X=35)]`

And we can find the individual probabilities and we got:

`P(X=34) = 35C34 (0.96)^(34) (1-0.96)^(35-34) =0.3494`

`P(X=35) = 35C35 (0.96)^(35) (1-0.96)^(35-35) =0.2396`

And after replace we got:

`P(X \leq 33) = 1-[0.3493+0.2396]= 0.4111`

b)
`z= (33-33.6)/(1.159)= -0.518`

And using the normal standard table or excel we got:

`P(X \leq 33)=P(z<-0.518) = 0.3022`

For this case the approximation is not exactly the exact answer but is an approximation, the difference is about 0.4111-0.3022= 0.1089.

Explanation:

Part a

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 now that:

`X \sim Binom(n=35, p=0.96)`

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=(n!)/((n-x)! x!)`

And for this case we want this probability:

`P(X \leq 33)`

And we can find this probability with the complement rule:

`P(X \leq 33)=1-P(X>33)= 1- P(X\geq 34) = 1-[P(X=34)+P(X=35)]`

And we can find the individual probabilities and we got:

`P(X=34) = 35C34 (0.96)^(34) (1-0.96)^(35-34) =0.3494`

`P(X=35) = 35C35 (0.96)^(35) (1-0.96)^(35-35) =0.2396`

And after replace we got:

`P(X \leq 33) = 1-[0.3493+0.2396]= 0.4111`

Part b

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

`E(X)=np=35*0.96=33.6`

`\sigma=√(np(1-p))=√(35*0.96(1-0.96))=1.159`

And we want this probability:

`P(X\leq 33)`

We find the z score given by:

`z= (33-33.6)/(1.159)= -0.518`

And using the normal standard table or excel we got:

`P(X \leq 33)=P(z<-0.518) = 0.3022`

For this case the approximation is not exactly the exact answer but is an approximation, the difference is about 0.4111-0.3022= 0.1089.