Question

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities. Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel region about 48% of strikes (while trolling) resulted in a catch. Suppose that on a given day a fleet of fishing boats got a total of 20 strikes. Find the following probabilities. (Round your answers to four decimal places.) (a)-12 or fewer fish were caught (b)-5 or more fish were caught (c)-between 5 and 12 fish were caught

Answer 1

a)
`P(X\leq 12)=0.8586`

b)
`P(X\geq 5)=0.9802`

c)
`P(5\leq X\leq 12)=0.8389`

Explanation:

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=20, p=0.48)`

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!)`

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

`np=20*0.48=9.6 \approx 10 \geq 10`

`n(1-p)=20*(1-0.48)=10.4 \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=20*0.48=9.6`

`\sigma=√(np(1-p))=√(20*0.48(1-0.48))=2.234`

Part a

We want this probability:

`P(X\leq 12)`

We can use the z score given by this formula
`Z=(x-\mu)/(\sigma)`.

`P(X\leq 12)=P((X-\mu)/(\sigma)\leq (12-9.6)/(2.234))=P(Z\leq 1.074)=0.8586`

Part b

We want this probability:

`P(X\geq 5)`

We can use again the z score formula and we have:

`P(X\geq 5)=1-P(X<5)=1-P((X-\mu)/(\sigma)< (5-9.6)/(2.234))=1-P(Z<- 2.059)=0.9802`

Part c

We want this probability:

`P(5\leq X\leq 12)=P((5-9.6)/(2.234)\leq (X-\mu)/(\sigma)\leq (12-9.6)/(2.234))=P(-2.059\leq Z \leq 1.074)`

`=P(Z<1.074)-P(Z<-2.059)=0.8586-0.0197=0.8389`