Question

A company is in the business of finding addresses of long-lost friends. The company claims to have a 70% success rate. Suppose that you have the names of nine friends for whom you have no addresses and decide to use the company to track them. (a) Make a histogram showing the probability of r = 0 to 9 friends for whom an address will be found. Maple Generated Plot Maple Generated Plot Maple Generated Plot Maple Generated Plot (b) Find the mean and standard deviation of this probability distribution. What is the expected number of friends for whom addresses will be found? (Round your answers to two decimal places.) μ = friends σ = friends (c) How many names would you have to submit to be 97% sure that at least two addresses will be found? (Enter your answer as a whole number.) names

Answer 1

Answer:

a) Figure attached

b) $E(X) =\mu= np = 9*0.7=6.3$

$Sd(X) =\sigma= \sqrt{np(1-p)}= \sqrt{9*0.7*(1-0.7)}=1.375$

c) For the case n= 6

$P(X \leq 2) = 0.9891$

For the case n= 5

$P(X \leq 2) = 0.9692$

So then we need at least  n=5 or n=6 to satisfy the condition required.

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:

$X \sim Binom(n=9, p=0.7)$

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

For this case we can use the following R code:

> x <- seq(0,9,by = 1)

> y <- dbinom(x,9,0.7)

> plot(x,y,main="Histogram",type = "h")

And we can see on the figure attached the solution.

We see that the higher probabilities are from 4 to 9

Part b

The expected value is given by:

$E(X) =\mu= np = 9*0.7=6.3$

The variance is given by:

$Var (X) =\sigma^2= np(1-p) = 9*0.7*(1-0.7)= 1.89$

And the standard deviation is:

$Sd(X) =\sigma= \sqrt{np(1-p)}= \sqrt{9*0.7*(1-0.7)}=1.375$

Part c

First we can find the probability that at least two addresses will be found in the list of 9 that we have like this:

$P(X \geq 2)$

We can use the complement rule and we have:

$P(X \geq 2) = 1-P(X$

We find the indicidual probabilities:

$P(X=0)=(9C0)(0.7)^0 (1-0.7)^{9-0}=0.00001968$

$P(X=1)=(9C1)(0.7)^1 (1-0.7)^{9-1}=0.000413$

$P(X \geq 2) = 1-[0.00001968+0.000413]=0.9996$

If we use the case of n=8 and we find $P(X\leq 2)$, we got:

$P(X \leq 2) = 0.9987$

For the case n= 7

$P(X \leq 2) = 0.9962$

For the case n= 6

$P(X \leq 2) = 0.9891$

For the case n= 5

$P(X \leq 2) = 0.9692$

So then we need at least  n=5 or n=6 to satisfy the condition required.