Question

USA Today reported that about 20% of all people in the United States are illiterate. Suppose you take eight people at random off a city street. (a) Make a histogram showing the probability distribution of the number of illiterate people out of the eight people in the sample. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot Incorrect: Your answer is incorrect. (b) Find the mean and standard deviation of this probability distribution. (Round your answers to two decimal places.) μ = people σ = people Find the expected number of people in this sample who are illiterate. (Round your answer to two decimal places.) people (c) How many people would you need to interview to be 97% sure that at least seven of these people can read and write (are not illiterate)? people Need Help? Read It Watch It

Answer 1

Answer:

a) Figure and code attached

b) $E(X)=\mu = np = 8*0.2= 1.6$

$Var(X) =\sigma^2= np(1-p) = 8*0.2*(1-0.2) = 1.28$

$Sd(X)=\sigma= \sqrt{1.28}= 1.131$

c) $P(X\geq 7) =0.97$

And we can calculate this with the complement rule.

$P(X \geq 7) = 1-P(X<7) = 1-P(X\leq 6) =0.97$

So then we have:

$P(X \leq 6) = 0.03$

And we are interested on the valueof n who satisfy this expression.

And for this we can verify this with the following code:

"=BINOM.DIST(6,E54,0.2,TRUE)"

And as we can see on the second figure attached the value who satisfy the condition would be n = 60.

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:

$X \sim Binom(n=8, p=0.2)$

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:

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

Part a

We can use the following R code to generate the histogram for this case:

> x <- seq(0,8,by = 1)

> y <- dbinom(x,8,0.2)

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

And as we can see we got the result on the figure attached. And the distribution seems to be skewed to the right.

Part b

For this case the expected value is given by:

$E(X)=\mu = np = 8*0.2= 1.6$

The variance is given by:

$Var(X) =\sigma^2= np(1-p) = 8*0.2*(1-0.2) = 1.28$

And the standard deviation would be:

$Sd(X)=\sigma= \sqrt{1.28}= 1.131$

Part c

For this case we have the following inequality:

$P(X\geq 7) =0.97$

And we can calculate this with the complement rule.

$P(X \geq 7) = 1-P(X<7) = 1-P(X\leq 6) =0.97$

So then we have:

$P(X \leq 6) = 0.03$

And we are interested on the valueof n who satisfy this expression.

And for this we can verify this with the following code:

"=BINOM.DIST(6,E54,0.2,TRUE)"

And as we can see on the second figure attached the value who satisfy the condition would be n = 60.

Answer 2

Binomial distribution is distribution tracking number of successes in a set of Bernoulli trials. The figures needed for given context are:

• Mean = 1.6
• Standard deviation = 1.13 approx
• Expected number of people = 2
• Number of people needed to interview to be 97% sure that at least seven of them can read is 60

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

The probability that out of n trials, there'd be x successes is given by

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

$E(X) = np\\Var(X) = np(1-p)$

For the given case, let the random variable X tracks the number of illiterates (success) in sample of 8 ( = n )  people taken from US.

Chance of a person being illiterate = 20% = 0.2 = probability of success = p

Thus, $X \sim B(8, 0.2)$

Thus, mean = expectation = E(X) = np = 0.2 times 8 = 1.6

Standard deviation of X = $\sqrt{Var(X)} = \sqrt{np(1-p)} = \sqrt{8\times 0.2\times (0.8)} \approx 1.13$

Expected value is mean itself. Thus, expected number of people = 1.6 $\approx 2$ (since number of people needs to be whole number)

Let we have same success probability(as it was given for whole US, not for those 8 people), and let we select n people such that at least 7 of them are literate can be said with being 97% sure.

Then

$X \sim B(n, 0.2)$

P(at least 7 out of n people are literate) = P( at max n-7 people are illiterate) = P(X ≤ n-7) = 1 - P(X > n-7) = 0.97

or

P(X = n -6) + P(X = n-5) + P(X = n-4) + P(X = n-3) + P(X = n-2) + P(X = n-1) + P(X = n)] = 0.03

From calculator, we get n = 60

The figures needed for given context are:

• Mean = 1.6
• Standard deviation = 1.13 approx
• Expected number of people = 2
• Number of people needed to interview to be 97% sure that at least seven of them can read is 60

Learn more about binomial distribution here:

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