Question

The Acme Candy Company claims that 8% of the jawbreakers it produces actually result in a broken jaw. Suppose 9571 persons are selected at random from those who have eaten a jawbreaker produced at Acme Candy Company. Would it be unusual for this sample of 9571 to contain 801 persons with broken jaws?

Answer 1

`E(X) = 9571*0.08=765.68`

And the deviation:

`\sigma = √(np(1-p))=√(9571*0.08*(1-0.08))= 26.54`

If we use the rule of thurm and the normal approximation we can find the limits usual for the random variable X, and we got:

`Lower= 765.68 -2*26.54 = 712.60`

`Upper= 765.68 +2*26.54 = 818.76`

So then since the value of 801 i contained on the range of usual values then is not unusual obtain a value of 801 for the sample

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 "number of people with broken jaws", on this case we now that:

`X \sim Binom(n=9571, p=0.08)`

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!)`

The expected value is given by:

`E(X) = 9571*0.08=765.68`

And the deviation:

`\sigma = √(np(1-p))=√(9571*0.08*(1-0.08))= 26.54`

If we use the rule of thurm and the normal approximation we can find the limits usual for the random variable X, and we got:

`Lower= 765.68 -2*26.54 = 712.60`

`Upper= 765.68 +2*26.54 = 818.76`

So then since the value of 801 i contained on the range of usual values then is not unusual obtain a value of 801 for the sample

Answer 2

In would not be unusual for this sample of 9571 to contain 801 persons with broken jaws

Explanation:

For each person, there are only two possible outcomes. Either they have broken jaws, or they do not. The probability of a person having a broken jaw is independent of any other person. So we use the binomial probability distribution to solve this question.

aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Values that are smaller than
`E(X) - 2.5√(V(X))` or greather than
`E(X) + 2.5√(V(X))` are considered unusual.

In this problem, we have that:

`n = 9571, p = 0.08`

So

`E(X) = np = 9571*0.08 = 765.68`

`√(V(X)) = √(np(1-p)) = √(9571*0.08*0.92) = 26.54`

Range of usual values:

`E(X) - 2.5√(V(X)) = 765.68 - 2.5*26.54 = 699.33`

`E(X) + 2.5√(V(X)) = 765.68 + 2.5*26.54 = 832.03`

801 is in the range of usual values

In would not be unusual for this sample of 9571 to contain 801 persons with broken jaws