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?

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

Step-by-step 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:

The standard deviation of the binomial distribution is:

Values that are smaller than or greather than are considered unusual.

In this problem, we have that:

So

Range of usual values:

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

explanation:

remove parenthesis:

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apply rule

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multiply fractions:

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divide:

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mordancy.

d is the only accurate one

step-by-step explanation: