Determine if the outcome is unusual. Consider as unusual any result that differs from the mean by more than 2 standard deviations. That is, unusual values are either less than $\mu - 2 \sigma$ or greater than $\mu + 2 \sigma$.
-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 792 persons with broken jaws?
A)No
B)Yes

Question

Quizplus · Accepted Answer

The expected number of broken jaws, given the 8% rate, is $0.08 \times 9571 = 765.68$. The standard deviation for a binomial distribution is $\sqrt{n p (1-p)} = \sqrt{9571 \times 0.08 \times (1-0.08)} \approx 25.5$. The range for usual outcomes is $765.68 \pm 2 \times 25.5$, or approximately $714.68$ to $816.68$. Since 792 is within this range, it is not unusual.

Determine If the Outcome Is Unusual μ−2σ\mu - 2 \sigmaμ−2σ Or Greater Than

Determine If the Outcome Is Unusual $\mu - 2 \sigma$ Or Greater Than