AP Statistics

Decorate. A store sells ornaments for Christmas trees. The prices of the ornaments are roughly normally distributed with a mean of $7.65 and a standard deviation of $1.45.

A)What is the probability that a randomly selected ornament will cost more than $10?

B)If eight ornaments are randomly selected, what is the probability that exactly 3 of them cost over $10?

A) For this problem, we will need to use a normal calculation, in that we find the z-score and the area to the right using Table A.

z = (10 - 7.65) / 1.45

z = 1.62

area to the left for a z-score of 1.62 = 0.9474

area to the right for a z-score of 1.62 = 0.0526

The probability that a randomly selected ornament will cost more than $10 is 0.0526 or 5.26%.

B) For this problem, we will use the binomial probability formula since the problem is asking for the probability that exactly 3 ornaments cost over $10. There are two forms of this equation. One is nCr x p^r x q^n-r and the other is (n r) x p^r x (1 - p)^n-r. I will show both formulas below.

8C3 x 0.0526^3 x 0.9474^5

(8 3) x 0.0526^3 x 0.9474^5

With both equations, the answer is the same. Whichever you are more familiar or comfortable with is the one I would recommend you use.

The probability that exactly 3 of the 8 ornaments cost over $10 is 0.00622 or 0.622%.

Hope this helps!! :)

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12 cups   is the answer

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