Question

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?

Answer 1

To find the probability that a randomly selected ornament will cost more than $10, calculate the z-score and find the area to the right of that z-score on a standard normal distribution table. The probability that exactly 3 of the 8 randomly selected ornaments cost over $10 can be found using the binomial probability formula.

Step-by-step explanation:

A) To find the probability that a randomly selected ornament will cost more than $10, we need to calculate the z-score and find the area to the right of that z-score on a standard normal distribution table. The formula to calculate the z-score is (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation. In this case, x = $10, μ = $7.65, and σ = $1.45. The z-score is (10 - 7.65) / 1.45 = 1.655. Using a standard normal distribution table, the area to the right of a z-score of 1.655 is 0.0495. Therefore, the probability that a randomly selected ornament will cost more than $10 is 0.0495 or 4.95%.

B) To find the probability that exactly 3 of the 8 randomly selected ornaments cost over $10, we need to use the binomial probability formula. The formula is P(X = k) = C(n, k) * p^k * (1-p)^(n-k), where P(X = k) is the probability of getting exactly k successes, n is the number of trials, p is the probability of success in each trial, and C(n, k) is the number of combinations of n items taken k at a time. In this case, n = 8, k = 3, and p = 0.0495 (probability of an ornament costing over $10). Plugging in these values, we get P(X = 3) = C(8, 3) * (0.0495)^3 * (1-0.0495)^(8-3) = 0.0034. Therefore, the probability that exactly 3 of the 8 randomly selected ornaments cost over $10 is 0.0034 or 0.34%.

Answer 2

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!! :)