Evaluate the expression 11) 11 C_4 A) 5040 B) 330 C) 3 D) 1980 Find the mean of

Question

Evaluate the expression 11) 11 C_4 A) 5040 B) 330 C) 3 D) 1980 Find the mean of the given probability distribution. The accompanying table shows the probability distribution for x_f the number that shows up when a loaded die is rolled. A) mu = 4.14 B) mu = 4.01 C) mu = 0.17 D) mu = 3.50 Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. Round to three decimal places. n = 30, x = 12, p = 0.20 A) 0.003 B) 0.014 C)0.108 D) 0.006 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 - 2sigma or greater than mu + 2sigma. 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 797 persons with broken jaws? A) Yes B) No

Explanation / Answer

(11) 11C4 = 11P4 / 4! = 11 * 10 * 9 * 8 / 24 = 330

Option B

(12)

Option A

(13) P(x) = nCx p^x q^(n - x)

P(12) = 30C12 0.2^12 0.8^(30 - 12) = 0.0064

Option D

(14) = np = 9571 * 0.08 = 765.68

= (npq) = (9571 * 0.08 * 0.92) = 26.54

- 2 = 765.68 - 2 * 26.54 = 712.6 and + 2 = 765.68 + 2 * 26.54 = 818.76

Since 712.6 < 797 < 818.76, the outcome is not unusual

Option B.

x P(x) x * P(x) 1 0.14 0.14 2 0.11 0.22 3 0.14 0.42 4 0.1 0.4 5 0.1 0.5 6 0.41 2.46 4.14

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