(choose the option that is the closest to your answer) Use Excel to answer each
ID: 3361501 • Letter: #
Question
(choose the option that is the closest to your answer) Use Excel to answer each of the following questions. The Problem: According to the American Council of Life Insurance, 40% of U.S. adults are covered by employee group life insurance. Suppose a sample of 30 U.S. adults is randomly selected. Assuming the claim is true, what is the probability that: n=30 1. Exactly 18 of them are covered by employee group life insurance? BINOMDIST(18,30,0.4,0) 0.0129383 1. a) 0.0129 b) 0.0001 c) 0.1299 d) 1.0600 2. 14 or fewer of them are covered by employee group life insurance? 2. a) 0.9000 b) 0.1000 c) 0.8246 d) 0.9946 3. Fewer than 12 of them are covered by employee group life insurance? 3. a)0.5555 b)0.4311 c) 0.4444 d) 0.0242 4. 9 or more of them are covered by emplove roup life nsurance? 4. a) 0.9060 b) 0.9999 c) 0.8888 d) 0.0111 5. More than 9 of them are covered by employee group life insurance? 5. a) 0.8237b) 0.9288 c) 0.7588 d) 0.6655 6. More than 8 but fewer than 12 of them are covered by employee group life insurance? 6. a) 0.7700 b) 0.9990 c) 0.3371 d) 0.4472 7. At least 18 of them are covered by employee group life insurance? 7.Explanation / Answer
2)
P(X <=14) = binomdist(14,30,0,4,1)
3)
P(X<12) = binomdist(12,30,0.4,1)
4)
P(X>=9) = 1 - P(X < 8)
= 1 - binomdist(8,30,0.4,1)
5)
P(X>9) = 1 - P(X<=9)
= 1 - binomdist(9,30,0.4,1)
6)
P(8<X<12) = P(X < 12) - P(X < 8)
= binomdist(11,30,0.4,1) - binomdist(7,30,0.4,1)
7)
P(X>=18) = 1 - P(X <18)
= 1 - binomdist(17,30,0.4,1)
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