Question 17-24 are related to the following The probability distribution for the

Question

Question 17-24 are related to the following The probability distribution for the number of automobiles sold during a day (x) at Bob Iron Motors is as follows. x f(x) 0 0.18 1 0.39 2 3 0.14 4 0.04 5 0.01 17 The probability that 2 automobiles will be sold is, a 0.28 b 0.26 c 0.24 d 0.22 18 The probability that less than 3 automobiles will be sold is, a 0.95 b 0.86 c 0.81 d 0.57 19 The probability that at most 3 automobiles will be sold is, a 0.99 b 0.95 c 0.81 d 0.57 20 The probability that at least 3 automobiles will be sold is, a 0.05 b 0.19 c 0.43 d 0.82 21 The expected value of the number of automobiles sold is, a 1.5 b 1.8 c 2.0 d 2.5 22 The variance of automobiles sold is, a 1.75 b 1.55 c 1.25 d 1.05 23 Suppose the gross profit per vehicle sold is $1,200. The expected value of gross profit is, a 1500 b 1800 c 2400 d 2600 24 If gross profit per vehicle sold is $1,200, the standard deviation of gross profit is, a 1351.86 b 1341.64 c 1303.39 d 1265.14 Question 17-24 are related to the following The probability distribution for the number of automobiles sold during a day (x) at Bob Iron Motors is as follows. x f(x) 0 0.18 1 0.39 2 3 0.14 4 0.04 5 0.01 17 The probability that 2 automobiles will be sold is, a 0.28 b 0.26 c 0.24 d 0.22 18 The probability that less than 3 automobiles will be sold is, a 0.95 b 0.86 c 0.81 d 0.57 19 The probability that at most 3 automobiles will be sold is, a 0.99 b 0.95 c 0.81 d 0.57 20 The probability that at least 3 automobiles will be sold is, a 0.05 b 0.19 c 0.43 d 0.82 21 The expected value of the number of automobiles sold is, a 1.5 b 1.8 c 2.0 d 2.5 22 The variance of automobiles sold is, a 1.75 b 1.55 c 1.25 d 1.05 23 Suppose the gross profit per vehicle sold is $1,200. The expected value of gross profit is, a 1500 b 1800 c 2400 d 2600 24 If gross profit per vehicle sold is $1,200, the standard deviation of gross profit is, a 1351.86 b 1341.64 c 1303.39 d 1265.14

Explanation / Answer

17

Sum of probabilities will be 1

1 = 0.18 + 0.39 + P(2) + 0.14 + 0.04 + 0.01

1 = P(2) + 0.76

P(2) = 0.24

18

P ( X < 3) = P(0) + P (1) + P(2)

= 0.18 + 0.39 + 0.24

= 0.81

19

P ( X < = 3) = P(0) + P (1) + P(2) + P(3)

= 0.18 + 0.39 + 0.24 +  0.14

= 0.95

20

P( at least 3 ) = P( 3) + P(4) + P(5)

= 0.14 + 0.04 +0.01

= 0.19

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