please solve the following: Consider a binomial experiment with n = 20 and pi =
ID: 1179074 • Letter: P
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please solve the following:
Consider a binomial experiment with n = 20 and pi = 0.35. Let x denote the number of successes. f(x = 5) = . 0.1387 0.1272 0.0982 0.0566 f(x 3) = f(x 3) = 0.0444 0.9556 0.0444 0.9879 0.0508 0.9492 0.0682 0.9318 E(x) = . 7 6 5 4 sd(x) = . 3.21 2.95 2.13 1.98 Suppose you are receiving a shipment of Gizmos in lots of 5,000. The manufacturing standards provide that 1% of Gizmos turn out defective. You randomly select a sample of n = 30 from each lot and return the shipment if more than 2 are found defective. The probability of returning a 0.0645 0.0457 0.0217 0.0033 The E270 common departmental final has 30 multiple questions. Each question has 5 choices. The instructor bases the test score from a scale of 100. If you guessed the answers for all 30 questions, what is the expected score? 26 24 20 18 John is a speed demon. The probability that he will get a speeding ticket while driving to and from work is 22 percent Answer the next three question based on a work week. What is the probability that John will get 2 tickets during the week (Monday-Friday)? 0.0879 0.1138 0.1712 0.2297 What is the probability that John will get at least one ticket during the work week? 0.7113 0.7464 0.7781 0.8196 If each ticket costs John $120 what is the average cost of speeding violation in the 5-day period? $144 S132 $120 $108Explanation / Answer
n = 20 p = 0.35
12 f(x=5) = 20 C5 0.35^5 0.65^15 = 0.1271991857097322 = 0.1272 option b is answer.
13. f(x<=3) = P(x=0) + P(x=1) + P(x=2) + P(x=3)
= 20 C0 0.35^0 0.65^20 + 20 C1 0.35^1 0.65^19 + 20 C2 0.35^2 0.65^18 + 20 C3 0.35^3 0.65^17
= 0.0443756034464517
= 0.0444
f(x>=3) = 1- f(x<3) = 1 - 0.0121177067103402 = 0.9878822932896598
option b is answer.
14) E(x) = mean = np= 20*0.35 = 7 option a is answer.
15) sd(x) = sqrt(npq) = sqrt(20*0.35*0.65) = 2.1330729007701542
option c is answer.
16) n = 30 p = 0.01
poisson distribution with mean = np = 30*0.01 =0.3
P(x>2) = 1 - p(x>=2) = 1- P(x=0 - p(x=1) - p(x=2)
= 1 - e^-0.3(1 + 0.3 + 0.3^2/2!)
=0.0035994931830894245
option d is answer.
17) n = 30
p = 1/5 = 0.2
expectation = np = 30*0.2 = 6 on scale of 100
expecation = 100/30*6 = 20
option c is answer.
18) p = 0.22
n = 5
P(x=2) = 5 C 2 0.22^2 (1-0.22)^3 = 10*0.22^2 * 0.78^3 = 0.2297
option d is answer.
19) P(x>=1) = 1- P(x<1) = 1-P(x=0)
= 1 - 5C0 0.22^0 0.78^5
= 0.7112825632
0.7113
option a is answer.
20) mean = 120*5*0.22 = 132 = ( cost * n * probablity)
option b is answer.
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