Question 1 (1 point) Fill in the blanks. Optionshouse.com tracked the performanc
ID: 3306623 • Letter: Q
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Question 1 (1 point)
Fill in the blanks. Optionshouse.com tracked the performance of their most active day traders and found that the probability of a winning call option pick was 0.4989. If in a day, 397 call options are picked by these traders, around __________ of them will be winners, give or take __________. Assume each pick is independent.
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Question 2 (1 point)
Fill in the blanks. Suppose the probability of a baseball player getting a hit in an at bat is 0.287. If the player bats 23 times during a week, his number of hits should be around __________, give or take __________. Assume each at bat is independent.
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Question 3 (1 point)
Fill in the blanks. Suppose the probability at a light bulb factory of a bulb being defective is 0.1065. If a shipment of 194 bulbs is sent out, the number of defective bulbs in the shipment should be around __________, give or take __________. Assume each bulb is independent.
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Question 4 (1 point)
When rolling a die 143 times, what is the probability of rolling a 6 between 25 and 28 (inclusively) times?
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Question 5 (1 point)
Imagine that somthing 1 for Statistics 2160 this term will have 52 questions. Each question has 5 multiple choice options, giving you a probability of 0.2 of getting each question right purely by guessing. Assuming that you guess on all questions, what is the probability that you get between 7 and 14 (inclusively) questions right on your somthing ?
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Question 6 (1 point)
Apple's market share in the smart phone market represents a proportion of 0.2853. If Apple conducts a survey nationwide of 500 smart phone users, what is the probability that at least 132 of the people are Apple users?
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Question 7 (1 point)
Pandora Radio has determined that the probability of a user liking a new song enough to give a "thumbs up" is 0.7504. Given that Pandora plays about 827 new songs per user per month on average, what is the probability that exactly 633 of those plays get a "thumbs up"?
1) 9.96 , 198.1 2) 397 , 9.96 3) 198.1 , 0.4989 4) 198.1 , 99.2000 5) 198.1 , 9.96Explanation / Answer
1. There is only two outcomes-success (winning a call option )and failure (not winning a call option). The probability of success p is 0.4989. The n=397 trials are independent, and probability of success is constant throughout the trials. This accounts for binomial distribution.The first and second blank consist of center and spread of the distribution. Blank1: Expected value, mu=np=397*0.4987=198.1, standard deviation, sigma=sqrt np(1-p)=sqrt [397*0.4987*(1-0.4987)]=9.96 Ans>5 [the rest of the answers have numerically incorrect values]
2.
There is only two outcomes-success (getting a hit )and failure (not getting a hit). The probability of success p is 0.287. The n=23 trials are independent, and probability of success is constant throughout the trials. This accounts for binomial distribution.The first and second blank consist of center and spread of the distribution. Blank1: Expected value, mu=np=23*0.287=6.601, standard deviation, sigma=sqrt np(1-p)=sqrt [23*0.287*(1-0.287)]=2.17 Ans>2 [the rest of the answers have numerically incorrect values]
3. There is only two outcomes-success (finding a defective bulb )and failure (not finding a defective bulb). The probability of success p is 0.1065. The n=194 trials are independent, and probability of success is constant throughout the trials. This accounts for binomial distribution.The first and second blank consist of center and spread of the distribution. Blank1: Expected value, mu=np=194*0.1065=20.661, standard deviation, sigma=sqrt np(1-p)=sqrt [194*0.1065*(1-0.1065)]=4.2966 Ans>4 [the rest of the answers have numerically incorrect values]
4. There are two outcomes, success (finding a 6) and failure (not finding a 6). The probability of success, p=1/6, and there are n=143 independent trials. The probability of success, is constant throughout the trials. Both np=143*1/6=23.83 and nq=143*(1-1/6)=119.16 are greater than 10. Use normal approximation to binomial distribution.
Mean, mu=np=23.83 and standard deviation, sigma=sqrt np(1-p)=sqrt [143*1/6*5/6]=4.46
P[25<=X<=28]=P[X<=28]-P[X<=25]=P[z<=(28-23.83)/4.46]-P[Z<=(25-23.83)/4.46]=P(Z<=0.93)-P(Z<=0.26)=0.8238-0.6026=0.2212. Probability cannot be negative and more than 1, hence option 2 and 3 are discaded. Option 1 and 4 are too far from the answer. Option 5 is the closest. The difference is due to rounding off mu and sigma.