iPad 9:12 PM /0.33 points WaneFM6 8.2.026. According to an article, 14.1% of Int
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Question
iPad 9:12 PM /0.33 points WaneFM6 8.2.026. According to an article, 14.1% of Internet stocks that entered the market in 1999 ended up trading below their initial offering prices. I you were an investor who purchased four Internet stocks at their initial offering prices, what was the probability that at least two of them would end up trading at or above their initial offering price? (Round your answer to four decimall places. My N 0.0.33 points WaneFM6 8.2.029 My N Your manufacturing plant produces air bags, and it is known that 20% of them are defective. Five air bags are tested. (a) Find the probability that four of them are defective. (Round your answer to four decimal places (b) Find the probability that at least two of them are defective. (Round your answer to four decimal places 11. 0.33 points WaneFM6 8.2.031 My N The probabiity that a randomly selected teenager watched a rented video at least once during a week was 0.67. What is the probability that at least 4 teenagers in a group of 6 watched a rented movie at least once last week? (Round your answer to four decimal places.) 12. -/037 points WaneFM6 8.2.032 My N The probability that a randomly selected teenager studied at least once during a week was only 0.69. What is the probability that less than haf of the students in your study group of 10 have studied in the last week? (Round your answer to two decimal places.) Home My AssignmentsExplanation / Answer
Solution9;
p=14.1%=0.141
n=4
binomial distribution with n,p
P(X>=2)
=P(X=3)+P(X=4)
=0.098
Solution10:
n=5
p=prob of success=20/100=0.2
binomial distribution
P(X=4)
=5C4(0.2)^4*(1-0.2)^5-4
=0.00064
P(X>=2)
=P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.2627
sOLUTION11:
p=0.67
n=6
x=4
P(X>=4)
=P(X=4)+P(X=5)+P(X=6)
=6C4*0.67^4*(1-0.67)^6-4+6C5*0.67^5*(1-0.67)^6-5+6C6*0.67^6*(1-0.67)^6-6
=0.6870
ANSWER::0.6870
Solution12
P(X<5)
n=10
x=5
p=0.69
P(X<5)=P(X<=4)
=0.0551096977
=0.06
ANSWER:0.06
used TI 93 CALC
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