34.You are working on a multiple choice test which consists of 25problems. Each
ID: 2950422 • Letter: 3
Question
34.You are working on a multiple choice test which consists of 25problems. Each of the problems has 5 answers, only one ofwhich is correct. If you are totally unprepared for the test,what is the probability that the first correct answer you mark ison the fourth problem?
A) 0.4096
B) 0.1024
C) 0.2410
D) 0.0016
In Melanie’s Styling Salon, the time to complete a simplehaircut is normally distributed with a mean of 25 minutes and astandard deviation of 4 minutes.
43. For a simple haircut, the middle 90 percent of the customerswill require
A) between 18.4 and 31.6 minutes.
B) between 19.9 and 30.1 minutes.
C) between 20.0 and 30.0 minutes.
D) between 17.2 and 32.8 minutes.
45.Assume that X is normally distributed with a mean m =$64. If P(X ³ $75) = 0.2981, then thestandard deviation of X is approximately equal to
A) $0.53
B) $20.75
C) $13.17
D) $4.84
The average number of patients arriving per hourat a certain outpatient clinic of a large hospital is 24. Thepatients arrive independently of one another and at a uniformaverage rate
33.The probability of zero patient arrivals in the next 4 minutesis
A) .2019
B) .0183
C) .1126
D) .0225
Explanation / Answer
A) 0.4096
B) 0.1024 (.8)3(.2)
C) 0.2410
D) 0.0016
In Melanie’s Styling Salon, the time to complete a simplehaircut is normally distributed with a mean of 25 minutes and astandard deviation of 4 minutes.
43. For a simple haircut, the middle 90 percent of the customerswill require
A) between18.4 and 31.6 minutes. P[-1.65 z 1.65] -----> P[18.4 x 31.6]
B) between 19.9 and 30.1 minutes.
C) between 20.0 and 30.0 minutes.
D) between 17.2 and 32.8 minutes.
45.Assume that X is normally distributed with a mean m =$64. If P(X ³ $75) = 0.2981, then thestandard deviation of X is approximately equal to
A) $0.53
B) $20.75 (75-64)/ =0.53 ---> = 20.75
C) $13.17
D) $4.84
The average number of patients arriving per hourat a certain outpatient clinic of a large hospital is 24. Thepatients arrive independently of one another and at a uniformaverage rate
33. Theprobability of zero patient arrivals in the next 4 minutes is
A) .2019 Poisson (1.60*e01.6)/0! = .2019
B) .0183
C) .1126
D) .0225
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.