MAT150 Exam 2 Review: Practice Questions

EXAM #2 REVIEW PRACTICE 1. A study advisory board consists of 22 members. Four members will be chosen to serve as the board’s chair, secretary, counselor and secretary. Each member is equally likely to serve in either of the positions. What is the probability of randomly selecting the four...

EXAM #2 REVIEW PRACTICE 1. A study advisory board consists of 22 members. Four members will be chosen to serve as the board’s chair, secretary, counselor and secretary. Each member is equally likely to serve in either of the positions. What is the probability of randomly selecting the four members who will be chosen for the board? 2. The manager of the accounting department wants to form a three-person advisory committee from the 20 employees in the department. In how many ways can the manager form this committee? 3. Fifty-five percent of candidates taking the college entrance exam pass the test. You randomly select six candidates. Find the probability that the number of candidates who pass is a) Exactly four b) At least three c) At most three d) More than three 4. Thirty-two percent of U.S. employees who are late for work blame the transportation. You randomly select 10 U.S. employees who are late for work and ask them whether they blame the transportation. The random variable represents the member who is late for work and blames the transportation. a) Find the mean of the binomial distribution for the given random variable. b) Find variance of the binomial distribution for the given random variable. c) Find the standard deviation of the binomial distribution for the given random variable. 5. The monthly utility bills in New York City are normally distributed, with a mean of $120 and a standard deviation of $14. Find the probability that a randomly selected utility bill is a) Less than 90. b) Between 110 and 140 c) More than 160 6. The mean amount of money that U.S. adults spend on gas in a week is $151 and the standard deviation is $49. Random samples of size 50 are drawn from this population and the mean of each sample is determined. a) Find the mean and standard deviation of the sampling distribution of sample means b) What is the probability that the mean amount spent on gas in a week for a certain sample is more than $160? c) What is the probability that the mean amount spent on gas in a week for a certain sample is between $135 and $150? d) What is the probability that the mean amount spent on gas in a week for a certain sample is less than $140?  7. A consumer price analyst claims that prices for liquid crystal display TV monitors are normally distributed, with a mean of $190 and a standard deviation of $48. What is the probability that a randomly selected TV monitor will cost more than 200? You randomly selected 10 TV monitors. What is the probability that their mean cost is more than 200? 8. The ages ( in years) of a random sample of students in a Mathematics class are listed below. Assume that the papulation standard deviation is 1.82. 19 20 17 19 17 21 23 21 17 17 19 19 17 20 23 18 18 18 18 19 a) Find the point estimate of the population mean. b) Construct a 95% confidence interval for the population mean. c) Determine the minimum sample size required to be 95% confident that the sample mean age is within 1/2 years of the population mean age. 9. The recovery times for high school football players that suffer from concussion are listed below 43 57 18 45 47 33 49 24 a) Find the sample mean and sample standard deviation C) Construct a 90% confidence interval for the population mean. 10. In a survey 1015 U.S. adults, 745 says that the energy situation in the U.S. is very serious. a) Find the point estimate for the population proportion. b) Construct a 90% confidence interval for the population proportion c) find the minimum sample size needed to estimate the population proportion at the 99% confidence level to ensure that the estimate is accurate, within 4% of the population proportion. 11. se the table below x 0 1 2 3 4 5 f 277 471 243 105 46 22 a) Construct a probability distribution b) Graph the probability distribution using a histogram c) Find the mean, variance, and standard deviation of the probability distribution d) Find the probability of randomly selecting at least four people. ANSWER KEYS 1. The probability of randomly selecting the four members is: 1 P(four members) = =0.000006 22 P 4 2. In how many ways can the manager form the committee 20! 20C 3= =1140ways ( 20−3 ) !3! 3. p=0.55 ,q=0.45 n=6 , x=0 ,1 , 2 , 3 , 4 , 5 , 6 4 2 6! A. P ( x=4 )=6 C 4 (0.55) (0.45) = (0.55)4 (0.45)2=0.28 ( 6−4 ) !4 ! B. P ( x ≤3 )=P ( 0 ) + P ( 1 )+ P ( 2 ) + P ( 3 )=0.0083+0.061+0.186+ 0.303=0.5583 C. P ( x ≥ 3 )=P ( 3 ) + P ( 4 ) + P ( 5 )+ P ( 6 ) =0.303+0.28+0.136+ 0.028=0.747 D. P ( x >3 )=1−P ( x ≤ 3 )=1−0.5583=0.442 4. The mean: np=10 ( 0.32 )=3.2 The variance: σ 2 =npq=10 ( 0.32∗0.68 )=2.176 The standard deviation: σ =√ npq σ =√ 2.176=1.48 5. Less than 90 90−120 z= =−2.14 14 A. P ( x 2.86) =1−0.9979=0.0021 6. Find the mean μ x =μ=151 σ 49 σ x= = =6.93 √n √ 50 (160−151) √50 B. =1.30 49 P ( x >160 )=P ( Z >1.3 ) =1−0.9032=0.0968 (135−151) √ 50 C. Z 1 = =−2.31 49 (150−151) √ 50 z2 = =−0.14 49 P ( 135< x< 150)=P (−2.13< Z ←0.14 )=0.4483−0.0104=0.4379 (140−151) √ 50 D. Z = =−1.59 49 P ( x 200 )=P ( z >0.21) =1−0.5832=0.4168 (200−190) √10 B) Z = =0.66 48 P ( x >200 )=P ( z >0.66) =1−0.7454=0.2546 8.  A. The point estimate: x= ∑ x = 380 =19 n 20 B. Construct a 95% confidence interval for the population mean. Z σ 1.96∗1.82 E= C = =0.80 √n √20 x−E