Please answer questions 5-25 5 The average speed of all vehicles on a highway is
ID: 3173008 • Letter: P
Question
Please answer questions 5-25
5 The average speed of all vehicles on a highway is µ = 64 mph with a standard deviation of = 5 mph. A random sample of 20 vehicles is clocked. What is the probability that the sample mean will exceed 66 mph? a 0.0525 b 0.0404 c 0.0367 d 0.0220 6 In the previous question what fraction of sample means from samples of n = 20 vehicles fall within ±1 mph from the population mean speed? a 0.6266 b 0.6528 c 0.6922 d 0.7286 7 If you doubled the sample size in the previous question, what fraction of the sample means would fall within ±1 mph from the population mean speed? a 0.7108 b 0.7924 c 0.8262 d 0.8558 8 In the previous question where n = 40, the middle interval which includes the 95% of the mean speed from samples of n = 40 vehicles is, a 61.63 66.37 b 62.45 65.55 c 63.05 64.95 d 63.37 64.63 Questions 4-8 are based on the following: The mean cost of getting a four-year college degree in a certain region of the country is $48,600 with a standard deviation of $8,100. Assume costs are normally distributed. 9 The fraction of costs for all graduates in this region that fall within ±$4,000 of the mean cost is? a 0.5284 b 0.4844 c 0.4246 d 0.3758 10 What fraction of sample means from samples of size n = 16 graduates fall within ±$4,000 from the population mean? a 0.8904 b 0.9198 c 0.9398 d 0.9522 11 In repeated sampling of n = 25 graduates, the interval which contains the middle 95% of sample mean costs is: x = ______, x = ______ a $41,615 $55,585 b $42,567 $54,633 c $43,837 $53,363 d $45,425 $51,775 12 In another region 5% of the x values from samples of size n = 25 are under $48,000 and 5% are over $57,000. From this sampling distribution information we can conclude that the population mean cost of a four-year college degree is = _______. µ a $50,500 b $51,500 c $52,500 d $53,500 13 In the previous question, the population standard deviation is = ______. a $12,980 b $13,720 c $14,240 d $14,760 Questions 14-17 are based on the following: The mean annual Medicare spending per enrollee is $13,800 with a standard deviation of $3,960. Answer questions 14-17 based on the sampling distribution of x for random samples of size n = 81 enrollees. 14 The fraction of sample means falling within ±$600 from the population mean is ______. a 0.9026 b 0.8849 c 0.8675 d 0.8262 15 95% of all x values from samples of size n = 81 deviate from the population mean of $13,800 by no more than ±$______. a $939 b $903 c $862 d $790 16 In repeated sampling of n = 81 enrollees, the middle interval which includes the middle 95% of sample mean spending is: x = ______, x = ______ a $12,380 $15,220 b $12,796 $14,804 c $12,938 $14,662 d $13,010 $14,590 17 In the previous question, to reduce the margin of error such that the middle 95% of all sample means deviate from the population mean by no more than ±$300, the minimum sample size is ______. a 625 b 670 c 744 d 805 The following binary data represent the students taking E270, where "1" is for students who are business majors and "0" for other majors. 1 0 1 1 1 0 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 0 1 1 1 1 1 1 1 0 0 1 0 1 1 1 1 1 1 1 1 0 1 1 0 1 0 1 1 1 1 1 1 1 1 1 18 If we take repeated samples of size n = 40 from this population of E270 students, the expected value of sample proportions would be ______. Use Excel!! a 0.86 b 0.82 c 0.78 d 0.75 Questions 19-22 are based on the following information Among all adult Indiana residents 84% are high school graduates. Answer questions 19-22 based on the sampling distribution of p for random samples of n = 500 Indiana adult residents. 19 What fraction of sample proportions fall under 0.82? a 0.0994 b 0.1112 c 0.1334 d 0.1601 20 The fraction of sample proportions obtained from samples of size n = 500 that fall within ±0.04 (4 percentage points) from the population proportion is _________. a 0.9498 b 0.9685 c 0.9854 d 0.9922 21 The lower and upper ends of the interval which contains the middle 95% of all sample proportions obtained from samples of size n = 950 are: p = _______, p = _______ a 0.806 0.855 b 0.817 0.863 c 0.788 0.872 d 0.779 0.881 22 In the previous question, in order to obtain a margin of error of ±0.02 (MOE = 0.02) for the middle interval that contains the middle 95% of all sample proportions, the minimum sample is: n = ______. a 1468 b 1401 c 1356 d 1291 Questions 23-25 are based on the following information Just before a mayoral election a local newspaper polls 450 voters in an attempt to predict the winner. Suppose that the candidate Johnny Comlately has 53% of the votes among all voters in a two-way race. 23 What is the probability that the newspaper’s sample will predict Johnny Comlately losing the election? a 0.1248 b 0.1003 c 0.0865 d 0.0695 24 In repeated polling of n = 450 voters, 95% of sample proportions would deviate from = 0.53, in either direction, by no more than ______ (or _____ percentage points). a 0.034 b 0.039 c 0.046 d 0.052 25 In order to make the probability of wrongly predicting loss at most 5%, the minimum number of voters to be included in the sample should be n = ______? a 625 b 698 c 745 d 940Explanation / Answer
Question 5
Here, we have to find P(Xbar > 66)
P(Xbar > 66) = 1 – P(Xbar < 66)
The z-score formula is given as below:
Z = (Xbar - µ) / [/sqrt(n)]
Z = (66 – 64) / [5/sqrt(20)]
Z = 1.788854
P(Xbar < 66) = P(Z< 1.788854) = 0.963181
P(Xbar > 66) = 1 – P(Xbar < 66)
P(Xbar > 66) = 1 – 0.963181 = 0.036819
Required probability = 0.0367
Question 6
Here, we have to find P(66 – 1 < Xbar < 66 + 1)
P(66 – 1 < Xbar < 66 + 1) = P(65<Xbar<66)
P(65<Xbar<66) = P(Xbar<66) – P(Xbar<64)
Z score for Xbar = 66
Z = (66 – 65)/[5/sqrt(20)] = 0.894427
P(Xbar<66) = P(Z< 0.894427) = 0.814453
Z score for Xbar = 64
Z = (64 – 65)/[5/sqrt(20)] =-0.894427
P(Xbar<64) = P(Z<-0.894427) = 0.185547
P(65<Xbar<66) = P(Xbar<66) – P(Xbar<64) = 0.814453 – 0.185547
Required probability = 0.6266
Question 7
Here, we have to find P(66 – 1 < Xbar < 66 + 1)
P(66 – 1 < Xbar < 66 + 1) = P(65<Xbar<66)
P(65<Xbar<66) = P(Xbar<66) – P(Xbar<64)
Z score for Xbar = 66
Z = (66 – 65)/[5/sqrt(40)] = 1.264911
P(Xbar<66) = P(Z< 1.264911) = 0.897048
Z score for Xbar = 64
Z = (64 – 65)/[5/sqrt(40)] = -1.26491
P(Xbar<64) = P(Z< -1.26491) = 0.102952
P(65<Xbar<66) = P(Xbar<66) – P(Xbar<64) = 0.897048 - 0.102952
Required probability = 0.7924
Question 8
For middle 95% area, the Z scores are -1.95996 and 1.95996.
X = mean + Z*(SD/sqrt(n))
X = 65 – 1.95996*(5/sqrt(40)) = 63.45052
X = 65 + 1.95996*(5/sqrt(40)) = 66.54948
Answer: 63.37 and 64.63
(There would be approximation error possible in the answers due to limitation of accuracy of z-table values. Most of the tables provide accurate values up to four digits.)