DOC PREVIEW
WCU ECO 251 - ECO 251 Third Hour Exam

This preview shows page 1-2-3-4 out of 13 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 13 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

TABLE 5-4 Not only is this almost identical to Grass2. The probabilities for x are still the same as on the last exam! Why are so many of you still trying to compute sample means and variances when there is no sample here?ANSWER: 0.90 = .20 + .45 + .15 + .05 + .05 = 1 - .10ANSWER: 2.0ANSWER: 1.18322 because and 1.18322 is the square root.ANSWER: 2.40ANSWER: .6795ANSWER:ANSWER: Binomial = 0.9898To find Note that zero successes correspond to 5 failures when . The probability of failure is .4 and or use the Binomial formula. , so251y0332 12/01/03 ECO251 QBA1 THIRD HOUR EXAMNov 24, 2003 Name: ____KEY_____________ Social Security Number: _____________________Class Time (Circle) 10am 11amPart I: (30+ points) Do all the following: All questions are 2 points each except as marked. Exam is normedon 50 points including take-home. (Showing your work can give partial credit on some problems!)1. In its standardized form, the normal distributiona) *has a mean of 0 and a standard deviation of 1.b) has a mean of 1 and a variance of 0.c) has an area equal to 0.5.d) cannot be used to approximate discrete probability distributions.2. For some value of z, the probability that a standardized normal variable is below z is 0.2090. The value of z isa) *– 0.81.b) – 0.31.c) 0.31.d) 1.96.Solution: Your diagram of the normal distribution should show a Normal curve with a mean at zero. 50% of the area is below zero, so a number with 20.9% below it must be below zero too. If the number has 20.9% below it, it must have 79.1% above it. The probability that a value of zis above zero is 50%, so the probability between the number we want and zero is 50% - 20.9% = 20.1%. According to the Normal table,  .2910.81.00 zP This means that  2910.081.0  zPtoo. So our value is -0.81.3. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will find a parking spot in the library parking lot in less than 3 minutes. Make a diagram!a) 0.3551b) *0.3085c) 0.2674d) 0.1915Solution:  1,5.3~ Nx      3085.1915.5.050.0050.015.333zPzPzPzPxPYour diagram for x should show a Normal curve with a mean at 3.5. 3 lies to the left of 3.5. Shade the area below 3. You must take the area below 3.5 and subtract the area between 3 and 3.5. If you make this diagram, put 3.5 in the middle, not zero!Your diagram for z should show a Normal curve with a mean at zero. -0.5 lies to the left of zero. Shade the area below -0.5. You must take the area below 0 and subtract the area between -0.5 and zero. Most people seem to be happier if they make the z diagram instead of thex diagram.251y0332 11/18/034. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot. Make a diagram!a) 0.0919b) 0.2255c) 0.4938d) *0.7745Solution:  1,5.3~ Nx      7745.3413.4332.00.10050.100.150.115.35.415.325.42zPzPzPzPxPYour diagram for x should show a Normal curve with a mean at 3.5. 2 lies to the left of 3.5, and 4.5 to the right of 3.5. Shade the area between 2 and 4.5. You must take the area between 2 and 3.5 and add the area between 3.5 and 4.5.Your diagram for z should show a Normal curve with a mean at zero. -1.5 lies to the left of zero, and 1 lies to the right of zero. Shade the area between -1.5 and 1. You must take the area between -1.5 and 0 and add the area between zero and 1.00.TABLE 5-4 Not only is this almost identical to Grass2. The probabilities for x are still the same as on the last exam! Why are so many of you still trying to compute sample means and variances when there is no sample here?The following table contains the joint probability distribution for x and y.Show your work neatly in questions 6-9.xy0 1 2 3 4 50 .10 .20 .25 0 0 06 0 0 .20 .15 .05 .05Total .10 .20 .45 .15 .05 .055. Referring to Table 5-4, the probability that x is at least 1 is ________.ANSWER: 0.90 = .20 + .45 + .15 + .05 + .05 = 1 - .10Work for questions 6-9 follows.2        20.1670.240.500.200.125.180.035.180.120.0025.20.45.90.20.005.05.15.45.20.10.20.16070.2045.55.05.05.15.20.0000025.20.10.6054321022xPxxxPxPyyPyyyPyPx3251y0332 11/18/03 To summarize  1xP,    00.2xxPxEx,  40.522xPxxE  1yP,    70.2yyPyEy and  20.1622yPyyE                                     8.75.12.17.24.2000000006505.6405.6315.6220.6106000500400300225.0120.0010.xyxyPxyE     400.270.200.28.7 yxxyxyExyCov,   40.100.240.52222xxxE and  91.870.220.162222yyyE. So that   6795.46176.91.840.1400.291.840.1400.22yxxyxy. (9850.2,1832.140.1 yx) Note that 11 xy always!6. Referring to Table 5-4, the mean or expected value of x is ________.ANSWER: 2.07. Referring to Table 5-4, the standard deviation of x is ________.ANSWER: 1.18322 because   40.100.240.52222xxxE and 1.18322 is the square root.8. Referring to Table 5-4, the covariance of x and y is ________. (4)ANSWER: 2.409. Referring to Table 5-4, the correlation between x and y is ________.ANSWER: .679510. If x has a binomial distribution with n = 4 and p = 0.3, then  1xP = ________ .ANSWER:    3483.6517.1111  xPxP11. Suppose that past history shows that 60% of college


View Full Document

WCU ECO 251 - ECO 251 Third Hour Exam

Documents in this Course
Load more
Download ECO 251 Third Hour Exam
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view ECO 251 Third Hour Exam and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view ECO 251 Third Hour Exam 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?