MATH 217 Test 2 Version A Name: KEY Sec Number: Answer all questions to the best of your ability. Note you should show as much work as is possible. For questions answered using Excel be sure to include the Excel code on your exam to ensure full credit. Complete credit will not be earned for just answers. 1. As part of its Silly Walks First Program the Ministry of Silly Walks monitors monthly Silly Walk time by individuals. The program identifies unusual Silly Walk use (time reported doing Silly Walks), and alerts individuals when their Silly Walk use may be considered to low or excessive. The following data represents the monthly Silly Walk use in minutes of a an individual monitored by the ministry. 346 345 489 358 471 442 466 505 466 372 442 461 515 549 437 480 490 429 470 516 (3 pts) What is the 43 percentile of the data set? P 43 = P ERCENT ILE(ARRAY, 0.43) = 461.85 Forty three percent of the data is less than or equal to 461.85. (6 pts) Write out a 5 number summary for the data set: The following table depicts a 5 number summary and how it is computed. Stat Value Code Min 345 =QUARTILE(ARRAY,0) Q1 435 =QUARTILE(ARRAY,1) Q2 466 =QUARTILE(ARRAY,2) Q3 489.25 =QUARTILE(ARRAY,3) Max 549 =QUARTILE(ARRAY,4) (6 pts) Between what two values does the middle 50 percent of the data span? Beyond what point would the ministry say that an individual has Silly Walked for an excessive amount of time during a given month? Here we Need the IQR and the upper and lower fences for the data set. IQR = Q3 Q1 = 489.25 435 = 54.25 Upper Fence = Q3 + 1.5(IQR) = 570.625 Lower Fence = Q1 1.5(IQR) = 353.625
Yielding all the information needed to create a box plot for the data set. A person would have excessively silly walked if they had done it for more than 570.625 minutes. (6 pts) Create a box plot for the Ministries data. (4 pts) The Ministry decides to use the lower fence as the cut-off point for the number of minutes at which point the individual should be contacted (for not Silly Walking enough). How much time should a person spend Silly Walking each month to avoid being contacted by the Ministry according to the data set provided? Was the individual in the data set ever contacted? The individual should spend at least 353.625 minutes doing Silly Walks to avoid being contacted. Lower Fence = Q1 1.5(IQR) = 353.625 There are two months in the data set where the Ministry should have notified the individual they are not Silly Walking enough. The months with 345 and 346 minuets of Silly Walking. (8 pts) Given the value 461 in row three column two of the provided data set. What is a z score corresponding to this value. Interpret the z score value you have obtained. In order to compute the z-score we need the sample mean and standard deviation. x = 20 i=1 x i 20 = 452.45 s = ST DEV (ARRAY ) = 57.99680118
461 x z 461 = = 0.147421924 s The value of 461 is approximately 0.147 standard deviations above the mean of the data set. 2. The following data set represents the relationship between the Credit Score and interest rates of a 36 month auto loan from the Flying Circus Industrial Bank. Credit Score Interest Rate (Percent) Credit Score Interest Rate (Percent) 545 18.982 675 8.612 595 17.967 705 6.680 640 12.218 750 5.150 (4 pts) Find the least-squares regression line treating the credit score as the explanatory variable and the interest rate as the response variable. Using the add a trend-line feature in Excel we can obtain: y = 0.0762x + 61.225 where y represents the predicted interest rate and x represents a credit score. (4 pts) Interpret the slope and y-intercept if appropriate. Note that credit scores have a range from (300-850). The slope value of -0.0762 may be interpreted as the percent by which the interest rate offered on a loan will decrease for every point the credit score improves. Thus if your credit score goes up one point your potential interest rate will come down -0.0762 percent. A credit score of zero is not appropriate and thus has no interpretation. (4 pts) Predict the interest rate a person with a credit score of 723. 0.0762(723) + 61.225 = 6.1324 The model predicts a person with a credit score of 723 would be offered a loan with an interest rate of 6.13%. (4 pts) Suppose that a Lumber Jack is offered an interest rate of 8.3% and his credit score is 680. Is this a good offer? Why? 0.0762(680) + 61.225 9.40899 Given the credit score of 680 would be offered a loan at approximately 9.4 %. An offer of 8.3 % is better than the model predicts it should be, so it is a solid offer.
(4 pts) What portion of the total variation in the interest rate is explained by the least squares regression? The value of r 2 = 0.9516 and thus the linear regression explains 95.1% of the variation in the interest rate. 3. (4 pts) Assuming the probability of a boy and a girl is equally likely. What is the probability that a family with exactly five children has 5 girls. P (5 girls) = (0.5) 5 = 0.03125 The probability of a family having 5 girls is 3.125 %. 4. The following table represents the employment status and gender of the civilian labor force ages 16-19 (in thousands). Male Female Employed 2328 2509 Unemployed 898 654 Not Seeking Employment 5416 5237 the civilian labor force is employed? P (employed) = (2328+2509)/(2328+2509+898+654+5416+5237) = 0.2838281 The probability of selecting an employed individual is 28.38%. the civilian labor force is male? P (male) = (2328+898+5416)/(2328+2509+898+654+5416+5237) = 0.5071001056 The probability of randomly selecting a male is 50.71%. the civilian labor force is employed and male? P (employed and male) = (2328)/(2328+2509+898+654+5416+5237) = 0.1366036850 Thus the probability of being employed and male is 13.66%.
(4 pts) What is the probability that a randomly selected 16-19 year old individual form the civilian labor force is employed or male? P (employed or male) = P (employed) + P (male) P (employed and male) = 0.2838 + 0.5071 0.1366 = 0.654299 The probability of selecting an individual who is employed or male is 65.43%.