#$&* course MTH 152 11/22/13 9:20 p.m. 018. ``q Query 18
.............................................
Given Solution: `aThe z-score is measured relative to the mean. The mean is 100, and you need to measure the z score of 115. 115 is 15 units from the mean, which gives you a z-score of 15 / 15 = 1. The table tells you that .341 of the distribution lies between the mean and z = 1. You want the proportion beyond 115. Since half the distribution lies to the right of the mean, and .341 of the distribution lies between the mean and z = 1, we conclude that .5 - .341 = .159 of the distribution lies to the right of z = 1. It follows that .159, or 15.9% of the distribution exceeds an IQ of 115. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Query problem 13.5.20 area between z=-1.74 and z=-1.14 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Z-score for -1.74 = .459 Z-score for -1.14 = .373 .459 - .373 = .086 = 8.6% confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aAccording to the table the z score for -1.74 is .373 and the z score for -1.14 is .459, meaning that .373 of the distribution lies between the mean and z = -1.73 and .459 of the distribution lies between the mean and z = -1.14. Since -1.74 and -1.14 both lie on the same side of the mean, the region between the mean and -1.74 contains the region between the mean and -1.14. The region lying between z = -1.14 and z = -1.74 is therefore that part of the .459 that doesn't include the .373. The proportion between z = -1.14 and z = -1.74 is therefore .459 - .373 = .086, or 8.6%. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay ------------------------------------------------ Self-critique Rating: Okay ********************************************* Question: `q Query problem 13.5.30 of 10K bulbs, mean lifetime 600 std dev 50, # between 490 and 720 **** How many bulbs would be expected to last between 490 and 720 hours? **** Describe in detail how you obtained your result. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 490 - 600 = - 110 110/50 = -2.2 720 - 600 = 120 120/50 = 2.4 - 2.2 = .486 2.4 = .492 .486 + .492 = 0.978 So 0.978*10,000= 9,780 light bulbs with a lifetime between 490 and 720. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aYou first calculate the z value for each of the given lifetimes 490 hrs and 720 hrs. You should then sketch a graph of the distribution so you can see how the regions are located within the distribution. Then interpret what the table tells you about the proportion of the distribution within each region and apply the result to the given situation. The details: The displacement from the mean to 490 is 490 - 600 = - 110 (i.e., 490 lies 110 units to the left of the mean). The z value corresponding to 490 hours is therefore z = -110/50 = -2.2. The area of the region between the mean and z = -2.2 is found from the table to be .486. Similarly 720 lies at displacement 720- 600 = 120 from the mean, giving us z = 120/50 = 2.4. The area of the region between the mean and z = 2.4 is shown by the table to be .492. Since one region is on the negative side and the other on the positive side of the mean, the region lying between z = -2.2 and z = 2.4 contains .486 + .492 = .978 of the distribution. Out of 10,000 bulbs we therefore expect that .978 * 10,000 = 9780 of the bulbs will last between 490 and 720 hours. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay ------------------------------------------------ Self-critique Rating: Okay ********************************************* Question: `q Query problem 13.5.48 A's for > mean + 3/2 s What percent of the students receive A's, and how did you obtain your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Formula = greater than x+ 1.5 s. Z =1.5 B grades: .5 / .500 - .433 = 0.067 6.7% Are As confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aA's are given for z scores greater than 1.5. The area between mean and z = 1.5 is given by the table as .433. To the right of z = 1.5, corresponding to the A's, we have .500 - .433 = .067 or 6.7% of the total area. So we expect that 6.7% of the group will receive A's. ** GENERAL ADVICE: To solve problems of the type covered in this section it is a good idea to follow a strategy something like the following: 1. Find the z-score(s) corresponding to the given values. 2. Look up the corresponding numbers on the table. 3. Sketch a graph of the normal distribution representing what the numbers in the table tell you. Be sure you understand that the table tells you the proportion of the distribution lying between the mean and the given z value. 4. Decide what region of the graph corresponds to the result you are trying to find. 5. Find the proportion of the total area lying within this region. 6. If necessary apply this proportion to the given numbers to get your final result. See how this procedure is applied in the given solutions. Then you should probably rework the section, being sure your answers agree with those given in the back of the text, and send me questions about anything you arent sure you understand. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Thanks! ------------------------------------------------ Self-critique Rating: Okay " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course MTH 152 9:20 p.m. 11/22/13 019. ``q Query 19
.............................................
Given Solution: `aThe equation is obtained by substituting the weights for y and the heights for x in the formula for the regression line. You get y = 3.35 x - 78.4. To predict weight when height is 70 you plug x = 70 into the equation: y = 3.35 * 70 - 78.4. You get y = 156, so the predicted weight for a man 70 in tall is 156 lbs. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay ------------------------------------------------ Self-critique Rating: Okay ********************************************* Question: `q Query problem 13.6.12 reading 83,76, 75, 85, 74, 90, 75, 78, 95, 80; IQ 120, 104, 98, 115, 87, 127, 90, 110, 134, 119 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Im confused on how to start this and how to find the sums. I will look at the given solution to help me out confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a n = 10 sum x = 811 sum x ^2 = 66225 sum y = 1104 sum y^2 = 124060 sum xy = 90437 a = [10(90437) - (811)(1104)] / [10(66225) - (811^2)] = 1.993 a = 1.99 b = [1104 - (1.993)(811) / 10 = -51.23 y' = 1.993x - 51.23 is the eqation of the regression line. ** STUDENT QUESTION How did you get sum x ^2 = 66225??? Is it not 811 * 811 = 657721? How did you come up with sum y^2 = 124060??? Is it not 1104 * 1104 = 1218816? I worked it out, can you tell me where I went wrong??? And I will try to rework the problem. INSTRUCTOR RESPONSE You didn't distinguish between sum x^2 and (sum x)^2. Sum x^2 means you figure out x^2 for every value of x, then add them. Remember that exponentiation precedes addition. (sum x)^2 means you add all the x values then square them. The same comment applies to sum y^2 vs. (sum y)^2. You didn't ask, but sum xy can also be confusing: Sum xy means multiply each x value by the corresponding y value, then add the products. This is order of operations: multiplication before addition &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This makes much more sense now ------------------------------------------------ Self-critique Rating: Okay ********************************************* Question: `q Query problem 13.6.15 years 0-5, sales 48, 59, 66, 75, 80, 89 What is the coefficient of correlation and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First find the sums of the values in the table The sum of all x values in the table equal 15. Square this value you get 55 The sum of y values in the table equal 418 The sum of x and y together equals 1186 Find that Ey^2 is equal to 16 N is equal to 6, as 6 is the number of pairs in the data Use all these sums to construct an equation and discover that the coefficient of correlation is .996. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a **STUDENT SOLUTION: X Y XY X^2 Y^2 0 48 0 0 2304 1 59 59 1 3481 2 66 132 4 4356 3 75 225 9 5626 4 80 320 16 6400 5 90 450 25 8100 Sums= 15 418 1186 55 30266 The coefficient of the correlation: r = .996 I found the sums of the following: x = 15, y = 418, x*y = 1186, x^2 = 55 n = 6 because there are 6 pairs in the data I also had to find Ey^2 = 30266 I used the following formula: r = 6(1186) - 15(418) / sq.root of 6(55) - (15)^2 * sq. root of 6(30266) - (418)^2 = 846 / sq. root of 105 * 6872 = 846 / sq. root of 721560 = 846 / 849.4 = .996 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay ------------------------------------------------ Self-critique Rating: Okay ********************************************* Question: `q Query problem 13.6.24 % in West, 1850-1990, .8% to 21.2% What population is predicted in the year 2010 based on the regression line? What is the equation of your regression line and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This seems to be a lot like 2 questions before this one. Im just lost when it comes to beginning these confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aSTUDENT SOLUTION: Calculating sums and regression line: n = 8 sum x = 56 sum x^2 = 560 sum = 77.7 sum y^2 = 1110.43 sum xy = 786.4 a = 1.44 b = -.39 r = .99 In the year 2010 the x value will be 16. y' = 1.44(16) - .39 = 22.65. There is an expected 22.65% increase in population by the year 2010. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This helps but I definitely need to study this more ------------------------------------------------ Self-critique Rating: Okay " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!