If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
018. ``q Query 18
Question: `q Query problem 13.5.12 percent above 115 IQ, mean 100 std dev 15
Your solution:
Confidence Assessment:
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):
Self-critique Rating:
Question: `q Query problem 13.5.20 area between z=-1.74 and z=-1.14
Your solution:
Confidence Assessment:
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):
Self-critique Rating:
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.
Your solution:
Confidence Assessment:
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):
Self-critique Rating:
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?
Your solution:
Confidence Assessment:
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 aren’t sure you understand.
Self-critique (if necessary):
Self-critique Rating: