Graded Assignment 4Name:KEY

251solngr4 4/4/06

Graded Assignment 4Name:KEY

(Open this document in 'Page Layout' view!)Class days and time:

Assume that. Do the following:

Assignment will not be accepted without diagrams! To get full credit keep it neat and stapled and don’t forget your class! All probabilities should be between zero and 1.

h. Find a symmetrical interval about the mean with 71% probability.

i. Find (the 71st percentile).

j. Find (What percentile is this?).

k. Find

Solution: Material in italics below is a description of the diagrams you were asked to make or a general explanation and will not be part of your written solution. The and diagrams should look similar. If you know what you are doing, you only need one diagram for each problem. General comment - I can't give you much credit for an answer with a negative probability or a probability above one, because there is no such thing!!! In all these problems we must find values of corresponding to our values of before we do anything else. A diagram for will help us because, if areas on both sides of zero are shaded, we will add probabilities, while, if an area on one side of zero is shaded and it does not begin at zero, we will subtract probabilities.

For make a Normal curve centered at 4 and shade the area from 2.2 to 6.3; for make a Normal curve centered at zero and shade the area from -0.14 to 0.18. Since the diagrams show an area on both sides of the mean, you add.

For make a Normal curve centered at 4 and shade the area from -1.4 to 1.4; for make a Normal curve centered at zero and shade the area from -0.42 to -0.20.Since the diagrams show an area on one side of the mean, you subtract.

Formake a Normal curve centered at 2 and shade the area below (left of) 1.17; for make a Normal curve centered at zero and shade the area below -0.18.Since the diagrams show an area on one side of the mean, you subtract.

Formake a Normal curve centered at 4 and shade the (entire!) area below (left of) 6.3; for make a Normal curve centered at zero and shade the area below 0.18. Don’t stop shading at zero, but shade all the way to the left corner of your graph.Since the diagrams show an area on both sides of the mean, you add.

Formake a Normal curve centered at 4 and shade the area between 4 and47.7- almost the entire area to the right of 4; for make a Normal curve centered at zero and shade the area between zero and 3.36.The diagrams show a shaded area on only one side of the mean, and might fool you into subtracting something from .4996, which, unless you subtract zero, is a very bad idea.

f. .1217.

For make a Normal curve centered at 4 and shade the area from 0 to 4; for make a Normal curve centered at zero and shade the area from -0.31 to0.The diagrams show a shaded area on only one side of the mean, and might fool you into subtracting something from .1217.

For make a Normal curve centered at 4 and shade the area from -6.08 to6.08; for make a Normal curve centered at zero and shade the area from -0.78 to 0.16.Since the diagrams show an area on both sides of the mean, you add.

In general, for the following problems remember that is a point with 2% above it and:

1) The values of you need here must come from the z-table (Table of the Standardized Normal Distribution), not the t-table because they aren't on the t-table. You can find values like or on the t-table, but I didn't ask for them.

2) Numbers like are values of , not probabilities; you can't find them by taking 0.24 or .76 on the part of the table and then reading a probability and claiming that it is a value of .

3) If the request is for an interval, two solutions are needed, but if the request is for a percentile, only one number is acceptable. An answer for a percentile including will not get full credit.

4) Note also that since, in the Normal distribution, the mean is the 50th percentile, a percentile below 50 is below the mean and a percentile above 50 is above the mean.

5) Once you have an appropriate value of z, use the formula to get the corresponding value of x.

h. Find a symmetrical interval about the mean with 71% probability.

Solution: Make a diagram. The diagram for will show a central area with a probability of 71%. It is split in two by a vertical line at zero into two areas with probabilities of 35.5%. The tails of the distribution each have a probability of 50% - 35.5% = 14.5%. From the diagram, we want two points and so that . The upper point, will have , and by symmetry . From the interior of the Normal table the closest we can come to .3550 is , which is slightly too high. The next best point would be 1.05 since. We can say Normally a second best point like would be acceptable, but here there is no question of the correct answer. So I will use , and our 71% symmetricalinterval for is -1.06 to 1.06.

Since , the diagram for (if we bother) will show 71% probability split in two 35.5% regions on either side of 4, with 14.5% above and 14.5% below . The interval for can then be written or -9.78 to 17.78.

To check this:

i. Find (the 71st percentile).

Solution:Make a diagram. The diagram for will show an area with a probability of 71% below. It is split by a vertical line at zero into two areas. The lower one has a probability of 50% and the upper one a probability of 71% - 50% =21%. The upper tail of the distribution above has a probability of 29%, so that the entire area above 0 adds to 21% + 29% = 50%. From the diagram, we want one point so that or . If we try to find this point on the Normal table, the closest we can come is , but is almost as close. Since .2100 is about halfway between these two probabilities, probably the best idea would be to use , though 0.55 or 0.56 would be acceptable.

Since , the diagram for would show71% probability split in two regions on either side of 4 with probabilities of 50% below 4 and 21% above 4 and below ,and with 29% above . , so thevalue of can then be written

To check this:

So 11.215 is about the 71st percentile.

j. Find (What percentile is this?).

Solution: Since 71% is above this point, 100% - 71% = 29% is below and it is the 29th percentile. Make a diagram. The diagram for will show an area in the left tail below with a probabilityof 29%. Above zero there is an area of 50%. So between and zero there is a probability of 50% - 29% = 21%. A diagram for would be pretty much the same, except that 4 would replace zero, so that we could clearly see that we are looking for a number below 4. From the diagram, we want one point so that . We can also see that . Because , . From the interior of the Normal table, the closest we can come is or . Though either of these would be acceptable, 0.555 is a better choice. So let’s say , so the value of can then be written

To check this:Or we could try

. So -3.215 is about the 29th percentile or .

k. Find

Solution:Make a diagram. The diagram for will show an area with a probability of 55% below . It is split by a vertical line at zero into two areas. The lower one has a probability of 50% and the upper one a probability of 55% - 50% = 05%. The upper tail of the distribution above has a probability of 45%, so that the entire area above 0 adds to 5% + 45% = 50%. From the diagram, we want one point so that . Up to this point, we are simply copying the answer to i). But at this point we realize that the subscript on is divisible by 5, so that we can find the number we need on the bottom of the t table. On the infinity line of the t table, the entry in the .45 column is 0.126. This tells us that

Since , the diagram for would show 55% probability split in two regions on either side of 4 with probabilities of 50% below 4 and 5% above 4 and below ,and with 45% above . so the value of can then be written

To check this:

So 5.638 is about the 55th percentile.