251Solngr4-082 6/25/08 (Open This Document in 'Page Layout' View!)

251Solngr4-082 6/25/08 (Open This Document in 'Page Layout' View!)

251solngr4-082 6/25/08 (Open this document in 'Page Layout' view!)

Graded Assignment 4

Name:KEY

Assume that . Do the following:

Assignment will not be accepted without diagrams!

a.

b.

c.

d.

e. One version of this seems to have had a 3 instead of a 2.

f.

g.

h. Find a symmetrical interval about the mean with 66% probability.

i. Find (the 66th 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.

a.

For make a Normal curve centered at 2 and shade the area from 0.5 to 4; for make a Normal curve centered at zero and shade the area from -0.14 to 0.18.Note that the diagrams below would not be acceptable on an exam because of the absence of a vertical line at 2 in the left diagram and the absence of a vertical line at zero in the right diagram. Also note that it is not worth the effort to do diagrams to scale on an exam.

b.

For make a Normal curve centered at 2 and shade the area from -0.21 to 0.21; for make a Normal curve centered at zero and shade the area from -0.20 to -0.16. Because this is to scale, you really can’t see much. This is where a little exaggeration helps.

c.

For make a Normal curve centered at 2 and shade the area below (left of) 0.04; for make a Normal curve centered at zero and shade the area below -0.18.Note that there is a rounding error in getting from x to z, so that the Minitab computation gives a more exact answer than you can get from the table.

d.

For make a Normal curve centered at 2 and shade the (entire!) area below (left of) 4; 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.

e.

from the lower part of the table. The diagrams would 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.

For make a Normal curve centered at 2 and shade the area from 0 to 2; for make a Normal curve centered at zero and shade the area from -0.18 to0.

g.

For make a Normal curve centered at 2 and shade the area from -4 to 4; for make a Normal curve centered at zero and shade the area from -0.64 to 0.09.

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 66% probability.

Solution: .Make a diagram. The diagram for will show a central area with a probability of 66%. It is split in two by a vertical line at zero into two areas with probabilities of 33%. The tails of the distribution each have a probability of 50% - 33% = 17%. 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 .3300 is or . Though either or is an acceptable answer, in fact 0.95 is closer, so I will use , and our interval for is -0.95 to 0.95.

Since ,the diagram for (if we bother) will show 66% probability split in two 33% regions on either side of 2, with 17% above and 17% below . The interval for can then be written or -8.45 to 12.45.

To check this:

i. Find (the 66th percentile).

Solution:Make a diagram. The diagram for will show an area with a probability of 66% 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 66% - 50% =16%. The upper tail of the distribution has a probability of 34%, so that the entire area above 0 adds to 50%. From the diagram, we want one point so that or . The closest we can come is . We can say though 0.42 would be acceptable.

Since ,the diagram for would show66% probability split in two regions on either side of 2 with probabilities of 50% below 2 and 16% above 2 and below ,and with 34% above . , so thevalue of can then be written

To check this:

So 6.51 is about the 66th percentile.

j. Find (What percentile is this?).

Solution: Since 66% is above this point, 100% - 66% = 34% is below this point and it is the 34th percentile. Make a diagram. The diagram for will show an area in the left tail with a probability of 34% to the left of. Above zero there is an area of 50%. So between and zero there is a probability of 50% - 34% = 16% (or 66% - 50% = 16%). A diagram for would be pretty much the same, except that 2 would replace zero, so that we could clearly see that we are looking for a number below 2. 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, 0.41 is closer, so let’s say or , so the value of can then be written

To check this:Or we could try

. So -3.39 is about the 34th percentile or .

k. Find

The bottom of the t table has the following.

.45 .40 .35 .30 .25 .20 .15 .10 .05 .025 .01 .005 .001

0.126 0.253 0.385 0.524 0.675 0.842 1.036 1.282 1.645 1.960 2.327 2.576 3.091

Note that the line gives critical values of the standardized Normal Distribution.

So and

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