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251solngr4-081 4/15/08

Graded Assignment 4Name:

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Assume that . Do the following:

Assignment will not be accepted without diagrams!

h. Find a symmetrical interval about the mean with 96% probability.

i. Find (the 98th 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 solution. General comment - I can't give you much credit for an answer with a negative probability because there is no such thing!!!

For make a Normal curve centered at 4 and shade the area from 4 to 11; for make a Normal curve centered at zero and shade the area from zero to one .Since this area starts at zero, you do not need to add or subtract – just look up the answer.

For make a Normal curve centered at 4 and shade the areas from 0 to 4 and 4 to 11; for make a Normal curve centered at zero and shade the areas from -0.57 to zero and zero to 0.43.Since you shaded areas are on both sides of the mean, you add.

For make a Normal curve centered at 4 and shade the area from 1 to 3 below 4, leaving blank spaces between 3 and 4 and above 4; for make a Normal curve centered at zero and shade the area below zero from -0.43 to -0.14, leaving the entire area above zero and the area between -0.14 and zero blank. Since the area is entirely on one side of the mean, you subtract.

For make a Normal curve centered at 4 and shade the area below 5, which includes the entire area below 4; for make a Normal curve centered at zero and shade the entire area below 0.14, which includes the area below zero.Since you shaded areas are on both sides of the mean, you add.

For make a Normal curve centered at 4 and shade the area below 3, which would leave a blank area between 3 and 4; for make a Normal curve centered at zero and shade the entire area below -0.14, which would leave a blank area between 3 and 4 and a blank area above zero. Because this area is on one side of the mean, you are subtracting.

Note that .4999 comes from the lower part of my table. For make a Normal curve centered at 4 and shade the areas from -22 to 4 and 4 to 11 ; for make a Normal curve centered at zero and shade the areas from -3.71 to zero and zero to 1.00. .Because this area is on both sides of the mean, you are adding.

For make a Normal curve centered at 4 and shade the areas from -10 to 4 and 4 to 10 ; for make a Normal curve centered at zero and shade the areas from -2.00 to zero and zero to 0.86.Because this area is on both sides of the mean, you are adding

In general, for the following problems: 1) The values of you need here must came from the z-table (Table of the Standardized Normal Distribution), not the t-table because they aren't on the t-table except for k. 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. 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.

h. Find a symmetrical interval about the mean with 96% probability.

Solution: . Make a diagram.The diagram for will show a central area with a probability of 96%. It is split in two by a vertical line at zero into two areas with probabilities of 48%. The tails of the distribution each have a probability of 50% - 48% = 2%. 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 .4800 is or . If we have to choose between them 2.05 is a little better, though 2.06 is acceptable. So we can say , and our interval for is -2.05 to 2.05.

Since , the diagram for (if we bother) will show 96% probability split in two 48% regions on either side of 4, with 2% above and 2% below . The interval for can then be written or -10.35 to 18.35. To check this

i. Find (the 98th percentile). Solution:Make a diagram.The diagram for will show an area with a probability of 98%. 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 48%. The upper tail of the distribution has a probability of 2%, so that the entire area above 0 adds to 50%. From the diagram, we want one point so that

or . We know from the previous problem that the closest we can come is .

Since , the diagram for would show 98% probability split in two regions on either side of 4 with probabilities of 50% below and 48% above and with 2% above . , so the

value of can then be written To check this

j. Find (What percentile is this?). Solution: Since 71% is above this point, 100% - 71% = 29% is below this point and it is the 29th percentile. Make a diagram.The diagram for will show an area in the left tail with a probability of 29% ending at . Above zero there is an area of 50%. So between and zero there is a probability of 50% - 29% = 71% - 50% = 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, I will compromise and use or , so the value of can then be written .

To check this

k. Find . Solution:Because this can be gotten from one of the values of z on the t-table, we go there. The last line of Table 18 says the following.

Significance Level

.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.

According to this .This implies that can then be written .

To check this

DIAGRAMS – Note that none of these diagrams would be acceptable on an exam. You must add a vertical line at zero to the z diagrams and, if you want to do the x diagrams, include a vertical line at the mean (4).

a. b.

c. d.

e. f.

g. h. Find a symmetrical interval about the mean with 96% probability.

i. Find (the 98th percentile).j. Find (What percentile is this?).

k. Find