QA 17

#$&*

course mth 152

I realized that I had not turned in QA 1711/15 11:45

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.

017. normal-curve models

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Question: `q001. Note that there are 8 questions in this assignment.

Sketch a histogram, i.e. a bar graph, showing the distribution of the number of ways to get 0, 1, 2, 3, 4 and 5 'heads' on a flip of 5 coins. Your histogram should show a bar for 0, 1, 2, 3, 4 and 5 'heads', and the height of a bar should represent the number of ways of getting that number of 'heads'.

Sketch also a histogram showing the probabilities of the different outcomes.

Describe both of your histograms.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I'm not sure what you looking for. They both look pretty much the same. They are both symmetrical. I am trying not to look ahead to your given solutions.

Ok, so I looked.

The ""x"" axis is labeled 0,1,2,3,4,5 with bars rising verticaly from each value. the bars extend to the values as follows; 0 to 1, 1 to 5, 2 to 10, 3 to 10, 4 to 5 and 5 to 1

for the the number of ways or c(n,r) of getting ""heads"" on a flip of 5 coins. If you draw a line from the outside corners of each bar to the bar beside it, moving from left to

right, it make a very syemmetical ""bell"" shape.

The second histogram looks almost identical with the values of the hieghtsof the bars being as followes from left to right: 0 to .0325, 1 to 2 to .1625, 2 to .325, 3 to .325, 4 to .1625 and 5 to .0325.

It to has the same syemmetricall ""bell"" shape.

confidence rating #$&*:

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Given Solution: Your first histogram should show 6 bars, one for each of the possible outcomes 0, 1, 2, 3, 4, 5. These bars should sit on top of a horizontal axis, like the x axis, and each should be labeled just below that axis with the outcome (0, 1, 2, 3, 4 and 5).

The heights of the bars will be 1, 5, 10, 10, 5 and 1, representing the numbers of possible ways for the six different outcomes to occur.

Your second histogram should have the same description as the first, except that the heights of the bars will be 1/32 = .0325, 5/32 = .1625, 10/32 = .325, 10/32 = .325, 5/32 = .1625 and 1/32 = .0325.

The vertical scales of the two histograms may of course be different, and both histograms may even look identical except for the labeling of the vertical axis.

Note that the bar representing, say, 2 will extend along the x axis from x = 1.5 to x = 2.5.

Note also that the distribution is symmetric about the central x value x = 2.5, which occurs on the boundary between the x = 2 and x = 3 bars of the graph. That is, the distribution to the left of x = 2.5 is a mirror image of the distribution to the right of x = 2.5.

Self-critique:

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Self-critique rating:

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Question: `q002. If we toss 64 coins, then the mean number of 'heads' is

mean = n * p = 64 * 1/2 = 32

and the standard deviation of the number of 'heads' is very close to

std dev = `sqrt( n * p * q ) = `sqrt( 64 * 1/2 * 1/2) = 4.

If we toss 64 coins a large number of times we expect that about 34% of the tosses will lie between 1 standard deviation lower than the mean and the mean, and that 34% of the tosses will lie between the mean and 1 stardard deviation higher than the mean.

What number is 1 standard deviation lower than the mean and what number is one standard deviation higher than the mean?

Out of 200 flips of 64 coins, how many would we expect to give us between 28 and 36 'heads' (use the percents given above, don't use combinations)?

How many would we expect to give less than 28 'heads' (again base your answer on the percents given above)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

32-4=28 is the st deviation lower than the mean and 34% of the tosses will land between 28 and 32

32+4=36 is the st deviation higher than the mean and 34% of the tossed will land between 32 and 36

34%+34%=68% will land between 28 and 36

200 flips * 68%= 136

136 out of 200 flips will lie between 28 and 36

If the mean is 32, then 50% of the out comes will be less than 32. If 34% percent will land between 1 standard dev of the mean

and the mean, then

50%-34%=16%

16% of 200=32 flips would expect to give less that 28 heads

confidence rating #$&*:

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Given Solution: The mean is 32 and the standard deviation is 4. An outcome one standard deviation lower than the mean is 32-4 = 28. An outcome one standard deviation higher than the mean will be 32 + 4 = 36. Note that we therefore expect that 34 percent of our outcomes will lie between 28 and 32, while another 34 percent lie between 32 and 36.

Out of 200 repetitions of the 64-flip experiment, we would therefore expect that 34% will lie between 28 and 32 while another 34% lie between 32 and 36. Thus a total of 68% lie between 28 and 36. Since 68% of 200 is .68 * 200 = 136, our expectation is that 136 of the 200 outcomes will lie between 28 and 32.

Since we expect that half of the outcomes, or 50%, will be less than the mean 32, then since 34% lie between 28 and 32, this leaves 16% of the outcomes falling below 28. Since 16% of 200 is 32, we expect that on the average 32 of 200 outcomes will lie below 28.

Note that we are 'fudging' a bit on this solution. If we had a histogram of this distribution, the bar representing 28 would actually extend from 27.5 to 28.5 on the x axis. Similarly the bar representing 32 would extend from 31.5 to 32.5, and the bar representing 36 would extend from 35.5 to 36.5. This needn't concern you much if the idea hasn't already occurred to you that there are nine, not eight outcomes from 28 through 36. The problem is resolved as follows

To represent the outcomes from 28 to 36 we would have to go from the middle of the bar representing 28 to the middle of the bar representing 36. The number of outcomes calculated here would therefore include only half of the outcomes 28 and 36, plus the remaining seven bars representing 29 - 35.

Self-critique:

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Self-critique rating:

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Question: `q003. A more detailed breakdown of proportions of 'normal' distributions (i.e., distributions based on the probabilities associated with large numbers of coin flips) which fall into various ranges is as follows:

std dev prop

0.25 0.099

0.50 0.191

0.75 0.273

1.00 0.341

1.25 0.394

1.50 0.433

1.75 0.460

2.00 0.477

The column 'std dev' stands for the number of standard deviations from the mean, and 'prop' stands for the proportion of all occurrences lying between the mean and the given number of standard deviations above the mean.

What proportion of a normal distribution is expected to lie between the mean and 1.25 standard deviations from the mean?

If a certain quantity is normally distributed, then given a sample of 200 instances how many would lie between the mean and 1.25 standard deviations above the mean?

Given a sample of 200 instances how many would lie between the mean and 0.25 standard deviations below the mean?

Given a sample of 200 instances how many would lie between the 1.25 standard deviations above the mean and 0.25 standard deviations below the mean?

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Your solution:

the proportion that is expected to lie between the mean and 1.25 st dev from the mean= .394

.394*200=78.8 lie between the mean and 1.25 std dev above the mean.

the proportion that is expected to lie between the mean and .25 st dev from the mean is .099

.099*200= 19.80 lie between the mean and .099 std dev below the mean.

78.8+19.80=98.60 instances that would lie between the 1.25 st dev above the mean and .25 st dev below the mean

confidence rating #$&*:

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Given Solution: According to the table, the proportion 0.394 of the distribution will lie between the mean and 1.25 standard deviations above the mean. This is consistent with the information given in the preceding problem, that 34% or 0.34 of the distribution should lie between the mean and one standard deviation above the mean.

Given 200 instances, we would therefore expect 0.394 * 200 = 78.6 of the outcomes to lie between the mean and one standard deviation above the mean.

The number of instances lying between the mean and 0.25 standard deviations below the mean should, because of the symmetry of the distribution, be the same as the number of instances between the mean and 0.25 standard deviations above the mean. According to the information given here, the portion of the distribution should account for 0.099 of the entire distribution. If there are 200 total occurrences, then 0.099 * 200 = 19.8 of the occurrences should lie between the mean and 0.25 standard deviations below the mean.

As we have seen here 78.6 of the outcomes should lie between the mean and 1.25 standard deviations above the mean, while 19.8 should lie between the mean and 0.25 standard deviations below the mean. There is no overlap between these regions, since one lies entirely below of the mean while the other lies entirely above the mean. So the total number lying between the two given extremes must be 78.6 + 19.8 = 98.4. Note that this corresponds to 0.394 + 0.099 of the distribution.

Self-critique:

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Question: `q004. If scores on a certain test are normally distributed with average 150 points standard deviation of 20 points, then how many standard deviations above or below the distribution is each of the following scores:

170, 120, 135, 155?

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Your solution:

170-150=20 20/20 st dev= 1 st dev above the mean

155-150=5 5/20 st dev= .25 st dev above the mean

150-135=15 15/20 st dev= .75 st dev below the mean

150-120=30 30/20 st dev= 1.5 st dev below the mean

confidence rating #$&*:

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Given Solution: Since 170 is 20 units above the mean of 150, and the standard deviation is 20, we see that 170 lies exactly one standard deviation above the mean.

We see that 120 lies 30 units below the mean of 150, which is 30/20 = 1.5 times the standard deviation 20. Thus 120 lies 1.5 standard deviations below the mean.

135 lies 15 units below the mean, or 15/20 = 0.75 of a standard deviation below the mean.

155 lies 5 units above the mean, or 5/20 = 0.25 of a standard deviation above the mean.

Note that we could use the proportion given in the preceding problem to determine what proportion of a distribution lie between the mean and each given value.

Self-critique:

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Self-critique rating:

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Question: `q005. The table from the previous problem is given again here:

std dev prop

0.25 0.099

0.50 0.191

0.75 0.273

1.00 0.341

1.25 0.394

1.50 0.433

1.75 0.460

2.00 0.477

If scores on a certain test are normally distributed with average 150 points standard deviation of 20 points, then out of 600 people taking the test how many are expected to score in each of the following ranges:

150 - 170

120 - 150

135 - 155

120-135?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

150 - 170= 1 stdv with the prop=.341*600=204

120 - 150= 1.5 stdv with the prop=.433*600= 260

135 - 155= .273 (.75 st dev of 135 below the mean) + .099 (.25 st dev of 155 above the mean)= .372*600=223

120-135?= .433 (1.5 st dev of 120 below the mean) - .273 (.75 st dev below the mean)= .16*600=96

confidence rating #$&*:

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Given Solution: We saw in the preceding problem that 170 lies one standard deviation above the mean. The proportion lying between the mean 150 and 170 is therefore 0.34 so that out of 600 outcomes, we would expect 0.34 * 600 = 204 to lie within the range 150 - 170.

As seen in the previous problem, 120 corresponds to an outcome 1.5 standard deviations below the mean. The range 120-150 consists of all outcomes lying between the mean and 1.5 standard deviations below the mean. The proportion lying between these values is seen from the table to be 0.433, we expect that of 600 outcomes we will have 0.433 * 600 = 260 in the range 120-150.

We have seen that 135 corresponds to an outcome 0.75 standard deviations below the mean, and that 155 corresponds to an outcome 0.25 standard deviations above the mean. Between 0.75 standard deviations below the mean and the mean we expect 0.273 of all outcomes, and between the mean and 0.25 standard deviations above the mean we expect another 0.099 of the outcomes. Since one range lies completely below and the other completely above the mean, there is no overlap and we expect that 0.273 + 0.099 = 0.372 of the 600 outcomes, or 0.372 * 600 = 223.2 of the outcomes will lie within the range 135-155.

We have seen that 135 corresponds to an outcome 0.75 standard deviations below the mean, and that 120 corresponds to an outcome 1.5 standard deviations below the mean. Between 0.75 standard deviations below the mean and the mean we expect 0.273 of all outcomes, and between 1.5 standard deviations below the mean and the mean we expect 0..477 of the outcomes. Since both ranges lie completely below the mean, they overlap and we therefore expect that 0.477 - 0.273 = 0.204 of the 600 outcomes, or .204 * 600 = 122.4 of the outcomes will lie within the range 120-135.

STUDENT COMMENT

I understood most of these except for the very last one.

INSTRUCTOR RESPONSE

You should try to be more specific about what you do and do not understand, so I can focus my answer appropriately. However the following might be helpful:

The solution basically says that 27.3% of outcomes lie between 135 and 150, while 47.7% lie between 120 and 135, so 47.7% - 27.3% = 20.4$ lie between 120 and 135.

Self-critique:

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Self-critique rating:

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Question: `q006. The table given in question `q005 can be used as a basis for answering the questions below.

A student clicks a mouse repeatedly, as fast as possible. The times between consecutive clicks has a mean of .132 second with a standarad deviation of .016 second.

Out of 100 clicks, how many would be expected to fall in each of the following ranges:

.132 second to .148 second

.148 second to .164 second

.116 second to .148 second

less than .116 second

between .128 second and .140 second

(Note: If you want to compare your performance to that of the student in this example, you may do so at http://vhcc2.vhcc.edu/dsmith/forms/ph1_timer_experiment.htm . This is intended only for your interest and you may ignore the part about submitting the document (though if you really want to you are welcome to submit all or part of it) ).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

std dev prop

0.25 0.099

0.50 0.191

0.75 0.273

1.00 0.341

1.25 0.394

1.50 0.433

1.75 0.460

2.00 0.477

.140-.132=.008 .008/.016=.5 std dev above the mean

.148-.132=.016 .016/.016=1 std dev above the mean

.164-.132=.032 .032/.016=2 std dev above the mean

.128-.132=.004 .004/.016=.25 std dev BELOW the mean

.116-.132=.016 .016/.016=1 std dev BELOW the mean

.132 second to .148 second= .341*100=34.1 clicks

.148 second to .164 second= .341-.477=.136*100=13.6 clicks

.116 second to .148 second= .341+.341=.682*100=68.2 clicks

less than .116 second= .341*100= 34.1-50 (100 clicks/2)=15.9 or 16 clicks

confidence rating #$&*:

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Self-critique Rating:

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Question: `q007. A runner runs a 100-meter race in an average of 11.52 seconds, with a standard deviation of .12 second.

In how many out of 100 races would the runner expect a time of 11.40 seconds or better?

In how many out of 100 races would the runner expect a time of 11.70 seconds or worse?

In how many out of 100 races would the runner expect a time between 11.40 seconds and 11.70 seconds?

What do you think is the fastest time the runner could expect in 100 races?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

std dev prop

0.25 0.099

0.50 0.191

0.75 0.273

1.00 0.341

1.25 0.394

1.50 0.433

1.75 0.460

2.00 0.477

11.52-11.40=.12/.12=1 st dev below the mean=.341*100=34.1-50=15.9 races

11.70-11.52=.18/.12=1.5 st dev above the mean=.433*100=43.3-50=6.7 races

.341+.433=.774*100=77.4 races

I'm not sure how to find the fastest time

confidence rating #$&*:

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Self-critique Rating:

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Question: `q008. What would you estimate to be the standard deviation of a distribution whose mean is 100, if 90% of the distribution lies between 75 and 125?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

100* 22.5/25 * 2.5/25= 100 * .90 * .1= 9 sq rt =3

confidence rating #$&*:

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Self-critique Rating:

this is a shot in the dark"

Self-critique (if necessary):

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Self-critique rating:

this is a shot in the dark"

Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Your work looks good. Let me know if you have any questions. &#

QA 17

#$&*

course mth 152

I realized that I had not turned in QA 1711/15 11:45

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.

017. normal-curve models

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Question: `q001. Note that there are 8 questions in this assignment.

Sketch also a histogram showing the probabilities of the different outcomes.

Describe both of your histograms.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I'm not sure what you looking for. They both look pretty much the same. They are both symmetrical. I am trying not to look ahead to your given solutions.

Ok, so I looked.

The ""x"" axis is labeled 0,1,2,3,4,5 with bars rising verticaly from each value. the bars extend to the values as follows; 0 to 1, 1 to 5, 2 to 10, 3 to 10, 4 to 5 and 5 to 1

for the the number of ways or c(n,r) of getting ""heads"" on a flip of 5 coins. If you draw a line from the outside corners of each bar to the bar beside it, moving from left to

right, it make a very syemmetical ""bell"" shape.

It to has the same syemmetricall ""bell"" shape.

confidence rating #$&*:

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The heights of the bars will be 1, 5, 10, 10, 5 and 1, representing the numbers of possible ways for the six different outcomes to occur.

Your second histogram should have the same description as the first, except that the heights of the bars will be 1/32 = .0325, 5/32 = .1625, 10/32 = .325, 10/32 = .325, 5/32 = .1625 and 1/32 = .0325.

The vertical scales of the two histograms may of course be different, and both histograms may even look identical except for the labeling of the vertical axis.

Note that the bar representing, say, 2 will extend along the x axis from x = 1.5 to x = 2.5.

Self-critique:

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Self-critique rating:

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Question: `q002. If we toss 64 coins, then the mean number of 'heads' is

mean = n * p = 64 * 1/2 = 32

and the standard deviation of the number of 'heads' is very close to

std dev = `sqrt( n * p * q ) = `sqrt( 64 * 1/2 * 1/2) = 4.

What number is 1 standard deviation lower than the mean and what number is one standard deviation higher than the mean?

Out of 200 flips of 64 coins, how many would we expect to give us between 28 and 36 'heads' (use the percents given above, don't use combinations)?

How many would we expect to give less than 28 'heads' (again base your answer on the percents given above)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

32-4=28 is the st deviation lower than the mean and 34% of the tosses will land between 28 and 32

32+4=36 is the st deviation higher than the mean and 34% of the tossed will land between 32 and 36

34%+34%=68% will land between 28 and 36

200 flips * 68%= 136

136 out of 200 flips will lie between 28 and 36

If the mean is 32, then 50% of the out comes will be less than 32. If 34% percent will land between 1 standard dev of the mean

and the mean, then

50%-34%=16%

16% of 200=32 flips would expect to give less that 28 heads

confidence rating #$&*:

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Self-critique:

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Self-critique rating:

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std dev prop

0.25 0.099

0.50 0.191

0.75 0.273

1.00 0.341

1.25 0.394

1.50 0.433

1.75 0.460

2.00 0.477

What proportion of a normal distribution is expected to lie between the mean and 1.25 standard deviations from the mean?

If a certain quantity is normally distributed, then given a sample of 200 instances how many would lie between the mean and 1.25 standard deviations above the mean?

Given a sample of 200 instances how many would lie between the mean and 0.25 standard deviations below the mean?

Given a sample of 200 instances how many would lie between the 1.25 standard deviations above the mean and 0.25 standard deviations below the mean?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

the proportion that is expected to lie between the mean and 1.25 st dev from the mean= .394

.394*200=78.8 lie between the mean and 1.25 std dev above the mean.

the proportion that is expected to lie between the mean and .25 st dev from the mean is .099

.099*200= 19.80 lie between the mean and .099 std dev below the mean.

78.8+19.80=98.60 instances that would lie between the 1.25 st dev above the mean and .25 st dev below the mean

confidence rating #$&*:

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Given 200 instances, we would therefore expect 0.394 * 200 = 78.6 of the outcomes to lie between the mean and one standard deviation above the mean.

Self-critique:

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Self-critique rating:

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170, 120, 135, 155?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

170-150=20 20/20 st dev= 1 st dev above the mean

155-150=5 5/20 st dev= .25 st dev above the mean

150-135=15 15/20 st dev= .75 st dev below the mean

150-120=30 30/20 st dev= 1.5 st dev below the mean

confidence rating #$&*:

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Given Solution: Since 170 is 20 units above the mean of 150, and the standard deviation is 20, we see that 170 lies exactly one standard deviation above the mean.

We see that 120 lies 30 units below the mean of 150, which is 30/20 = 1.5 times the standard deviation 20. Thus 120 lies 1.5 standard deviations below the mean.

135 lies 15 units below the mean, or 15/20 = 0.75 of a standard deviation below the mean.

155 lies 5 units above the mean, or 5/20 = 0.25 of a standard deviation above the mean.

Note that we could use the proportion given in the preceding problem to determine what proportion of a distribution lie between the mean and each given value.

Self-critique:

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Self-critique rating:

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Question: `q005. The table from the previous problem is given again here:

std dev prop

0.25 0.099

0.50 0.191

0.75 0.273

1.00 0.341

1.25 0.394

1.50 0.433

1.75 0.460

2.00 0.477

150 - 170

120 - 150

135 - 155

120-135?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

150 - 170= 1 stdv with the prop=.341*600=204

120 - 150= 1.5 stdv with the prop=.433*600= 260

135 - 155= .273 (.75 st dev of 135 below the mean) + .099 (.25 st dev of 155 above the mean)= .372*600=223

120-135?= .433 (1.5 st dev of 120 below the mean) - .273 (.75 st dev below the mean)= .16*600=96

confidence rating #$&*:

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STUDENT COMMENT

I understood most of these except for the very last one.

INSTRUCTOR RESPONSE

You should try to be more specific about what you do and do not understand, so I can focus my answer appropriately. However the following might be helpful:

The solution basically says that 27.3% of outcomes lie between 135 and 150, while 47.7% lie between 120 and 135, so 47.7% - 27.3% = 20.4$ lie between 120 and 135.

Self-critique:

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Self-critique rating:

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Question: `q006. The table given in question `q005 can be used as a basis for answering the questions below.

A student clicks a mouse repeatedly, as fast as possible. The times between consecutive clicks has a mean of .132 second with a standarad deviation of .016 second.

Out of 100 clicks, how many would be expected to fall in each of the following ranges:

.132 second to .148 second

.148 second to .164 second

.116 second to .148 second

less than .116 second

between .128 second and .140 second

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

std dev prop

0.25 0.099

0.50 0.191

0.75 0.273

1.00 0.341

1.25 0.394

1.50 0.433

1.75 0.460

2.00 0.477

.140-.132=.008 .008/.016=.5 std dev above the mean

.148-.132=.016 .016/.016=1 std dev above the mean

.164-.132=.032 .032/.016=2 std dev above the mean

.128-.132=.004 .004/.016=.25 std dev BELOW the mean

.116-.132=.016 .016/.016=1 std dev BELOW the mean

.132 second to .148 second= .341*100=34.1 clicks

.148 second to .164 second= .341-.477=.136*100=13.6 clicks

.116 second to .148 second= .341+.341=.682*100=68.2 clicks

less than .116 second= .341*100= 34.1-50 (100 clicks/2)=15.9 or 16 clicks

confidence rating #$&*:

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Self-critique Rating:

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Question: `q007. A runner runs a 100-meter race in an average of 11.52 seconds, with a standard deviation of .12 second.

In how many out of 100 races would the runner expect a time of 11.40 seconds or better?

In how many out of 100 races would the runner expect a time of 11.70 seconds or worse?

In how many out of 100 races would the runner expect a time between 11.40 seconds and 11.70 seconds?

What do you think is the fastest time the runner could expect in 100 races?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

std dev prop

0.25 0.099

0.50 0.191

0.75 0.273

1.00 0.341

1.25 0.394

1.50 0.433

1.75 0.460

2.00 0.477

11.52-11.40=.12/.12=1 st dev below the mean=.341*100=34.1-50=15.9 races

11.70-11.52=.18/.12=1.5 st dev above the mean=.433*100=43.3-50=6.7 races

.341+.433=.774*100=77.4 races

I'm not sure how to find the fastest time

confidence rating #$&*:

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Self-critique Rating:

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Question: `q008. What would you estimate to be the standard deviation of a distribution whose mean is 100, if 90% of the distribution lies between 75 and 125?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

100* 22.5/25 * 2.5/25= 100 * .90 * .1= 9 sq rt =3

confidence rating #$&*:

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Self-critique Rating:

this is a shot in the dark"

Self-critique (if necessary):

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Self-critique rating:

this is a shot in the dark"

Self-critique (if necessary):

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Self-critique rating:

#*&!