Jim Thompson HW may 19

Problem 1

 False

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Explanation:

It doesn't matter if the distribution is symmetric or not. The standard deviation is not affected by symmetry (or lack thereof). Instead, the standard deviation measures how spread out a data set is. If we had all points stacked on top of each other, forming a single tower of sorts, then the standard deviation would be zero. That doesn't happen with the Christchurch dot plot. Therefore, the standard deviation is some positive value. The more spread out the data is, the larger the standard deviation will be. The standard deviation is never negative.

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Problem 2

 True

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Explanation:

We have these two sets of data

S = {-1, 0,0,0,1,2,3,3,3,3,3,5,6,6,6,6,6,7,8}

C = {-4,-3,-3,-3,-2,-1,0,0,0,0,0,2,3,3,3,3,3,4,5}

where S and C represent St. Louis and Chicago respectively. We could calculate the standard deviation by hand; however, I find that to be tedious busy work. This is especially true if the data set is quite large. Instead, I recommend using a spreadsheet or calculator to compute the standard deviation. We want the sample standard deviation, since these data points are a representative sample of every day of the year.

You should find that the sample standard deviation of set S is roughly 2.716 while the sample standard deviation of set C is roughly 2.716

Therefore, the two standard deviations are equal.

Note that if we were to shift each dot of dot plot C exactly 3 units to the left, then they will line up perfectly with what dot plot S is showing. This shows that the data sets have the same level of spread.

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Problem 3

 True

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Explanation:

We could use a calculator like we did with problem 2. Simply enter the data values and crunch the numbers. However, I'll go with a visual approach.

Focus on the lower two dotplots. After looking closely at both, you should notice that they are almost identical graphs. The only difference is that the dot at "5" for Chicago has been moved to "8" for London. Every other dot is the same between data sets however.

So because that dot has been moved out further from the group, the data set for London is more spread out. Therefore, the standard deviation is higher here. The statement that Chicago has a lower standard deviation is true.

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Problem 4

 C) social sciences

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Explanation:

Make a spreadsheet as you see in the diagram labeled "problem 4" (see attached image below). The x values are the midpoints of each interval. The column f represents the frequencies. We multiply the x values and frequencies to get the x*f column. Adding up everything in that column gets us the value 4590 (shown in gray). Divide that over n = 100 to get the sample mean xbar = 45.9

The f*(x-xbar)^2 column consists of multiplying each frequency (f) with the expression (x-xbar)^2. Adding up everything in that column leads to the value shown in yellow. Divide that over n-1 = 99 to get 50.9495 approximately. Then apply the square root to get roughly 7.1379 which is the standard deviation.

To recap, we found these values

sample mean = 45.9sample standard deviation = 7.1379 (approximate)

So the mean salary is $45,900 and standard deviation is about $7,138. While these values don't perfectly match any of the given choices, we're fairly close to "social sciences", which is likely the answer.

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Problem 5

 A) business

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Explanation:

We use the same format and steps as problem 4. The difference of course is that we're dealing with a different data distribution. Refer to the table that says "problem 5" to see the spreadsheet.

The value in gray represents the sum of the x*f values. Divide that over n = 100 to get 5220/100 = 52.2

The value in yellow represents summing everything in the f*(x-xbar)^2 column (we don't sum in the yellow value itself or we'll have circular logic going on here). Divide that result in yellow over n-1 = 99 to get roughly 293.3434; then apply the square root to get roughly 17.1273 and this is the sample standard deviation.

To summarize so far, we have:

sample mean = 52.2sample standard deviation = 17.1273

This then corresponds to

mean salary = $52,200standard deviation of salaries = $17,127

Like before, those last two values don't match up perfectly with any of the three given professions; however, we're fairly close to the mean and standard deviation of the business majors. This is likely the answer.

b is your answer because I'm good at math and that is the answer