height VUD x = x 1 + x x N N 2 + (x 2 x) 2 + (x N x) 2. N

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Aron Gordon
6 years ago
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1 Math 3: CSM Tutorial: Probability, Statistics, and Navels Fall 2 In this worksheet, we look at navel ratios, means, standard deviations, relative frequency density histograms, and probability density functions. Whew! Here are the navel ratios, meaning the ratios height VUD (VUD stands for vertical umbilical displacement, meaning the height of your belly button), of the 3 students who were in CSM class this semester on day one. (We lost one of them; too bad for him/her.) Find the mean x and standard deviation s of the above data. (If your calculator has mean and standard deviation keys, go for it, but please write down on this paper the formulas you re using, with the appropriate numbers plugged in. You can use... if you want, but write down enough that I can tell what you re doing.) Write your answers to three decimal places. The formulas are: x = x + x x N N ; We compute: and x = r (x x) s = 2 + (x 2 x) 2 + (x N x) 2. N =.629 s = r ( )2 + ( ) ( ) 2 + (.4.629) 2 3 =.95. The sample size N in this case is 3. I get x =.629 and s =.97, roughly. They might get slightly di erent values because of roundo etc. (Also, some calculators will define the standard dev with an N in the bottom instead of an N.) As long as they get something close, it s OK. (over)

2 2. We are now going to construct a relative frequency density table for our navel ratio data, as follows. First, let s divide our data up into subranges of length.4, and starting at.4. That is: let s count how many of our navel ratios x satisfy.4 apple x <.44; how many satisfy.44 apple x <.48; how many satisfy.48 apple x <.52; and so on. Put these counts into the appropriate places in the Frequency column of the table below. Now, divide each of the counts you just found by both the length of the corresponding subrange (each subrange has length.4), and by the size N of our sample (which is 3). The numbers you get by doing this are called relative frequency densities. Put these relative frequency densities into the appropriate places in the Relative Frequency Density column of the table below. Subrange Frequency Relative Frequency Density.4<x< <x<.48.48<x<.52.52<x<.56.56<x<.6.6<x<.64.64<x<.68.68<x<.72.72<x<.76.76<x<.8.8<x<.84.84<x<.88.88<x<.92.92<x<.96.96<x<2. 2.<x< I won t fill in the table (though the students should); you can read the entries o of the histogram below.

3 3. Draw, on the axes below, a relative frequency density histogram depicting the information from the table in problem 2. LABEL YOUR AXES AND YOUR HISTOGRAM. PLEASE BE NEAT: use a straightedge to draw your boxes. Also mark, on the horizontal axis of your histogram, the values x, x 3s, and x + 3s x-3s x+3s In at most a few sentences, describe the overall shape of your above histogram. roughly bell-shaped, with some outliers. 5. It turns out that, in many many data sets, at least 99.7% of all the data points fall within 3 standard deviations of the mean. Is this true of the above data set? Explain. We get x 3s =.34385, and x + 3s = All but one of the data points falls within this range. That is, 29/3 = 96.7% of the data does. This is NOT at least99.7% of the data. So the outlier really is an aberration.

4 6. One theory says that, on average, in many populations, the navel ratio studied in the above problems is about equal to the golden ratio, which equals ( + p 5)/2.68. We are going to TEST this theory, using a hypothesis test often employed in statistical studies. (a) Let x be the sample mean you computed above, for our sample of 3 students, and let µ be the hypothesized population mean of (+ p 5)/2, which again is.68. Compute the quantity x µ, and write your answer in this space (to three decimal places):.. (b) Now compute the quantity 2.575s/ p N, which we ll need in part (c) below. Here s is the standard deviation you found above, and N = 3 is the size of your sample. Write your answer here (to three decimal places):.45. (c) Suppose, in general, you want to test whether the mean of some population equals some number µ. (In our case, we want to test whether the mean navel ratio in the overall population equals the golden ratio.) For reasons we won t get into just yet, the following hypothesis test is often used: compute the mean x of a sample, and compare this sample mean to the hypothesized population mean µ. If the di erence between x and µ is less than the number 2.575s/ p N, we ACCEPT the hypothesis that the population mean equals µ ; if this di erence is MORE than 2.575s/ p N, we reject this hypothesis. Based on the numbers above (that is, compare your answers in parts (a) and (b) above), should we accept or reject the hypothesis that the mean navel ratio, in the general population, equals the golden ratio? Explain. Since. <.45, we accept.

5 7. Let s go back to our relative frequency density histogram, and explore why we chose to plot relative frequency density (rather than just frequncy, say) on the vertical axis. (a) Consider the bar above the interval.6 apple x <.64, on the above histogram. What does the area of this bar tell you, in terms of what proportion, or percentage, of our data points lie in this interval? Please explain. Hint: the baselength of the bar is.4; its height is # of data points in the interval,.4 N where again, here, N = sample size = 3. The area of the bar is equal to the percentage of data that lies in the interval. Why? Because the area is base times height, which is.4 # of data points in the interval.4 N = # of data points in the interval, N which IS the proportion, or percentage, of data points that lie in this interval. (b) Consider the bar above the interval.64 apple x <.68, on the above histogram. What does the area of this bar tell you, in terms of what proportion, or percentage, of our data points lie in this interval? Please explain. Same as in (a). (c) Consider the four bars above the interval.6 apple x <.76, on the above histogram. What does the total area of these four bars tell you, in terms of what proportion, or percentage, of our data points lie in the interval.6 apple x <.76? Please explain. Again, the area equals the proportion, since the proportion between.6 and.76 is the sum of the proportions between.6 and.64; between.64 and.68; between.68 and.72; and between.72 and.76.

6 8. Imagine a very very large data set, consisting of data points taking on some range of values. Let s subdivide that range up into many many subintervals, each of which is very very thin, and draw a relative frequency density histogram corresponding to this subdivision. If the subintervals are small enough, we can imagine that the tops of the bars meld together into something that looks not like a jagged skyline, but rather like a smooth curve, or in other words, like the graph of a function y = f(x). This function y = f(x) is called the probability density function often abbreviated p.d.f. for the population (set of data points) under consideration. Who knows what such a pdf might look like. It might have a bell shape, it might not. Let s say it looks like the function f sketched below. y a b x Now here s the point. Reflect on everything you ve studied above, concerning relative frequency density histograms; the connection between area of the bars over a certain interval with the proportion of data points that are in that interval; and the process by which histograms meld into smooth curves. Using all of this, answer this question: If a and b are two x values as depicted on the above graph, then what percentage of the data represented there lies in the interval a apple x < b? Let s call this percentage P (a apple x < b). You should be able to express your answer in two ways: (a) in terms of area under the curve (go ahead and shade in, in the above picture, the area that equals P (a apple x < b)); it s the area under the graph of f between x = a and x = b; (b) as a definite integral. R b f(x) dx. a

7 9. Why is it called a probability density function, and what does this have to do with probability, anyway? Easy. Suppose we select, at random, an individual (that is, a data point) from the above population. What is the probability that this data point x will satisfy a apple x < b (in terms of quantities already discussed)? Please explain. Intuitively, the probability that a randomly chosen data point x will satisfy a apple x < b is just the proportion of the data in this range; but this proportion is P (a apple x < b) = R b f(x) dx. a. A population is said to be normally distributed, with mean and standard deviation, if its pdf f(x) is given by the formula f(x) = e x2 /2 p 2. Given such a population, write down an integral that would express the probability that a randomly selected data point from this population is at least but less than 4. What obstacles do you see to actually evaluating this integral? How might you overcome these obstacles? This integral is p 2 Z 4 e x2 /2 dx. The major obstacle is that e x2 /2 doesn t have a nice antiderivative. We could overcome this by approximating the integral with Riemann sums.

Density Curve (p52) Density curve is a curve that - is always on or above the horizontal axis.

Density Curve (p52) Density curve is a curve that - is always on or above the horizontal axis. 1.3 Density curves p50 Some times the overall pattern of a large number of observations is so regular that we can describe it by a smooth curve. It is easier to work with a smooth curve, because the histogram