Continuous Random Variables Consider the following table of sales, divided into intervals of 1000 units each, interval (0,1000] (1000,2000] (2000,3000] (3000,4000] (4000,5000] (5000,6000] (6000,7000] and the relative frequency of each interval. interval (0,1000] (1000,2000]

(2000,3000] (3000,4000] (4000,5000] (5000,6000] (6000,7000] relative freq. 0 0.05 0.25 0.30 0.25 0.10 0.05 1.00

Were going to divide the relative frequencies by the width of the cells (which here is 1000). This will make the graph have an area of 1. interval (0,1000] (1000,2000] (2000,3000] (3000,4000] (4000,5000] (5000,6000] (6000,7000] relative f(x) relative freq. freq. cell width 0 0.05

0.25 0.30 0.25 0.10 0.05 0 0.00005 0.00025 0.00030 0.00025 0.00010 0.00005 Graph interval f(x)

relative freq. cell width f(x) = p(x) 0.00030 (0,1000] 0 (1000,2000] 0.00005 (2000,3000] 0.00025

(3000,4000] 0.00030 0.00015 (4000,5000] 0.00025 0.00010 (5000,6000] 0.00010 (6000,7000]

0.00005 0.00025 0.00020 0.00005 0 0 1000 2000 3000 4000 5000 6000 7000 sales The area of each bar is the frequency of the category, so the total area is 1.

Graph interval f(x) relative freq. cell width f(x) = p(x) 0.00030 (0,1000] 0 (1000,2000] 0.00005

(2000,3000] 0.00025 (3000,4000] 0.00030 0.00015 (4000,5000] 0.00025 0.00010 (5000,6000]

0.00010 (6000,7000] 0.00005 0.00025 0.00020 0.00005 0 0 1000 2000 3000 4000 5000 6000 7000 sales

Here is the frequency polygon. If we make the intervals 500 units instead of 1000, the graph would probably look something like this: f(x) = p(x) The height of the bars increases and decreases more gradually. sales If we made the intervals infinitesimally small, the bars and the frequency polygon would become smooth, looking something like this: f(x) = p(x)

This what the distribution of a continuous random variable looks like. This curve is denoted f(x) or p(x) and is called the probability density function. sales pmf versus pdf For a discrete random variable, we had a probability mass function (pmf). The pmf looked like a bunch of spikes, and probabilities were represented by the heights of the spikes. For a continuous random variable, we have

a probability density function (pdf). The pdf looks like a curve, and probabilities are represented by areas under the curve. Pr(a < X < b) f(x) = p(x) a b sales A continuous random variable has an infinite number of possible values & the probability of any one particular value is zero. If X is a continuous random variable,

which of the following probabilities is largest? (Hint: This is a trick question.) 1. 2. 3. 4. Pr(a < X < b) Pr(a X < b) Pr(a < X b) Pr(a X b) Theyre all equal. They differ only in whether they include the individual values a and b, and any one particular value has zero probability! Properties of

probability density functions (pdfs) 1. f(x) 0 for values of x This means that when we draw the pdf curve, while it may be on the left side of the vertical axis (have negative values of x), it can not go below the horizontal axis, where f would be negative. Pr( - < X < ) = 1 The total area under the pdf curve, which corresponds to the total probability, is 1. Example f(x) = 2 f(x) = 0 if 1 x 1.5 otherwise

and This function satisfies both the properties of pdfs. f(x) First, its never negative. Second, the total area under the curve is (1/2) (2) = 1. 2.0 0 1.0 1.5 x

Cumulative Distribution Function for a Continuous Random Variable F(x) = Pr(X x) = area under the f(x) curve up to where X=x. Rectangle Example: What is F(1.2)? F(1.2) = Pr(X 1.2) = the area under the pdf up to where x is 1.2. f(x) 2.0 0 1.0

1.5 x Rectangle Example: What is F(1.2)? F(1.2) = Pr(X 1.2) = the area under the pdf up to where x is 1.2. = (0.2) (2.0) f(x) = 0.4 2.0 0 1.0 1.2

1.5 x Continuous Uniform Distributions. Distributions like our rectangle example are called uniform distributions. Notice that there are both discrete uniform distributions that we discussed earlier and continuous uniform distributions. The continuous uniform distribution has the following form: 1 f ( x) b a if a x b

f ( x) 0 otherwise f ( x) Notice that the area of the rectangle will always be 1 because 1 b a 0 area = length width 1 (b a) 1 b a

a b x Mean, variance, and standard deviation of the continuous uniform distribution mean: a b 2 variance: 2 (

b a ) 2 12 standard deviation: (b a)2 12 The most famous distribution is the Normal or Gaussian distribution. Its probability density function (pdf) is f ( x)

1 2 e 1 [( x ) / ]2 2 for - x is the mean of the distribution, is the standard deviation, e 2.718, and 3.14 . It is sometimes denoted N (, 2), which means the normal

distribution with a mean of and a variance of 2. If you have three normal distributions with the same standard deviation (same spread), but different means (different averages), they would look like this: 1 2 3 If you had the same mean but different standard deviations, it would look like this: large standard deviation

If you had the same mean but different standard deviations, it would look like this: medium standard deviation large standard deviation If you had the same mean but different standard deviations, it would look like this: small standard deviation middle standard deviation largest standard deviation Keep in mind that the areas are all the same, since they all equal 1.

Recall that if a random variable has mean and standard deviation , then (X-)/ has mean 0 and standard deviation 1. If X is normally distributed, then (X-)/ will be standard normal, N( normal with mean 0 and variance 1. This theorem is extremely useful. It means that we dont need to use the messy normal formula. We can standardize any normal distribution and look up probabilities in tables for the standard normal distribution. Using the standard normal table is not difficult, but it takes practice to get accustomed to it. The table in your textbook gives probabilities that

the standard normal (often called Z) is less than a particular number, that is Pr(Z a). Some tables are set up differently, so you need to notice how a table is computed when you use it. For example, the book we were using when I prepared most of these lecture notes gives probabilities that Z is between zero and a positive number, that is, Pr(0 Z a). Given any setup, you can always calculate the probabilities that you need. Z table: You get the integer part & the 1st decimal from the left column & the second decimal from the top row. Example: Pr(Z 2.63) = 0.9957 z

3.0 ? 0 2.63 Z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.1 0.2 2.6 0.9957

.9957 Do not memorize a lot of rules. You just need to remember 2 easy facts. 1. The graph is symmetric about 0. 2. The total area under the curve is 1. 0 Z Example Pr(Z < 1.85) 0 1.85

Z Example Pr(Z < 1.85) = 0.9678 0.9678 0 1.85 Z Example Pr(0 < Z < 1.85) 0.9678

0 1.85 Z Example Pr(0 < Z < 1.85) = 0.9678 0.5 = 0.4678 0.5 0.4678 0.9678 0

1.85 Z Example Pr(Z > 1.85) 0.9678 0 ? 1.85 Z Example

Pr(Z > 1.85) = 1 - 0.9678 = 0.0322 0.9678 0 0.0322 1.85 Z Example Pr(Z < -1.85) There are two ways to do this. 0.9678

-1.85 0 0.0322 1.85 Z Pr( Z < -1.85) = Pr(Z > 1.85) = 0.0322 Example The second way to determine Pr(Z < -1.85) is directly from the negative part of the Z table. -1.85

0 Pr( Z < -1.85) = 0.0322 Z Example Pr(Z > -1.85) -1.85 0 1.85 Z Example

Pr(Z > -1.85) = Pr(Z < 1.85) = 0.9678 -1.85 0 1.85 Z Example Pr(-1< Z < 2) -1.00 0 2.00

Z Example Pr(-1< Z < 2) = 0.9772 - 0.1587 = 0.8185 0.158 7 -1.00 0.9772 0 2.00 Z

Example Pr(1< Z < 2) ? 0 1 2 Z Example Pr(1< Z < 2) = 0.9772 - 0.8413 = 0.1359 0.841 3 0.9772

0 1 2 Z Example Pr( Z < 7) 0 Z 7 Example

Pr( Z < 7) = 1.0000 (to 4 decimal places) 0 Z 7 Example Pr( Z > 7) 0 Z 7 Example

Pr( Z > 7) = 0.0000 (to 4 decimal places) 0 Z 7 Example Pr(0 < Z < 7) 0 Z 7 Example

Pr(0 < Z < 7) = 0.5000 0 Z (to 4 decimal places) 7 Example What is the value of a such that Pr(Z < a) = 0.9207 ? 0.9207 0 a

Z Example What is the value of a such that Pr(Z < a) = 0.9207 ? 0.9207 0 1.41 Z Example What is the value of b such that Pr(Z > b) = 0.0250 ? 0.0250

0 b Z Example What is the value of b such that Pr(Z > b) = 0.0250 ? 1 0.0250 = 0.9750 0 0.0250 b Z

Example What is the value of b such that Pr(Z > b) = 0.0250 ? 1 0.0250 = 0.9750 0 0.0250 1.96 Z Example What is the value of k such that Pr(0 < Z < k) = 0.4750 ? 0.4750

0.5 + 0.4750 = 0.9750 0 k Z Example What is the value of k such that Pr(0 < Z < k) = 0.4750 ? 0.4750 0.5 + 0.4750 = 0.9750 0 1.96

Z Example: If X is N(2, 9), determine Pr(X 5). Pr(X 5) X- 5 Pr X- 5 2 Pr 3

Pr( Z 1) 0.8413 0.8413 0 1.00 Z Useful Fact The distribution of the individual observation is the same as the distribution of the population from which it was drawn. For example, if the mean height of a population of men is 70 inches, then the expected value or mean of a randomly selected man will be 70

inches. Also, if 5% of the population of men is over 78 inches, then the probability that a randomly selected man will be over 78 inches tall is 5%. Example: Suppose that womens heights are normally distributed with mean 64 inches & standard deviation 3 inches. What is the probability that a randomly selected woman is under five feet tall? Pr(X 60) X - 60 Pr X - 60 64

Pr 3 Pr( Z 1.33) 0.0918 -1.33 0 Z Sample Distribution of X the probability distribution of all possible values of that could occur when a sample of size n is taken from some specified population

x Example: Suppose we have a population of chips, 1/3 of which have a 1 on them, 1/3 have a 2, & 1/3 have a 3. (a) Show in table form the distribution of the sample mean (with n=2, sampled with replacement). (b) Graph the distribution of the sample mean. (c) Graph the distribution of the original population of chips. (d) What are the mean & variance of the original population? (e) What are the mean & variance of the sample mean? Show in table form the distribution of the sample mean (with n=2, sampled with replacement). sample 1,1

1,2 2,1 1,3 3,1 2,2 2,3 3,2 3,3 x sample mean 1.0 1.5 1.5 2.0 2.0 2.0

2.5 2.5 3.0 x sample mean 1.0 1.5 2.0 2.5 3.0 probability 1/9 2/9 3/9 2/9

1/9 Graph the distribution of the sample mean. x sample mean 1.0 1.5 2.0 2.5 3.0 probability 1/9 2/9 3/9 2/9 1/9

Probability 3/9 2/9 1/9 0 0.5 1.0 1.5 2.0 2.5 sample mean

x 3.0 Graph the distribution of the original population of chips. Since there are three equally likely values (1, 2, and 3), the distribution looks like this. Probability 1/3 0 1 2

3 x What are the mean & variance of the original population? x 1 2 3 p(x) 1/3 1/3 1/3 xp(x)

1/3 2/3 3/3 =6/3=2 What are the mean & variance of the original population? x 1 2 3 p(x) 1/3 1/3 1/3

xp(x) 1/3 2/3 3/3 =6/3=2 x2 x2p(x) 1 1/3 4 4/3 9 9/3 E(X2) =14/3 V(X) = E(X2)- [E(X)]2 = 14/3 22 = 2/3

What are the mean & variance of the sample mean X ? x 1.0 1.5 2.0 2.5 3.0 p( x ) 1/9 2/9 3/9 2/9 1/9 xp( x )

1/9 3/9 6/9 5/9 3/9 E ( X ) 18 / 9 2 What are the mean & variance of the sample mean X ? x 1.0 1.5 2.0 2.5 3.0 p( x )

1/9 2/9 3/9 2/9 1/9 xp( x ) 1/9 3/9 6/9 5/9 3/9 E ( X ) 18 / 9 2 x 2

1.00 2.25 4.00 6.25 9.00 2 x p( x ) 2 0.111 0.500 1.333 1.389 1.000

E ( X ) 4.333 In our chip example, we found that E ( X ) 2 E ( X ) 2 2 V (X ) 3 1 V (X ) 3 The expected value (or mean) of the original population

and the sample mean are the same. However, the variance of the sample mean is smaller than the variance of the original population. In general, E ( X ) E ( X ) 2 V ( X ) 2 V (X ) n

The mean of the original population and the mean of the sample are the same. However, as long as the sample size is more than one, the variance of the sample mean is smaller than the variance of the original population. Intuition: Suppose that each person in the class randomly sampled 50 men from a population of men whose average height is 70 inches. Each student then calculated his/her sample mean. If you averaged together all the sample means, youd expect to get something very close to 70 inches. The expected value or mean of the original population and the expected value of the sample mean are the same. Why does the sample mean have a smaller variance than the original population?

Suppose 5% of the population is taller than 78 inches (66). Then theres a 0.05 probability of selecting at random an individual with a height over 66 inches. It is much less likely that you will select at random 50 men whose average is over 66. You get a mixture of tall guys & short guys, so your average tends to be pretty close to the average for the population. So the distribution of the sample mean clusters more tightly about the average than does the distribution of the original population. That is, the variance of the sample mean is smaller than the variance of the original population. Central Limit Theorem As the sample size n increases, the distribution of the sample mean X of a random sample from a population (not necessarily normal) with mean and variance 2 approaches normal with mean

and variance 2/n. Example: Suppose the grades of a large class have a mean of 72 and a standard deviation of 9. a. What is the probability that the average grade of a random sample of 25 students will be above 77? b. If the population is normal, what is the probability that an individual student drawn at random will have a grade over 77? Notice that in part b, we need to assume normality but we didnt in part a. This is because the central limit theorem assures us of an approximately normal distribution for the sample mean of a reasonably large sample, but not for the distribution of a single observation. To be assured of that, we need to know that the distribution of the original population was normal.

What is the probability that the average grade of a random sample of 25 students will be above 77? Pr( X 77) X 77 72 Pr 9 n 25

Pr Z 2.78 = 1 - 0.9973 0.9973 0 2.78 Z = 0.0027 If the population is normal, what is the probability that an individual student drawn at random will have a grade over 77?

X 77 72 Pr( X 77) Pr 9 Pr Z 0.56 = 1 - 0.7123 0.7123 0 0.56 Z = 0.2877

Notice The probability of selecting an individual at random who has a grade five points above the class mean is much greater than the probability of randomly selecting a sample of 25 students who have an average that is five points above the class mean. (0.2877 versus 0.0027) Problem Suppose we are sampling (without replacement) and we sample the entire population. Then our sample mean will always be the same as the population mean. No spread. Zero variance. If we dont sample the entire population but do sample a large part of it, our variance will not be zero but it will be very small.

Thus, as the sample size n approaches the population size N, the variance of the sample mean approaches zero. Our formula for the variance of the sample mean was 2/n. 2/n approaches zero as n approaches infinity, not as n approaches N. So we need to make some adjustment when we sample a large part of our population. Finite Population Correction Factor When we sample a substantial part of our population (more than 5%), we need to multiply the standard deviation of our sample mean by this factor: N n N1

So the standard deviation of the sample mean which was: becomes: n n N n N1 The standard deviation of the sample mean adjusted for a large sample n

N n N1 Notice that if you sample just one observation (n=1), the square root part becomes 1, and the formula becomes the old formula: n If you sample the entire population (n=N), the formula has a value of 0. So the adjustment works the way it should. So our standardized X is now X n

N n N 1 Example: What is the probability that the mean of a sample of 36 observations, from a population of 300, will be less than 14, if the population mean & population standard deviation are 15.30 & 4.10 respectively? If we sample more than 5% of our population, we need to use the finite population correction factor, and here we are sampling 12%. Pr( X 14.00) X 14.00 15.30

Pr N n 4.10 300 36 300 1 36 n N1

Pr Z 2.02 = 0.0217 -2.02 0 Z Up until now, as long as our sample was not too large, we have used this standardization formula: X Z n

However, what if we dont know what the population standard deviation is? The next best thing is the sample standard deviation: But when we use s instead of , the result is not a normally distributed variable. It has what is called a t or students t distribution. n s (X X ) i 1

i n 1 X t s n 2 Our t distribution looks very similar to the Z distribution. Both are symmetric, bell-shaped, & have a mean of zero. But while the Z has a variance of 1, the t has a variance of (n-1)/(n-3), which is greater than one. So the t is wider than the Z.

So, the numbers are different & we need to use a different table. The t distribution is tabulated based on the number of degrees of freedom. The number of degrees of freedom, is denoted by dof, df, or the Greek letter nu: In this context, the number of degrees of freedom is n-1, where n is the number of observations. For each number of degrees of freedom, there is a different t distribution. The degrees of freedom are often indicated as a subscript on the t. For example, t15 is a t distribution with 15 degrees of freedom.

When the sample size is large, the Z and t distributions are virtually indistinguishable. When we have at least 100 observations, we will be able to use values from the Z table for our t. Some people use the Z to approximate the t when there are 30 or more observations. When were working with s, I prefer to use the t until 100 observations. When you take Intermediate Statistics (EC252), make sure you know what your instructor uses. The t distribution is set up differently from the Z table. The table in your text book gives probabilities that the t is greater than a specified positive number, that is, Pr(t a). Some tables are set up differently, so you need to

notice how a table is computed when you use it. Example: For 6 observations or 5 df, Pr(t5 > 3.3649) = 0.01 0 0.10 0.05 0.025 t5 0.01

d.f. = 1 2 3 4 5 6 100 3.3649 3.3649 0.005 Example: What is the probability that the sample mean is less than 800, if the population mean is

843.3419 and the sample standard deviation is 105, based on a sample of 25. Pr( X 800) X 800 843.3419 Pr 105 s n 25 -2.0639

0 t24 Pr( t24 2.0639 ) 0.025 Using Excel to solve t distribution problems On an Excel spreadsheet, you can get the t distribution as follows: Either click the fx (insert function) button OR click the formulas tab, and then click insert function select statistical as the category of function, scroll down to the t.dist or t.inv function, and click on it fill in the information in the dialog box .

Using Excel to solve t distribution problems Were going to look at Excels two-tailed t functions. There are also some other t functions on Excel that you can explore. t.dist.2t - you provide the number x on the horizontal axis and the degrees of freedom, and it then gives you the area or probability in the two tails. (Note that the number you provide must be positive.) t.inv.2t - you provide the probability or area that you want in the two tails and the degrees of freedom, and it then gives you the number on the horizontal axis. Example: Suppose that you wanted to use Excel to find the probability that a t with 24 degrees of freedom is less than -2.0639. Note that the area to the left of

-2.0639 is the same as the area to the right of +2.0639. Select t.dist.2t . Specify 2.0639 as x, 24 for degrees of freedom. -2.0639 0 2.0639 t24 Excel will provide you the probability value of 0.05. Since you only want the left tail, the number you want is half of 0.05 or 0.025 .

Example: Suppose instead you wanted to use Excel to find out for what value of k Pr(t24 < -k or t24 > k) is equal to 0.05. Select t.inv.2t . Specify 0.05 for probability, and 24 for degrees of freedom. -k 0 k t 24 Excel will provide you the value on the horizontal axis of

2.0639. Normal Approximations Recall that in the section on discrete distributions, we discussed the binomial distribution, which we used when we were considering the number of successes (x) on n independent trials with a probability of success on any given trial . We said that the binomial distribution is approximately symmetric if n 5 and n( 5 . We also said that the mean of the binomial distribution is = n and the variance is 2 = n(), so the standard deviation is n (1- ) . It is sometimes easier to use the normal distribution to approximate a binomial probability than to calculate the precise binomial probability.

The normal distribution provides a reasonable approximation for the binomial if n 5 and n( 5. However, because the binomial is discrete and the normal is continuous, we make an adjustment which we call the continuity correction. Recall that for discrete probability distributions, the probability that the variable takes on a particular value x is the height of a spike at that point. However, for a continuous probability distribution, the probability is the area under the probability density curve. To adjust for that difference, we include an extra half a unit on the ends of our intervals when we calculate normal probabilities as approximations for binomial probabilities. Examples (1) To estimate binomial probability Pr(X = 20), calculate normal probability Pr(19.5 X 20.5). (2) To estimate binomial probability Pr(3 X 7),

calculate normal probability Pr(2.5 X 7.5). (3) To estimate binomial probability Pr(X 10), calculate normal probability Pr(X 9.5). (4) To estimate binomial probability Pr(X < 15), calculate normal probability Pr(X < 15.5). Example: Suppose that the probability that any individual in a given population supports a particular proposition is 0.20. Consider a group of 500 people. Determine the probability that at least 75 people support the proposition. In this problem, n = 500 and So n = 100 5 and n( = 400 5 , and this should be an excellent approximation. The mean is n = 100 and the standard deviation is n (1- ) = 500(0.20)(0.80) 80 8.9443 . For the continuity correction, estimate the binomial

probability Pr(X 75) using the normal probability Pr(X 74.5). We need to standardize the normal by subtracting off the mean and dividing by the standard deviation. So we have this: Pr(X 74.5) = Pr[(X- )/ (74.5 100)/8.9443] = Pr(Z -2.85) 0.9978 -2.85 0 Z = 1- Pr(Z < -2.85)

= 1 - 0.0022 = 0.9978 So its extremely likely that at least 75 people in a sample of 500 will support the proposition.