Statistics for Business and Economics: Random Variables:Continuous STT 315: Section 107 Acknowledgement: I d like to thank Dr. Ashoke Sinha for allowing me to use and edit the slides. Murray Bourne (interactive math) and digitalfirst.bfwpub.com for applets
Probability Distribution The histogram of the random variable The plot probability density or relative frequency (p(x)) against the value of random variable (x). 2
Probability distribution Let X be any random variable with real number values. a and b are fixed numbers. P( a X b) = Area under the histogram of Y between, including a and between, including b Case discrete: X takes discrete values like 0,1,2,3,4,.10. X ~ Bin(10,0.5) P(1 X 2) =Area of rectangle_1 + area of rectangle_2 =height of rectangle_1* width rectangle_1 + height of rectangle_2* width rectangle_2 3
= binompdf(10,0.5,1)*width of rectangle 1 + binompdf(10,0.5,2)*width of rectangle 2 = 0.009765 * 1 + 0.043945 * 1 = 0.05371 = Binomcdf(10,0.5,2) - Binomcdf(10,0.5,0) 4
Case Continuous: Let Y be a continuous random variable eg: Y follows normal distribution P( 1 X 2)= 2 dx. 1 f x = = Integral of f(x) i.e probability density function from 1 to 2. = 1 2 1 e (x μ)2 2σ 2 σ 2π dx
Normal Distribution 7
Normal random variable A normal random variable X is a continuous random variable has a probability distribution which is bell-shaped, i.e., unimodal, symmetric. In many data-sets, the histogram is bell-shaped. These data-sets can be modeled using normal distribution. Example :Height of population, shoe size, intelligence, error in measurement 8
Normal distribution Normal distribution is identified by its mean (μ) and standard deviation (σ). The form of the normal curve [probability density function] is defined for all real x, i.e. < x <, f x = 1 (x μ) 2 σ 2π e 2σ 2, where π = 3.1415 and e = 2.71828 A normal random variable with μ = 0, σ = 1 is called a standard normal random variable. If X is normal with mean mean (μ) and standard deviation σ, then Z = X μ is standard normal. σ 9
Normal graph effect of mean and standard deviations http://www.intmath.com/counting-probability/normal-distribution-graph-interactive.php Go to above applet for interactive demo
Computing normal probabilities Since normal random variable is continuous P X = x = 0 for all x. Thus for any two numbers a and b, P a < X < b = P a < X b = P a X < b = P a X b. We shall use TI 83/84 for computation. We generally face two type of problems: To compute P a < X < b [use normalcdf]; Given the value of p finding x such that p = P X x [use invnorm]. 11
Approximately what percent of U.S. women do you expect to be between 66 in and 67 in tall? Heights of adult women are normally distributed with mean of 63.6 in, standard deviation of 2.5 in. Use TI 83/84 Plus. Press [2nd] & [VARS] (i.e. [DISTR]) Select 2: normalcdf Format of command: normalcdf(lower bound, upper bound, mean, std.dev.) For this problem: normalcdf(66, 67, 63.6, 2.5) = 0.0816. i.e. about 8.2% of adult U.S. women have heights between 66 in and 67 in. 12
P(1 X 2) = P(X=1) + P(1<X<2) +P(X=2) = Area of red + Area of yellow + Area of green = 0 + P(1<Y<2) + 0 Area of red, green rectangle = Height *( width= 0) =0
Approximately what percent of U.S. women do you expect to be less than 64 in tall? Heights of adult women are normally distributed with mean of 63.6 in, standard deviation of 2.5 in. Note that here upper bound is 64, but there is no mention of lower bound. So it is infinity (negative) So take a very small value for lower bound of your choice, say -10^10 which represents infinity For this problem normalcdf(-10^10, 64, 63.6, 2.5) = 0.5636. i.e. about 56.4% of adult U.S. women have heights less than 64 in. 14
Approximately what percent of U.S. women do you expect to be more than 58 in tall? Heights of adult women are normally distributed with mean of 63.6 in, standard deviation of 2.5 in. Note that here lower bound is 58, but there is no mention of upper bound. So it is + infinity (positive) So take a very high value for upper bound of your choice, say 10^10 which represents + infinity For this problem normalcdf(58, 10^10, 63.6, 2.5) = 0.987. i.e. 98.7% of adult U.S. women have heights more than 58 in. 15
What about men s height? Heights of adult men are normally distributed with mean of 69 in, standard deviation of 2.8 in. normalcdf(60, 10^10, 69, 2.8) = 0.999. Hence 99.9% adult male will have height more than 60 in. normalcdf(64, 10^10, 69, 2.8) = 0.963. So 96.3% adult male will have height more than 64 in. Thus for U.S. Army height restriction for women is more restrictive compared to men. But for U.S. Marine height restriction for men is more restrictive compared to women. 16
Below what height 80% of U.S. men do have their heights? Heights of adult men are normally distributed with mean of 69 in, standard deviation of 2.8 in. The question is to find the height x such that {Percent of men s height < x} = 80% = 0.8. Use TI 83/84 Plus. Press [2nd] & [VARS] (i.e. [DISTR]) Select 3: invnorm Format of command: invnorm(fraction, mean, std.dev.) For this problem: invnorm(0.8, 69, 2.8) = 71.36. i.e. 80% of U.S. men have heights less than 71.36 in. 17
Remark: invnorm invnorm only considers percentage or fraction in the lower tail of normal distribution. For example, suppose the question is Above what height 10% of U.S. men do have their heights? Notice here the question is find the height x such that {Percent of men s height > x} = 10% = 0.1. This means {Percent of men s height < x} = (100-10)% = 90% = 0.9. For this problem: invnorm(0.9, 69, 2.8) = 72.59. i.e. 90% of U.S. men have heights less than 72.59 in, i.e. 10% of U.S. men have heights more than 72.59 in. 18
Normal approximation of binomial distribution Suppose X~Bin n, p. Hence μ = np, σ X = npq. If n is very large, then the probability distribution of X can be approximated by normal distribution with μ = np, σ X = npq. However, X being binomially distributed is a discrete random variable, whereas normal distribution is continuous. So we need a continuity correction. If n is very large, then we compute as follows: P Bin n, p r. v. x P N np, npq r. v. x + 0. 5. 19
How large n should be? To apply normal approximation to binomial distribution n should be so large that the interval np ± 3 npq should lie in the range 0 and n. Eg: n=1000 p=0.7 then np ± 3 npq =700 ± 3*14.49 Here 656.52, 743.47 lie between 0 and n=1000 The applet demo : http://digitalfirst.bfwpub.com/stats_applet/stats_applet_2_cltbinom.html 20
Example Let X be binomially distributed with n = 50, p = 0.2. Here μ = np = 10, σ = npq = 2.828. Thus np ± 3 npq = 1.515, 18.49, which lies in the range 0 and n = 50. Hence n is large enough. P X 16 P N 10, 2.828 16.5 = normalcdf 100, 16.5, 10, 2.828 = 0.9892. P X < 16 = P(X 15) P N 10, 2.828 15.5 = normalcdf 100, 15.5, 10, 2.828 = 0.9741. P X = 16 = P X 16 P X < 16 0.9892 0.9741 = 0.0151. P X > 12 = 1 P X 12 1 PሾN 10, 2.828 12.5 ሿ = 1 normalcdf 100, 12.5, 10, 2.828 = 0.1883. P X 12 = 1 P X < 12 = 1 P X 11 1 normalcdf 100, 11.5, 10, 2.828 = 0.2979. 21
Example Suppose X is normally distributed with mean μ and standard deviation σ. What is the probability that the value of X will be within 1.5 standard deviation from the mean? i.e. P μ 1.5σ X μ + 1.5σ =? Solution: Remember that Z = X μ The z-score of μ 1.5σ is = the z-score of μ + 1.5σ is = σ μ 1.5σ μ σ μ+1.5σ μ σ is standard normal. = 1.5 and = 1.5. So, P μ 1.5σ X μ + 1.5σ = P( 1.5 Z 1.5 ) =normalcdf(-1.5,1.5,0,1)=0.8664. 22
Sum of independent random variables 23
Combining Random Variables Let X and Y be two random variables. Then E(X ± Y) = E(X) ± E(Y). If further X and Y are independent, then V(X ± Y) = V(X) + V(Y). Notice that for variance both have a plus sign on the right hand side. For expectation, independence assumption is not necessary, but for the above variance formula it is required. Variance for dependent case will not be treated in this course. 24
Example Suppose X and Y are two independent random variables with E(X) = 4, V(X) = 2, E(Y) = -3, V(Y) = 4. Then E(X+Y) = E(X)+E(Y) = 4+(-3) = 1. V(X+Y) = V(X)+V(Y) = 2+4 = 6. σ(x+y) = std. dev. of (X+Y) = V(X+Y) = 6 = 2.45. E(X-Y) = E(X)-E(Y) = 4-(-3) = 7. V(X-Y) = V(X)+V(Y) = 2+4 = 6. σ(x-y) = std. dev. of (X-Y) = V(X Y) = 6 = 2.45. 25
Example Suppose X and Y are two independent random variables with Then E(3X-2Y) = E(3X) - E(2Y) = 3E(X) - 2E(Y) = 3 4-2 (-3) = 12 + 6 = 18. E(X) = 4, V(X) = 2, E(Y) = -3, V(Y) = 4. 26
Example Suppose X and Y are two independent random variables with E(X) = 4, V(X) = 2, E(Y) = -3, V(Y) = 4. Then V(3X-2Y) = V(3X) + V(2Y) = 3 2 V(X) + 2 2 V(Y) = 9 2 + 4 4 = 18 + 16 = 34. σ(3x-2y) = std. dev. of (3X-2Y) = V(3X-2Y) = 34 = 5.83. 27
Example These formulas can be extended to more than two random variables. Suppose we have the following information about random variables X, Y and Z. X, Y and Z are independent, and Random variables Expectations E(X + Y - Z) = E(X) + E(Y) E(Z) = (-4) + 2 9 = -11. V(X + Y - Z) = V(X) + V(Y) + V(Z) = 2 + 6 + 4 = 12. σ(x + Y - Z) = 12 = 3.464. Variances X -4 2 Y 2 6 Z 9 4 28
Another Example Suppose X, Y and Z are independent, and Random variables Expectations Standard deviations X 0 1 Y 0.2 3 Z 2.4 5 Notice that here we are given the standard deviations (not the variances). E(Y Z X) = E(Y) E(Z) E(X) = 0.2 2.4 0 = -2.2. V(Y Z X) = V(X) + V(Y) + V(Z) = 1 2 + 3 2 + 5 2 = 35. σ(y Z X) = 35 = 5.916. 29
Sum of independent normal random variables Suppose X is normal with mean μ 1 and variance σ 1 2, Y is normal with mean μ 2 and variance σ 2 2, and X and Y are independent of each other. Then X + Y is normally distributed with mean μ 1 + μ 2 and variance σ 1 2 + σ 2 2. That means E X + Y = μ 1 + μ 2. var X + Y = σ 1 2 + σ 2 2. σ X + Y = σ 1 2 + σ 2 2. 30
Example Suppose the monthly revenue in investment A is normally distributed with mean $25 and std.dev. $8, and that in investment B is normally distributed with mean $31 and std.dev. $10. If you have both investments, what is the probability that your total monthly revenue will be more than $75? The total monthly revenue will be normally distributed with mean $(25+31)=$56, and std. dev. 8 2 + 10 2 = $12.806. So probability that your total monthly revenue will be more than $75 is = normalcdf(75,1000,56,12.806) = 0.069. 31
Example Suppose the monthly revenue in investment A is normally distributed with mean $25 and std.dev. $8, and that in investment B is normally distributed with mean $31 and std.dev. $10. Above what value 80% of total monthly revenue will lie? The total monthly revenue will be normally distributed with mean $(25+31)=$56, and std. dev. 8 2 + 10 2 = $12.806. If x is the value above which 80% of total monthly revenue will lie, then 20% of total monthly revenue will lie below x. Thus x = invnorm(0.2,56,12.806) = $45.83. 32
Uniform distribution 33
Uniform distribution A continuous random variable X is uniformly distributed in the interval ሾa, bሿ if its probability density function is f x = 1 b a, for a x b. In this case, if a l < u b o P l < X < u = u l b a, o E X = a+b 2, o σ X = b a 12. 34
Example If price of gas (X) in East Lansing has uniform distribution in the interval $[3.45, 3.95] per gallon. Probability that gas price will be between $3.50 and $3.60 = P 3.50 X 3.60 = 3.60 3.50 3.95 3.45 = 0.2. Probability that gas price will be less than $3.70 = P( X < 3.70 ) = P 3.45 X < 3.70 = 3.70 3.45 = 0.5. 3.95 3.45 Probability that gas price will be more than $3.90 = P( X > 3.90 ) = P 3.90 < X 3.95 = 3.95 3.90 = 0.1. 3.95 3.45 Probability that gas price will be $3.82 = P X = 3.82 = 0, because X is a continuous random variable. 35
Example If price of gas (X) in East Lansing has uniform distribution in the interval $[3.45, 3.95] per gallon. The expected gas price in East Lansing is = 3.45+3.95 2 $3.70. = The standard deviation of gas price in East Lansing is = $0.144. = 3.95 3.45 12 36
Sampling distributions 37
Remember Population is the complete set of all items that we are interested in studying. Parameters are the values we calculate from the population data. Population mean (for quantitative variables), population proportion (categorical variables) etc. are the examples of parameters. A sample is a subset of the population. Statistics are values we compute from sample data. Sample mean, sample proportion etc. are the examples of statistics. Our goal is to make inference on parameters based on relevant statistics. 38
An example Consider a population with 10 individuals with the following smoking habit: Individual #: 1 2 3 4 5 6 7 8 9 10 Smoking habit: N N N N S S N N S N where S = smoker, and N = non-smoker. So 3 out of 10 people in the population is smoker. Here the population proportion of smoker is: p 3 10 0.3. 39
An example Suppose we decide to estimate population proportion on the basis of a sample proportion. Suppose simple random samples of size 4 (with replacement) are considered. Individuals selected Smoking habit Sample proportion (2, 4, 4, 9) (N, N, N, S) 1/4 = 0.25 (4, 7, 8, 10) (N, N, N, N) 0/4 = 0 (5, 6, 8, 8) (S, S, N, N) 2/4 = 0.5 Notice that the sample proportion s value depends on the sample selected, but the population proportion s value is fixed. 40
Few questions Can we justify the use of sample proportion as an estimator of population proportion? What can we expect about the value of sample proportion when population proportion (p) is 0.3? Does this behavior depend on the value of p? What is the margin of error, if we estimate p with sample proportion? (To be answered in a later lecture.) As sample proportion is a variable, what is its distribution? 41
Few questions Does it matter how the sample is selected? Does the sample size matter? Is this a problem of population proportion only? Or do we face it for other parameters also? This is a problem for all parameters, which are fixed in value for a particular population. The value of any statistic changes with the sample selected. 42
Sampling Distribution As any statistic s value changes with the selected sample, so statistic is a itself a random variable. The probability distribution of a sample statistic is called the sampling distribution of the statistic. In this course we shall study sampling distributions of sample proportion and sample mean. 43
Sampling method and sample size Samples must be independent. Simple random sampling with replacement ensures independence. Holds (approximately) also for without replacement sampling as long as the sample size is smaller than 10% of the population size. Sample size must be large enough. What is large enough depends on the statistic we are considering, i.e. different rules of large enough for sample proportion and sample mean. It is the sample size what is important, NOT what fraction of population is sampled. 44
Sampling distribution of sample proportion ( p) 45
Sampling distribution of sample proportion Consider in a population a categorical variable with two categories: success and failure. e.g., smoking habit variable the level smoker can be considered as success, and non-smoker as failure. Let p be the population proportion of success. A random sample from the population is drawn. Observations in the sample are independent. Sample size is n. Let x be the number of success in the sample. Then sample proportion of success is p Ƹ = x n. 46
Sampling distribution of sample proportion The expected value of pƹ is equal to p, i.e. E The standard deviation of pƹ is σ p = p(1 p) p Ƹ = p. If we repeatedly simulate the selection of samples from the population with large enough sample size, the distribution of the sample proportions we found in the samples will be roughly normally distributed and the distribution will be N p, n p(1 p) n.. 47
Sampling distribution of sample proportion When is sample large enough for the last result to hold? If n is so large that np > 9, and n(1 p) > 9. This is covered if the number of successes and failures are both at least 10. 48
Example One Of all the cars on the highway, about 80% exceed the speed limit. If we clock the next 50 cars that pass, what might we expect to find? Is the independence condition met? Most likely NO, because the cars moving at the same time may influence each others behavior. Suppose we randomly select 50 cars that pass. Is the independence condition met? Yes. 49
Example One Of all the cars on the highway, about 80% exceed the speed limit. Suppose we randomly select 50 cars that pass. Is sample size large enough condition met? a) Yes b) No Because, np = 50 0.8 = 40 > 9, and n(1-p) = 50 0.2 = 10 > 9. 50
Example One Of all the cars on the highway, about 80% exceed the speed limit. Suppose we randomly select 50 cars that pass. What is the expected proportion of cars in the sample to exceed the speed limit? A. 20% B. 80% C. 2.83% D. 0.057% 51
Example One Of all the cars on the highway, about 80% exceed the speed limit. Suppose we randomly select 50 cars that pass. What is the standard deviation of the sample proportion of cars exceeding the speed limit? A. 20 B. 80 C. 2.83 D. 0.057 52
Example One Of all the cars on the highway, about 80% exceed the speed limit. Suppose we randomly select 50 cars that pass. What is the chance that more than 90% of cars in the sample exceeded the speed limit? A. 0.80 B. 0.20 C. 0.057 D. 0.039 E. 0.961 normalcdf(0.9,100,0.8,0.057) = 0.039. 53
Sampling distribution of sample mean (ഥx) 54
Sampling distribution of sample mean Suppose the mean of population distribution is µ and standard deviation σ. A random sample from the population is drawn. Observations in the sample are independent. Sample size is n. Let the sample mean be xҧ The expected value of xҧ is equal to µ. The standard deviation of xҧ is σ. n 55
Central Limit Theorem (CLT) If we repeatedly simulate the selection of samples from the population with large enough sample size, the distribution of the sample mean in random sampling roughly follows a normal model and the distribution will be N(μ, σ n ). The larger the sample size, the closer to normal the distribution will be. But how large is large enough to apply CLT? We can use CLT if n 30. 56
Example Two At birth, babies average 7.8 pounds, with a standard deviation of 2.1 pounds. A random sample of 34 babies born to mothers living near a factory that might be polluting the air and water shows a mean birth-weight of only 7.2 pounds. What is the expected value of sample mean? A. 34 lb B. 7.2 lb C. 7.8 lb D. 2.1 lb 57
Example Two At birth, babies average 7.8 pounds, with a standard deviation of 2.1 pounds. A random sample of 34 babies born to mothers living near a factory that might be polluting the air and water shows a mean birth-weight of only 7.2 pounds. What is the standard deviation of sample mean? A. 1.23 lb B. 7.2 lb C. 0.36 lb D. 2.1 lb 58
Example Two At birth, babies average 7.8 pounds, with a standard deviation of 2.1 pounds. A random sample of 34 babies born to mothers living near a factory that might be polluting the air and water shows a mean birth-weight of only 7.2 pounds. What is the chance that the sample mean is lower than 7.2 lbs? A. 0.952 B. 0.388 C. 0.612 D. 0.048 normalcdf(-100, 7.2, 7.8, 0.36) = 0.048. 59
Example If price of gas (X) in East Lansing has uniform distribution in the interval $[3.45, 3.95] per gallon. Remember that μ = E X = $3.70, σ X = $0.144. Suppose we collect gas prices from 35 gas stations of East Lansing. The expected sample average of gas prices is E തX = μ = $3.70. The standard deviation of sample average of gas prices is σ തX = σ n = 0.144 35 = $0.024. Since n > 30, we have തX~N 3.70, 0.024. The chance that the sample average will be less than $3.65 is P തX < 3.65 = normalcdf 100, 3.65, 3.70, 0.024 = 0.0186. 60