Analysis of Economic Data Dr. Ka-fu Wong ECON1003 Analysis of Economic Data

The Continuous Probability Distribution Chapter Five The Continuous Probability Distribution GOALS Uniform probability distribution Normal probability distribution. Define and calculate z values. Determine the probability an observation will lie between two points using the standard normal distribution. Determine the probability an observation will be above or below a given value using the standard normal distribution. Compare two or more observations that are on different probability distributions. Use the normal distribution to approximate the binomial probability distribution. l

Continuous Probability Distributions The curve f(x) is the continuous probability distribution (or probability curve or probability density function) of the random variable x if the probability that x will be in a specified interval of numbers is the area under the curve f(x) corresponding to the interval. Properties of f(x) f(x)  0 for all x The total area under the curve of f(x) is equal to 1

Mean and Variance for Continuous Probability Distributions Mean and Variance for Discrete Probability Distributions

The Uniform Distribution If c and d are numbers on the real line, the probability curve describing the uniform distribution is The mean and standard deviation of a uniform random variable x are

The Uniform Probability Curve

The Normal Probability Distribution The normal probability distribution is defined by the equation and  are the mean and standard deviation,  = 3.14159 … and e = 2.71828 is the base of natural or Naperian logarithms.

Importance of Normal Distribution Describes many random processes or continuous phenomena Basis for Statistical Inference

Characteristics of a Normal Probability Distribution bell-shaped and single-peaked (unimodal) at the exact center of the distribution.

Characteristics of a Normal Probability Distribution Symmetrical about its mean. The arithmetic mean, median, and mode of the distribution are equal and located at the peak. Thus half the area under the curve is above the mean and half is below it.

Characteristics of a Normal Probability Distribution The normal probability distribution is asymptotic. That is the curve gets closer and closer to the X-axis but never actually touches it.

The density functions of normal distributions with zero mean value and different standard deviations Bell-shaped Symmetric Mean=median = mode Unimodal Asymptotic

Difference Between Normal Distributions x (a) x (b) x (c)

Normal Distribution Probability Probability is the area under the curve! A table may be constructed to help us find the probability f(X) X c d

Infinite Number of Normal Distribution Tables Normal distributions differ by mean & standard deviation. f(X) Each distribution would require its own table. X

The Standard Normal Probability Distribution The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is also called the z distribution. A z-value is the distance between a selected value, designated X, and the population mean , divided by the population standard deviation, . The formula is:

Working with mean and Standard Deviation Set Data Mean St Dev (1) 19 20 21 20.00 0.82 (2) -1 1 0.00 (3) 0.71 (4) 38 40 42 40.00 1.63 (5) 57 60 63 60.00 2.45 (2) = (1) – mean(1): Mean(2)=0; Stdev(2)=Stdev(1) (3) = (1) + mean(1) Mean(3)=Mean(1); Stdev(3)<Stdev(1). (4) = (1)*2; (5) = (1)*3 Mean(4)=mean(1)*2; mean(5)=mean(1)*3 Stdev(4)=stdev(1)*2; stdev(5)=stdev(1)*3

Transform to Standard Normal Distribution A numerical example Any normal random variable can be transformed to a standard normal random variable x x- (x-)/σ x/σ -2 -1.4142 1 -1 -0.7071 0.7071 2 1.4142 3 2.1213 4 2.8284 Mean std

The Standard Normal Probability Distribution Any normal random variable can be transformed to a standard normal random variable Suppose X ~ N(µ,  2) Z=(X-µ)/  ~ N(0,1) P(X<k) = P [(X-µ)/  < (k-µ)/  ]

Standardize the Normal Distribution Standardized Normal Distribution   = 1 z  X  = 0 Z Z Because we can transform any normal random variable into standard normal random variable, we need only one table!

Standardizing Example Normal Distribution Standardized Normal Distribution  = 10  = 1 Z  = 5 X  = 0 .12 Z 6.2 Z

Obtaining the Probability Standardized Normal Probability Table (Portion) .02 Z .00 .01  = 1 Z 0.0 .0000 .0040 .0080 0.0478 0.1 .0398 .0438 .0478 0.2 .0793 .0832 .0871  = 0 0.12 Z Z 0.3 .1179 .1217 .1255 Shaded Area Exaggerated Probabilities

Example P(3.8  X  5) Normal Distribution Standardized Normal Distribution  = 10  = 1 Z 0.0478 3.8  = 5 X -0.12  = 0 Z Z Shaded Area Exaggerated

Example (2.9  X  7.1) X   2 . 9  5 Z     . 21  10 X   7 . Normal Distribution Z    . 21 Standardized Normal Distribution  10  = 10  = 1 Z .1664 .0832 .0832 2.9 5 7.1 X -.21 .21 Z Shaded Area Exaggerated

Example (2.9  X  7.1) X   2 . 9  5 Z     . 21  10 X   7 . Normal Distribution Z    . 21 Standardized Normal Distribution  10  = 10  = 1 Z .1664 .0832 .0832 2.9 5 7.1 X -.21 .21 Z Shaded Area Exaggerated

Example P(X  8) X   8  5 Z    . 30 Normal Distribution  10 Standardized Normal Distribution  = 10  = 1 Z .5000 .3821 .1179  = 5 8 X  = 0 .30 Z Z Shaded Area Exaggerated

Example P(7.1  X  8) X   7 . 1  5 Z    . 21  10 X   8  5 Normal Distribution Z    . 30 Standardized Normal Distribution  10  = 10  = 1 Z .1179 .0347 .0832  = 5 7.1 8 X  = 0 Z .21 .30 z Shaded Area Exaggerated

Normal Distribution Thinking Challenge You work in Quality Control for GE. Light bulb life has a normal distribution with µ= 2000 hours &  = 200 hours. What’s the probability that a bulb will last between 2000 & 2400 hours? less than 1470 hours?

Solution P(2000  X  2400) Normal Distribution = P[(X-µ)/  < (2400-µ)/  ] – P [(X-µ)/  < (2000-µ)/  ] = P[(X-µ)/  < (2400-µ)/  ] – 0.5 Normal Distribution Standardized Normal Distribution X   2400  2000 Z    2 .  200  = 200  = 1 Z .4772  = 2000 2400 X  = 0 2.0 Z Z

Solution P(X  1470) X   1470  2000 Z     2 . 65  200 P(X<1470) = P [(X-µ)/  < (1470-µ)/  ] X   1470  2000 Z     2 . 65  200 Normal Distribution Standardized Normal Distribution  = 200  = 1 Z .5000 .0040 .4960 X Z 1470  = 2000 -2.65  = 0 Z

Finding Z Values for Known Probabilities What Is Z Given P(Z) = 0.1217? Standardized Normal Probability Table (Portion)  = 1 .01 Z .00 .02 Z .1217 0.0 .0000 .0040 .0080 0.1 .0398 .0438 .0478  = 0 .31 Z 0.2 .0793 .0832 .0871 Z 0.3 .1179 .1217 .1255 Shaded Area Exaggerated

Finding X Values for Known Probabilities Normal Distribution Standardized Normal Distribution  = 10  = 1 Z .1217 .1217 ?  = 5 X  = 0 .31 Z Z Shaded Area Exaggerated

EXAMPLE 1 The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $2,000 and a standard deviation of $200. What is the z-value for a salary of $2,200?

EXAMPLE 1 continued What is the z-value of $1,700 ? A z-value of 1 indicates that the value of $2,200 is one standard deviation above the mean of $2,000. A z-value of –1.50 indicates that $1,700 is 1.5 standard deviation below the mean of $2000.

Areas Under the Normal Curve About 68 percent of the area under the normal curve is within one standard deviation of the mean.  ±  P( -  < X <  + ) = 0.6826 About 95 percent is within two standard deviations of the mean.  ± 2  P( - 2  < X <  + 2 ) = 0.9544 Practically all is within three standard deviations of the mean.  ± 3  P( - 3  < X <  + 3 ) = 0.9974

Areas Under the Normal Curve Between: ± 1  - 68.26% ± 2  - 95.44% ± 3  - 99.74% µ-2σ µ µ+2σ µ-3σ µ-1σ µ+1σ µ+3σ

EXAMPLE 2 The daily water usage per person in New Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons. About 68 percent of those living in New Providence will use how many gallons of water? About 68% of the daily water usage will lie between 15 and 25 gallons.

EXAMPLE 3 What is the probability that a person from New Providence selected at random will use between 20 and 24 gallons per day? P(20<X<24) =P[(20-20)/5 < (X-20)/5 < (24-20)/5 ] =P[ 0<Z<0.8 ]

Example 3 continued The area under a normal curve between a z-value of 0 and a z-value of 0.80 is 0.2881. We conclude that 28.81 percent of the residents use between 20 and 24 gallons of water per day. See the following diagram.

x EXAMPLE 3 P(0<z<.8)=.2881 0<x<.8 -4 -3 -2 -1 0 1 2 3 4

How do we know P(0<z<0.8)=0.2881 P(z<c) Computer software (such as MegaStat in Excel), based on an numerical integration of the standard normal density function. Numerical table for “Standard normal” c P(0<z<c) c

How do we find P(0<z<0.8) P(0<z<c) P(z<c) c c P(0<z<0.8) = 0.2881 P(0<z<0.8) = P(z<0.8) – P(z<0) =0.7881 – 0.5 =0.2881

EXAMPLE 3 continued What percent of the population use between 18 and 26 gallons of water per day? Suppose X ~ N(µ,  2) Z=(X-µ)/  ~ N(0,1) P(X<k) = P [(X-µ)/  < (k-µ)/  ]

How do we find P(-0.4<z<1.2) P(0<z<c) P(z<c) c c P(-0.4<z<1.2) = P(z<1.2) - P(z<-0.4) = P(z<1.2) - P(z>0.4) = P(z<1.2) – [1- P(z<0.4)] =0.8849 – [1- 0.6554] =0.5403 P(-0.4<z<1.2) = P(-0.4<z<0) + P(0<z<1.2) =P(0<z<0.4) + P(0<z<1.2) =0.1554+0.3849 =0.5403 P(-0.4<z<0) =P(0<z<0.4) because of symmetry of the z distribution.

EXAMPLE 4 Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A. What is the lowest score a student can earn and still receive an A?

Example 4 continued Those with a score of 77.2 or more earn an A. To begin let k be the score that separates an A from a B. 15 percent of the students score more than k, then 35 percent must score between the mean of 72 and k. Write down the relation between k and the probability: P(X>k) = 0.15 and P(X<k) =1-P(X>k) = 0.85 Transform X into z: P[(X-72)/5) < (k-72)/5 ] = P[z < (k-72)/5] P[0<z < s] =0.85 -0.5 = 0.35 Find s from table: P[0<z<1.04]=0.35 Compute k: (k-72)/5=1.04 implies K=77.2 Those with a score of 77.2 or more earn an A.

The Normal Approximation to the Binomial The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n. The normal probability distribution is generally a good approximation to the binomial probability distribution when n  and n(1-  ) are both greater than 5. Why can we approximate binomial by normal? Because of the Central Limit Theorem.

The Normal Approximation continued Recall for the binomial experiment: There are only two mutually exclusive outcomes (success or failure) on each trial. A binomial distribution results from counting the number of successes. Each trial is independent. The probability is fixed from trial to trial, and the number of trials n is also fixed.

The Normal Approximation binomial

Continuity Correction Factor Because the normal distribution can take all real numbers (is continuous) but the binomial distribution can only take integer values (is discrete), a normal approximation to the binomial should identify the binomial event "8" with the normal interval "(7.5, 8.5)" (and similarly for other integer values). The figure below shows that for P(X > 7) we want the magenta region which starts at 7.5.

Continuity Correction Factor Example: If n=20 and p=.25, what is the probability that X is greater than or equal to 8? The normal approximation without the continuity correction factor yields z=(8-20 × .25)/(20 × .25 × .75)0.5 = 1.55, P(X ≥ 8) is approximately .0606 (from the table). The continuity correction factor requires us to use 7.5 in order to include 8 since the inequality is weak and we want the region to the right. z = (7.5 - 20 × .25)/(20 × .25 × .75)0.5 = 1.29, P(X ≥ 7.5) is .0985. The exact solution from binomial distribution function is .1019. The continuity correct factor is important for the accuracy of the normal approximation of binomial. The approximation is quite good.

EXAMPLE 5 A recent study by a marketing research firm showed that 15% of American households owned a video camera. For a sample of 200 homes, how many of the homes would you expect to have video cameras? What is the mean? What is the variance? What is the standard deviation?

EXAMPLE 5 continued What is the probability that less than 40 homes in the sample have video cameras? “Less than 40” means “less or equal to 39”. We use the correction factor, so X is 39.5. The value of z is 1.88.

Example 5 continued From Standard Normal Table the area between 0 and 1.88 on the z scale is .4699. So the area to the left of 1.88 is .5000 + .4699 = .9699. The likelihood that less than 40 of the 200 homes have a video camera is about 97%.

EXAMPLE 5 P(z<1.88) =.5000+.4699 =.9699 z=1.88 z 0 1 2 3 4

The Exponential Distribution If  is positive, then the exponential distribution is described by the probability density function mean mx=1/l standard deviation sx=1/l

Example: Computing Exponential Probabilities Given mx=3.0 or l=1/3=.333, 0.05 0.1 0.15 0.2 0.25 0.3 0.35 5 9 l=0.333 x mx

Chapter Five The Continuous Probability Distribution - END -