Lecture 8: The Normal Distribution

The normal distribution is the most important distribution in business and economics.

Continuous Distributions

Recall: the manufacturing plant that produces pieces of metal with two holes. The distance between the holes varies from one piece to another.

Construct a histogram in which the area of the bar representing each class interval is equal to the proportion of all pieces within this class interval. (i.e. the height of the bar equals the proportion of all pieces within the interval divided by the width of the interval).

The area under this histogram must equal 1. This is called a density scale.

Now say we increase the number to 1,000,000 sampled, rather than 1,000. We would now be able to divide the class intervals more finely.

Finally, taken to the limit, the histogram would become a smooth curve.

The probability that the distance in hole centers lies within a particular range is equal in value to the area under the smooth curve in this range.

Probability Density Function of a Continuous Random Variable 

In the limit, as more and more observations are gathered concerning a continuous random variable, and as class intervals become narrower and more numerous, the histogram (using a density scale) of the variable becomes a smooth curve called a Probability Density Function (PDF). The total area under any PDF must equal 1. The probability that a random variable will assume a value between any  two points, a and b, equals the area under the random variable's PDF over the interval from a to b.

NOTE: in Calculus terms

The Normal Distribution

The PDF of a Normal Random Variable is given by:

The normal distribution is a family of distributions that depend on and .

Common Characteristics of normal distributions

1. Symmetrical and bell-shaped

2. Probability that a value will lie within k standard deviations of the mean ±1 = 68.3% ±2 = 95.4% ±3 = 99.7%

3. Location and shape determined entirely by and .

The normal distribution is important because

1. It is a reasonably good approximation to many populations (especially in nature).

2. The probability distribution of the sample mean should be close to the normal distribution.

3. The normal distribution can be used in many instances to approximate the binomial distribution.

The Standard Normal Curve

If one expresses any normal random variable as a deviation from its mean, and measures these deviations in units of its standard deviation, the resulting random variable is called a Standard Normal Random Variable. Its probability distribution is called the Standard Normal Curve.

The Standard Normal Random Variable

NOTE: If any normal variable is expressed in standard units, its probability distribution is given by the Standard Normal Curve.

of a standard normal random variable is equal to 0.

of a standard normal random variable is equal to 1.

Calculating Normal Probabilities

The Probability that a Normal Random Variable lies between two points (in our example: J. Crew --> height of male between 72 and 76 inches) can be calculated in two steps:

1.	Find the points on the Standard Normal Distribution corresponding to these two points.
	(72 - 70) / 2 = 1	(76 - 70) / 2 = 3 }
2.	Determine the area under the Standard Normal Curve between the two points. { the area under the standard normal between 1 & 3 }

Using the Table of the Standard Normal Distribution (Appendix Table II)

1.	Area between zero and a positive value. The table gives the area under the standard normal curve from zero to a positive value, so just read it off the table. (row 1.1 column 0.0 --> 0.3643)
2.	Area between zero and a negative value. symmetrical so do the same as above.
3.	Area to the left of some positive value. look in the table and add 0.5 (because the probability to the left of 0 is 0.5). ( 0.500 + 0.3643 = 0.8643 )
4.	Area to the right of some negative value. same as #3. ( 0.500 + 0.4778 = 0.9778 )
5.	Area to the right of some positive value. area to the right of 0 = 0.5 so we subtract. ( 0.500 + 0.4505 = 0.0495 )
6.	Area to the right of some negative value. same as #5. ( 0.500 + 0.3531 = 0.1469 )
7.	Area between two positive values. difference between Area(1.82) and Area(1.10). ( 0.4656 + 0.3643 = 0.1013 )
8.	Area between two negative values. same as #7. ( Area(-1.60) - Area(-0.92) = 0.4452 + 0.3212 = 0.1240 )
9.	Area between negative and positive values. sum the two probabilities in the tables. ( 0.2257 + 0.1879 = 0.4136 )

Example 3 An analyst for a competing tire company tells reporters that 90% of the tires have a diameter of 36.0020 inches or less and that 10% have a diameter of more than 36.0020 inches. The analyst is incorrect. What figure should be substituted for 36.0020 to make his statement true? solution 1. find the number which the value of the standard normal random variable will exceed with probability 0.10. Since the probability must be 0.50 - 0.10 = 0.40, look in Appendix Table II for the value of Z corresponding to a probability of 0.40. This is 1.28. 2. Find the diameter of a tire corresponding to this value of the standard normal variable. Z = (X - ) / --> 1.28 = (X - 36) / 0.001 X = + Z* --> X = 36 + 1.28(0.001) = 36.00128 inches. So the probability is 0.10 that a tire will exceed 36.00128 inches in diameter.

The Normal Distribution as an Approximation to the Binomial Distribution

If n (the number of trials) is large and (the probability of success) is not to close to 0 or 1, the probability distribution of the number of successes occurring in n Bernoulli trials can be approximated by a normal distribution. Experience indicates that the approximation is fairly accurate as long as n > 5 when <=1/2 and n(1-) > 5 when > 1/2.

The Continuity Correction

Continuity Correction

A correction due to the fact that a discrete probability distribution is being approximated by a continuous one. In general, to find the probability that a binomial variable equals at least c but no more than d (where c < d), we find the probability that a normal variable ( with mean n and standard deviation [n(1-)]^1/2 ) lies between (c - 1/2) and (d + 1/2).