Chapter 6 ~ Normal Probability Distributions

Chapter Goals Learn about the normal, bell-shaped, or Gaussian distribution How probabilities are found How probabilities are represented How normal distributions are used in the real world

6.1 ~ Normal Probability Distributions The normal probability distribution is the most important distribution in all of statistics Many continuous random variables have normal or approximately normal distributions Need to learn how to describe a normal probability distribution

Normal Probability Distribution 1. A continuous random variable 2. Description involves two functions: a. A function to determine the ordinates of the graph picturing the distribution b. A function to determine probabilities 3. Normal probability distribution function: This is the function for the normal (bell-shaped) curve f x e ( ) = - 1 2 s p m 4. The probability that x lies in some interval is the area under the curve

The Normal Probability Distribution

Probabilities for a Normal Distribution Illustration

Notes The definite integral is a calculus topic We will use the TI83/84 to find probabilities for normal distributions We will learn how to compute probabilities for one special normal distribution: the standard normal distribution We will learn to transform all other normal probability questions to this special distribution Recall the empirical rule: the percentages that lie within certain intervals about the mean come from the normal probability distribution We need to refine the empirical rule to be able to find the percentage that lies between any two numbers

Percentage, Proportion & Probability Basically the same concepts Percentage (30%) is usually used when talking about a proportion (3/10) of a population Probability is usually used when talking about the chance that the next individual item will possess a certain property Area is the graphic representation of all three when we draw a picture to illustrate the situation

6.2 ~ The Standard Normal Distribution There are infinitely many normal probability distributions They are all related to the standard normal distribution The standard normal distribution is the normal distribution of the standard variable z (the z-score)

Standard Normal Distribution Properties: The total area under the normal curve is equal to 1 The distribution is mounded and symmetric; it extends indefinitely in both directions, approaching but never touching the horizontal axis The distribution has a mean of 0 and a standard deviation of 1 The mean divides the area in half, 0.50 on each side Nearly all the area is between z = -3.00 and z = 3.00 Notes: Table 3, Appendix B lists the probabilities associated with the intervals from the mean (0) to a specific value of z Probabilities of other intervals are found using the table entries, addition, subtraction, and the properties above

Table 3, Appendix B Entries The table contains the area under the standard normal curve between 0 and a specific value of z

Example Example: Find the area under the standard normal curve between z = 0 and z = 1.45 z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 1.4 0.4265 A portion of Table 3: .

Using the TI 83/84 To find the area between 0 and 1.45, do the following: 2nd DISTR 2 which is normalcdf( Enter the lower bound of 0 Enter a comma Then enter 1.45 Close the parentheses if you like or hit “Enter” The value of .426 is shown as the answer! Interpretation of the result: The probability that Z lies between 0 and 1.45 is 0.426

Example Example: Find the area under the normal curve to the right of z = 1.45; P(z > 1.45) Area asked for

Using the TI 83/84 To find the area between 1.45 and ∞, do the following: 2nd DISTR 2 which is normalcdf( Enter the lower bound of 1.45 Enter a comma Then enter 1 2nd EE 99 Close the parentheses if you like or hit “Enter” The value of .074 is shown as the answer! Interpretation of result: The probability that Z is greater than 1.45 is 0.074

Example Example: Find the area to the left of z = 1.45; P(z < 1.45)

Using The TI 83/84 To find the area between - ∞ and 1.45, do the following: 2nd DISTR 2 which is normalcdf( Enter the lower bound of -1 2nd EE 99 Enter a comma Then enter 1.45 Close the parentheses if you like or hit “Enter” The value of 0.926 is shown as the answer! Interpretation of result: The probability that Z is less than 1.45 is 0.926

Notes The addition and subtraction used in the previous examples are correct because the “areas” represent mutually exclusive events The symmetry of the normal distribution is a key factor in determining probabilities associated with values below (to the left of) the mean. For example: the area between the mean and z = -1.37 is exactly the same as the area between the mean and z = +1.37. When finding normal distribution probabilities, a sketch is always helpful

Example Example: Find the area between the mean (z = 0) and z = -1.26 Area asked for

Using the TI 83/84 Find the area to the left of z = -0.98 Use -1E99 for - ∞ and enter 2nd DISTR Normalcdf (-1e99, -0.98) which gives .164 - 0. 98 Area asked for

Example Example: Find the area between z = -2.30 and z = 1.80 P z ( . ) - < = + 2 30 1 80 4893 4641 9534

Using the TI 83/84 Find the area between z = -2.30 and z = 1.80 Enter 2nd DISTR, normalcdf (-2.3, 1.80) and press enter .953 is given as the answer. Remember, the function normalcdf is of the form: Normalcdf(lower limit, upper limit, mean, standard deviation) and if you’re working with distributions other than the standard normal (recall mean = 0, stddev = 1), you must enter the values for mean and standard deviation

Normal Distribution Note The normal distribution table may also be used to determine a z-score if we are given the area (working backwards) Example: What is the z-score associated with the 85th percentile?

Using the TI 83/84 There is another function in the DISTR list that is used to find the value of z (or x) when the probability is given. For the previous problem, we are actually asking what is the value of z such that 85% of the distribution lies below it.

Using the TI 83/84 Use 2nd DISTR invNorm( to calculate this value 2nd DISTR invNorm(.85) “ENTER” gives us a value of 1.036 which is shown

Example Example: What z-scores bound the middle 90% of a standard normal distribution?

Using the TI 83/84 The TI 83/84 calculates areas from -∞ to the value of z we are interested in. Therefore, we must get a little creative to solve some problems. Using the idea that the total area equals one comes in very handy here! For the example given, where we are interested in the value of z that bounds the middle 90%, the tails therefore represent a total of 10%. Divide this in two since it is symmetric and this gives 5% in each tail.

Using the TI 83/84 Now use the 2nd DISTR invNorm with .05 in the argument like this: Which gives an answer of -1.645 Since the distribution is symmetric, the upper limit is 1.645, so 90% of the distribution lies between (-1.645, 1.645)

Using the TI 83/84 Now let’s work the problems on page 279

6.3 ~ Applications of Normal Distributions Apply the techniques learned for the z distribution to all normal distributions Start with a probability question in terms of x-values Convert, or transform, the question into an equivalent probability statement involving z-values

Standardization Suppose x is a normal random variable with mean m and standard deviation s The random variable has a standard normal distribution

Example Example: A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of 0.02. Assume the amount of fill is normally distributed and a bottle is selected at random: 1) Find the probability the bottle contains between 32.00 oz and 32.025 oz 2) Find the probability the bottle contains more than 31.97 oz When x z = - 32.00 32.0 0.00 ; 0.02 m s Solutions: 1) When x z = - 32 025 32.025 32.0 1 25 . ; 0.02 m s

Solution Continued Area asked for P x z ( . ) 0. 32.0 32 025 02 1 25 1 25 3944 < = - æ è ç ö ø ÷

Example, Part 2 P x z ( . ) > = - æ è ç ö ø ÷ + 31 97 32.0 0. 02 1 2) 32.0 - 1 50 . P x z ( . ) > = - æ è ç ö ø ÷ + 31 97 32.0 0. 02 1 50) 5000 4332 9332

Notes The normal table may be used to answer many kinds of questions involving a normal distribution Often we need to find a cutoff point: a value of x such that there is a certain probability in a specified interval defined by x Example: The waiting time x at a certain bank is approximately normally distributed with a mean of 3.7 minutes and a standard deviation of 1.4 minutes. The bank would like to claim that 95% of all customers are waited on by a teller within c minutes. Find the value of c that makes this statement true.

Solution P x c z ( ) 0. . £ = - æ è ç ö ø ÷ 95 3 7 1 4

Example Example: A radar unit is used to measure the speed of automobiles on an expressway during rush-hour traffic. The speeds of individual automobiles are normally distributed with a mean of 62 mph. Find the standard deviation of all speeds if 3% of the automobiles travel faster than 72 mph.

Solution P x ( ) . > = 72 03 P z ( . ) > = 1 88 03 - 72 62 s 03 P z ( . ) > = 1 88 03 - 72 62 s 1.88 = x - m ; z = s 1 . 88 s = 10 / . = 10 1 88 5 32 s

Notation If x is a normal random variable with mean m and standard deviation s, this is often denoted: x ~ N(m, s) Example: Suppose x is a normal random variable with m = 35 and s = 6. A convenient notation to identify this random variable is: x ~ N(35, 6).

6.4 ~ Notation z-score used throughout statistics in a variety of ways Need convenient notation to indicate the area under the standard normal distribution z(a) is the algebraic name, for the z-score (point on the z axis) such that there is a of the area (probability) to the right of z(a)

Illustrations z(0.10) represents the value of z such that the area to the right under the standard normal curve is 0.10 z(0.10) z(0.80) represents the value of z such that the area to the right under the standard normal curve is 0.80 z(0.80)

Example z(0.10) = 1.28 Example: Find the numerical value of z(0.10): 0.10 (area information from notation) Table shows this area (0.4000) z(0.10) z(0.10) = 1.28

Example z(0.80) = -0.84 Example: Find the numerical value of z(0.80): Look for 0.3000; remember that z must be negative z(0.80) Use Table 3: look for an area as close as possible to 0.3000 z(0.80) = -0.84

Notes The values of z that will be used regularly come from one of the following situations: 1. The z-score such that there is a specified area in one tail of the normal distribution 2. The z-scores that bound a specified middle proportion of the normal distribution

Example Example: Find the numerical value of z(0.99): 0.01 z(0.99) Because of the symmetrical nature of the normal distribution, z(0.99) = -z(0.01)

Example Example: Find the z-scores that bound the middle 0.99 of the normal distribution: z(0.995) -z(0.005) z(0.005) z(0.005) = 2.575 and z(0.995) = -z(0.005) = -2.575

6.5 ~ Normal Approximation of the Binomial Recall: the binomial distribution is a probability distribution of the discrete random variable x, the number of successes observed in n repeated independent trials Binomial probabilities can be reasonably estimated by using the normal probability distribution

Background & Histogram Background: Consider the distribution of the binomial variable x when n = 20 and p = 0.5 Histogram: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 The histogram may be approximated by a normal curve

Notes The normal curve has mean and standard deviation from the binomial distribution: Can approximate the area of the rectangles with the area under the normal curve The approximation becomes more accurate as n becomes larger

Two Problems 1. As p moves away from 0.5, the binomial distribution is less symmetric, less normal-looking Solution: The normal distribution provides a reasonable approximation to a binomial probability distribution whenever the values of np and n(1 - p) both equal or exceed 5 2. The binomial distribution is discrete, and the normal distribution is continuous Solution: Use the continuity correction factor. Add or subtract 0.5 to account for the width of each rectangle.

Example Example: Research indicates 40% of all students entering a certain university withdraw from a course during their first year. What is the probability that fewer than 650 of this year’s entering class of 1800 will withdraw from a class? Let x be the number of students that withdraw from a course during their first year x has a binomial distribution: n = 1800, p = 0.4 The probability function is given by:

Solution Use the normal approximation method:

Random Number Generation With each rand execution, the TI-84 Plus generates the same random-number sequence for a given seed value. The TI-84 Plus factory-set seed value for rand is 0. To generate a different random-number sequence, store any nonzero seed value to rand. To restore the factory-set seed value, store 0 to rand or reset the defaults (Chapter 18). Note: The seed value also affects randInt(, randNorm(, and randBin( instructions.