INTRODUCTORY STATISTICS Chapter 6 The Normal Distribution CHAPTER

INTRODUCTORY STATISTICS Chapter 6 The Normal Distribution

CHAPTER 6: THE NORMAL DISTRIBUTION 6. 1 The Standard Normal Distribution 6. 2 Using the Normal Distribution

CHAPTER OBJECTIVES By the end of this chapter, you should be able to: • Recognize the normal probability distribution and apply it appropriately. • Recognize the standard normal probability distribution and apply it appropriately. • Compare normal probabilities by converting to the standard normal distribution.

THE NORMAL DISTRIBUTION The normal, a continuous distribution, is the most important of all the distributions. Its graph is bell-shaped. You see the bell curve in almost all disciplines. Some of these include psychology, business, economics, the sciences, nursing, and, of course, mathematics. Some of your instructors may use the normal distribution to help determine your grade. Most IQ scores are normally distributed. Often real-estate prices fit a normal distribution. The normal distribution is extremely important, but it cannot be applied to everything in the real world. In this chapter, you will study the normal distribution, the standard normal distribution, and applications associated with them.

I. Review Material A) Activity: Where do I stand? 1. ) Let’s create a dotplot showing the height distribution of all the students in this class.

2. ) Calculate the 5 -number summary, the mean and standard deviation: Min= Q 1 = Med= Q 3= Max= Mean= Standard Deviation =

B) Percentiles Definition: The pth percentile of a distribution is the value with p percent of the observations less than or equal to it. Example 1: Mrs. Munson is concerned about how her daughter’s height and weight compare with those of other girls her age. She uses an online calculator to determine that her daughter is at the 87 th percentile for weight and the 67 th percentile for height. Explain to Mrs. Munson what this means.

Example 2: Peter is a star runner on the track team, and Molly is one of the best sprinters on the swim team. Both athletes qualify for the league championship meet based on their performance during the regular season. a) In the track championships, Peter records a time that would fall at the 80 th percentile of all his race times that season. But his performance places him at the 50 th percentile in the league championship meet. Explain how this is possible. b) Molly swims a bit slowly for her in the league swim meet, recording a time that would fall at the 50 th percentile of all her meet times that season. But her performance places Molly at the 80 th percentile in this event at the league meet. Explain how this could happen.

C) Standard Score or z-score: From standard deviation, we can assign a piece of data a “Standard Score, ” which allows us to compare that piece of data to other pieces in the set. This is calculated with the following formula: where “z” is the standard score and “x” is the piece of data you are assessing

Because of the subtraction, it IS POSSIBLE to get a standard score that is negative, which means it is to the left (or below) the mean. A standard score of 1 means you are 1 standard deviation above the mean, 2 means 2 above, etc. Example 1: Calculate the standard scores for each x-value if the mean of the data is 10 and the standard deviation is 2. 1. x=2, z=_____ 2. x=12, z=_____ 3. x=9, z=_____ 4. x=15, z=______

Example 2: Sophia scored a 660 on the SAT Math test. Jim took the ACT Math test and he received a 26. SAT scores have a mean of 500 and a standard deviation of 100. ACT scores are normally distributed with mean 18 and standard deviation 6. Assuming that both tests measure the same kind of ability, who did better?

Example 3: Francine, who is 25 years old, has her bone density measured using DEXA (dual-energy X-ray absorptiometry test). Her results indicate a bone density in the hip of 948 g/cm 2, which converts to a standardized score of z = -1. 45, In the population of 25 -year-old women like Francine, the mean bone density in the hip is 956 g/cm 2. a) Francine has not taken a statistics class in a few years. Explain to her in simple language what the standardized score tells her about her bone density. b) Use the information provided to calculate the standard deviation of bone density in the reference population.

c) Francine’s friend, Lisa, who is 35 years old, has her bone density measured using DEXA. Hers is reported as 948 g/cm 2, but her standardized score is z = 0. 50. The mean bone density in the hip for the reference population of 35 -year-old women is 944 g/cm 2. Whose bones are healthier- Francine’s or Lisa’s? Justify your answer. Complete the next set of questions in your packet.

6. 1 THE NORMAL DISTRIBUTION A) Density Curves Definition: A density (also called probability) curve is a curve that is always on or above the horizontal axis, and has area exactly 1 underneath it (that is, the area between the curve and the x−axis). A density curve describes the overall pattern of a distribution. The area under the curve and above any range of values is the proportion of all observations that fall in that range. Example 1: The histogram represents the scores of 947 seventh-grade students in Gary, Indiana on the Iowa Test of Basic Skills.

B) Normal Distributions Just about anything you measure turns out to be normally distributed, or at least approximately so. That is, usually most of the observations cluster around the mean, with progressively fewer observations out towards the extremes Although most variables are normally distributed, it is not the case that all variables are normally distributed.

As examples, consider the following: 1) Values of a dice roll 2) Flipping a coin 3) Salary distributions Make a sketch of each example below: 1) 2) The first two distributions are called 3)

Normal curves are symmetric, single-peaked, and bell-shaped. All normal distributions have similar shapes and are determined solely by their mean µ and standard deviation σ. The points at which the curve changes concavity are located a distance σ on either side of µ. We will use the area under these curves to represent a percentage of observations. (These areas correspond to integrals, for those of you with some experience with calculus. )

The normal distribution has two parameters (two numerical descriptive measures) 1) the mean (μ) and 2) the standard deviation (σ). If X is a quantity to be measured that has a normal distribution with mean (μ) and standard deviation (σ), we designate this by writing X ~ N(μ, σ)

Some basic properties of a normal distribution: 1) The curve is symmetrical about a vertical line drawn through the mean, μ. 2) In theory, the mean is the same as the median, because the graph is symmetric about μ. 3) The normal distribution depends only on the mean and the standard deviation. 4) The area under the curve must equal one. 5) A change in the standard deviation, σ, causes a change in the shape of the curve; the curve becomes fatter or skinnier depending on σ. 6) A change in μ causes the graph to shift to the left or right. This means there an infinite number of normal probability distributions. One of special interest is called the standard normal distribution.

THE STANDARD NORMAL DISTRIBUTION • The standard normal distribution is a normal distribution of standardized values called z-scores. • A z-score is measured in units of the standard deviation.

EXAMPLE 1 Suppose X ~ N(5, 6). This says that x is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then: z=2 Therefore, x = 17 is two standard deviations (2σ) above or to the right of the mean μ = 5. Now suppose x = 1. z = - 0. 67 Therefore x = 1 is 0. 67 standard deviations (– 0. 67σ) below or to the left of the mean μ = 5.

What is the z-score of x, when x = 1 and X~N(12, 3)? μ = 12 σ=3 z = - 11/3 = - 3. 67

C) Applications of the Normal Curve Three reasons we are interested in normal distributions: 1. Normal distributions are good descriptions for some distributions of real data. Examples include test scores and characteristics of biological populations (such as height or weight). 2. Normal distributions are good approximations to the results of many kinds of chance outcomes. An example is the proportion of heads in a repeatedly tossed coin experiment. 3. Many statistical inference procedures based on normal distributions work well for other roughly symmetric distributions.

THE EMPIRICAL RULE If X is a random variable and has a normal distribution with mean μ and standard deviation σ, then the Empirical Rule says the following: • About 68% of the x values lie between – 1σ and +1σ of the mean μ (within one standard deviation of the mean). • About 95% of the x values lie between – 2σ and +2σ of the mean μ (within two standard deviations of the mean). • About 99. 7% of the x values lie between – 3σ and +3σ of the mean μ (within three standard deviations of the mean). Notice that almost all the x values lie within three standard deviations of the mean.

The empirical rule is also known as the 68 -95 -99. 7 rule.

EXAMPLE 1 The mean height of 15 to 18 year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6. 28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18 -year-old male from Chile in 2009 to 2010. Then X~N(170, 6. 28). a. Suppose a 15 to 18 -year-old male from Chile was 168 cm tall from 2009 to 2010. The z-score when x = 168 cm is z =_______. This z-score tells you that x =168 is____standard deviations to the____(right or left) of the mean _____ (What is the mean? ). b. Suppose that the height of a 15 to 18 -year-old male from Chile from 2009 to 2010 has a z-score of z = 1. 27. What is the male’s height? The z-score (z=1. 27) tells you that the male’s height is ____ standard deviations to the _____ (right or left) of the mean.

EXAMPLE 2 From 1984 to 1985, the mean height of 15 to 18 -year-old males from Chile was 172. 36 cm, and the standard deviation was 6. 34 cm. Let Y= the height of 15 to 18 -year-old males in 1984 to 1985. Then Y~N(172. 36, 6. 34). a. About 68% of the y values lie between what two values? These values are________. The z-scores are ________, respectively. b. About 95% of the y values lie between what two values? These values are________. The z-scores are ________ respectively. c. About 99. 7% of the y values lie between what two values? These values are ________. The z scores are ________, respectively.

The scores on a college entrance exam have an approximate normal distribution with mean, µ = 52 points and a standard deviation, σ= 11 points. a. About 68% of the y values lie between what two values? These values are ________. The z-scores are ________, respectively. b. About 95% of the y values lie between what two values? These values are ________. The z-scores are ________, respectively. c. About 99. 7% of the y values lie between what two values? These values are ________. The z-scores are ________, respectively.

6. 2 USING THE NORMAL DISTRIBUTION The shaded area in the following graph indicates the area to the left of x. This area is represented by the probability P(X < x). Normal tables, computers, and calculators provide or calculate the probability P(X < x). Shaded area represents P(X < x) = Area to the left of the vertical line through x. P(X > x) = Area to the right of the vertical line through x.

Shaded area represents P(X < x) The area to the right is then P(X > x) = 1 – P(X < x). The area to the left is then P(X < x) = 1 – P(X > x) P(X < x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions.

CALCULATIONS OF PROBABILITIES Example The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five. a) Find the probability that a randomly selected student scored more than 65 on the exam? Let X = a score on the final exam. Then X ~ N(63, 5) where μ = 63 and σ = 5. Draw a graph.

b. Find the probability that a randomly selected student scored less than 85. 1) Draw a graph. 2) Find P(x < 85), and shade the graph. 3) Using a calculator, find P(x < 85) 4) normalcdf(0, 85, 63, 5) = 1 (rounds to one) The probability that one student scores less than 85 is approximately one (or 100%).

c. Find the 90 th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k). 1) Draw a graph. (For each problem or part of a problem, draw a new graph). Draw the x-axis. Shade the area that corresponds to the 90 th percentile. 2) Let k = the 90 th percentile. P(x < k) is the area to the left of k. The 90 th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and ten percent are the same or higher. The variable k is often called a critical value. 3) The 90 th percentile is 69. 4. This means that 90% of the test scores fall at or below 69. 4 and 10% fall at or above.