Chapter 2 – Day 1.

Chapter 2 – Day 1

Density Curve: Mathematical model that describes a set of data. Above x-axis Total area under curve = 1

Percentile: Percent of the observations at or below a value

a. Verify that this is a density curve.
Example #1 Using the following uniform density curve, answer the question: a. Verify that this is a density curve. A = bh A = (8)(0.125) A = 1

Example #1 Using the following uniform density curve, answer the question: b. What is the probability that the random variable has a value less than 3? A = bh A = (3)(0.125) A = 0.375

Example #1 Using the following uniform density curve, answer the question: c. What is the probability that the random variable has a value between 3 and 5? A = bh A = (2)(0.125) A = 0.25

d. What is the percentile for the variable that has a value of 6?
Example #1 Using the following uniform density curve, answer the question: d. What is the percentile for the variable that has a value of 6? A = bh A = (6)(0.125) A = 0.75

e. What value for the variable is in the 25th percentile?
Example #1 Using the following uniform density curve, answer the question: e. What value for the variable is in the 25th percentile? A = bh 0.25 = (b)(0.125) 2 = b

In a density curve: The mean, median, and quartiles can be located by eye.

The mean, , is the balance point of the curve, if it were made of solid material.

The median, M, divides the area under the curve in half.
The quartiles with the median divide the curve into quarters.

The mean and the median are the same only if the distribution is symmetrical. The median is a measure of center that is resistant to skew and outliers. The mean is not. Mean Median

Mean Median Mean Median

Example #2 A group of 78 third-grade students in a Midwestern elementary school took a “self-concept” test that measured how well they felt about themselves. Higher scores indicate more positive self-concepts. A histogram for these students’ self-concept scores are given below. Draw an appropriate density curve for summarizing the histogram on the graph above. How would you describe the shape of this density curve?

Example #2 A group of 78 third-grade students in a Midwestern elementary school took a “self-concept” test that measured how well they felt about themselves. Higher scores indicate more positive self-concepts. A histogram for these students’ self-concept scores are given below. Draw an appropriate density curve for summarizing the histogram on the graph above. How would you describe the shape of this density curve? Skew Left

A = mode A = A = mean B = median B = n/a B = median C = mean C = n/a
Example #3 Label A, B, and C as either the mean, median, or mode for each picture. A = mode A = A = mean Mean, median, mode B = median B = n/a B = median C = mean C = n/a C = mode

a. The mean and median are equal.
Example #4 For the density curve below, which of the following is true? a. The mean and median are equal. b. The mean is greater than the median. c. The mean is less than the median d. The mean could be either greater than or less than the median e. The mean is 0.5

Normal Curves: Density curve that is symmetric, unimodal, and bell-shaped It is used to compare information from different populations or to find the percentile of a certain value.

N(,) where  = mean and  = standard deviation of the population.
The mean is the center and the standard deviation is the distance from the mean.

The NORMAL(or BELL-SHAPED) DISTRIBUTION describes many different data sets, like:
Scores on a midterm exam Weights of people Calorie consumption per day Lengths of pregnancies Heights of people IQ The # of M&M’s in a 1lb bag

68% of the observations fall within  of the mean , or 
The Rule: In the normal distribution with mean  and standard deviation : 68% of the observations fall within  of the mean , or  95% of the observations fall within 2 of , or 2 99.7% of the observations fall within 3 of , or 3

Find the area between each of the deviations.
Example #5 Find the area between each of the deviations. 34% 34% 13.5% 13.5% 2.35% 2.35% .0015% .0015%

Example #6 The mean GPA for Taft HS is 2.9 with the standard deviation of 0.5. What GPA score will place a student in the top 16% of the class?

Example #6 The mean GPA for Taft HS is 2.9 with the standard deviation of 0.5. What GPA score will place a student in the top 16% of the class?  = 0.5 0.16  = 2.9

One standard Deviation above mean
Example #6 The mean GPA for Taft HS is 2.9 with the standard deviation of 0.5. What GPA score will place a student in the top 16% of the class? One standard Deviation above mean 3.4

Example #7 According to Nielsen Media Research, people watch television an average of 7 hours per day. Assume that these times are normally distributed with a standard deviation of 4 hours. What value would place you in the 84th percentile?  = 4 0.84  = 7 11

Assessing Normality: Looks can be deceiving! Method #1: Make a histogram 1. Find the mean and sd 2. Measure the intervals for 1, 2, and 3 sd 3. Count how many observations fall between these standard deviations 4. Compare to the Rule

Method #2: Normal Probability Plot 1. Plug data into a list 2. Make a Normal Probability Plot 3. If the data is normal then it will make a straight line

Calculator Tip: Normal Probability Plot Statplot – Normal Probability Plot

Example #9: Here are the lengths in feet of 44 great white sharks:
Draw a histogram of the data using your calculator. Describe the distribution of the lengths of the sharks.

b. Compare the mean and the median
b. Compare the mean and the median. Does this comparison support your assessment of the shape of the distribution in a? Explain. Median of Shark Length = Mean of Shark Length = About the same, median higher

c. Make a Normal Probability Plot. What does the graph tell you?

Example #10 A Normal probability plot for the amount of lactic acid in a sample of 30 pieces of cheese is shown below. Is the lactic acid distribution approximately Normal? Justify your answer.

Example #11 The plot shown at the right is a Normal probability plot for a set of data. The data value is plotted on the x axis, and the standardized value is plotted on the y axis. Which statement is true for these data? (a) The data are clearly Normally distributed. (b) The data are approximately Normally distributed. (c) The data are clearly skewed to the left. (d) The data are clearly skewed to the right. (e) There is insufficient information to determine the shape of the distribution.

Day 2 – Standardizing Values

NOTATION: Sample Population Mean x  “mu” Standard Deviation Sx or s  “sigma” Note: Never use x on the calc

The Standard Normal Curve:
Standardizing is a technique to help us determine percentages, proportions, probabilities, or area under the curve for any set of data. The requirement is that the data must me normally distributed. We will use TABLE A. (In the text, you have TABLE A in the front 2 pages of the book.)

Convert the standard deviation to 1.
To standardize: Convert the mean to 0. Convert the standard deviation to 1. = 1 = 0

The observations are given new values called z-scores.
The Z-score represents how many standard deviations an observation is away from the mean. Formula: x –   Z =

Example #1 The distribution of the duration of human pregnancies (i.e. the number of days between conception and birth) has been found to be approximately normal with mean  = 266 and the  = 16. a. What is the Z-Score for a human pregnancy of X = 266? x –   266 – 266 16 0 . 16 Z = = = =

x –   250 – 266 16 -16 16 Z = = = = -1 Example #1
The distribution of the duration of human pregnancies (i.e. the number of days between conception and birth) has been found to be approximately normal with mean  = 266 and the  = 16. b. What is the Z-Score for a human pregnancy of X = 250? x –   250 – 266 16 -16 16 Z = = = = -1

Example #2 Let’s us TABLE A to find the following: Draw the picture first, shade the region you want and look up the X in Table A to find the proportion, probability (percentage) to the left of that z-score. The proportion is also known as probability that the value of a particular member of a population will fall in the given interval. P (Z < -2.20) =  = 1 =0

Example #2 Let’s us TABLE A to find the following: Draw the picture first, shade the region you want and look up the X in Table A to find the proportion, probability (percentage) to the left of that z-score. The proportion is also known as probability that the value of a particular member of a population will fall in the given interval. P (Z < -2.20) =  = 1 Z -2.20 =0

Example #2 Let’s us TABLE A to find the following: Draw the picture first, shade the region you want and look up the X in Table A to find the proportion, probability (percentage) to the left of that z-score. The proportion is also known as probability that the value of a particular member of a population will fall in the given interval. P (Z < -2.20) = 0.0139  = 1 Z -2.20 =0

Example #2 Let’s us TABLE A to find the following: Draw the picture first, shade the region you want and look up the X in Table A to find the proportion, probability (percentage) to the left of that z-score. The proportion is also known as probability that the value of a particular member of a population will fall in the given interval. P (Z > 2.20) =  = 1  = 0 Z 2.20

1 – P(Z < 2.20) = P (Z > 2.20) = Example #2
Let’s us TABLE A to find the following: Draw the picture first, shade the region you want and look up the X in Table A to find the proportion, probability (percentage) to the left of that z-score. The proportion is also known as probability that the value of a particular member of a population will fall in the given interval. P (Z > 2.20) = 1 – P(Z < 2.20) =  = 1  = 0 Z 2.20

P (Z > 2.20) = 1 – P(Z < 2.20) = 1 – 0.9861 = 0.0139 Example #2
Let’s us TABLE A to find the following: Draw the picture first, shade the region you want and look up the X in Table A to find the proportion, probability (percentage) to the left of that z-score. The proportion is also known as probability that the value of a particular member of a population will fall in the given interval. P (Z > 2.20) = 1 – P(Z < 2.20) = 1 – = 0.0139  = 1  = 0 Z 2.20

P(Z > -0.95) = 1 – P(Z <-0.95) = Example #2
Let’s us TABLE A to find the following: Draw the picture first, shade the region you want and look up the X in Table A to find the proportion, probability (percentage) to the left of that z-score. The proportion is also known as probability that the value of a particular member of a population will fall in the given interval. P(Z > -0.95) = 1 – P(Z <-0.95) =  = 1  = 0 Z -0.95

P(Z > -0.95) = 1 – P(Z <-0.95) = 1 – 0.1711 = 0.8289 Example #2
Let’s us TABLE A to find the following: Draw the picture first, shade the region you want and look up the X in Table A to find the proportion, probability (percentage) to the left of that z-score. The proportion is also known as probability that the value of a particular member of a population will fall in the given interval. P(Z > -0.95) = 1 – P(Z <-0.95) = 1 – = 0.8289  = 1  = 0 Z -0.95

P (Z < 1.25) =  = 1  = 0 Z 1.25

P (Z < 1.25) = 0.8944  = 1  = 0 Z 1.25

P (-1.04 < Z < 3.01) = P(Z < 3.01) – P(Z < –1.04)  = 1 Z
 = 0 Z 3.01

P (-1.04 < Z < 3.01) = P(Z < 3.01) – P(-1.04)
= – =  = 1 Z -1.04  = 0 Z 3.01

P (0.15 < Z < 1.41) = P(Z < 1.41) – P(Z<0.15)  = 1  = 0

P (0.15 < Z < 1.41) = P(Z < 1.41) – P(Z<0.15)
= – =  = 1  = 0 Z 1.41 Z 0.15

Calculator Tip: Z-score Probabilities 2nd Dist – Normalcdf, (lowerbound, upperbound, mean, sd)

x –   29 – 34 3.5 -5 3.5 Z = = = = -1.4285 Example #3
Suppose that the fuel efficiency (in miles per gallon) of a Beetle varies with each tank of gas according to a normal distribution with  = 34 and standard deviation  = 3.5 miles per gallon. a. What proportion of all tanks would get 29 miles per gallon or less? x –   29 – 34 3.5 -5 3.5 Z = = = =

P (Z < ) =  = 3.5  = 1 =0 x 29 =34 Z

P (Z < ) = 0.0764  = 3.5  = 1 =0 x 29 =34 Z

Example #3 Suppose that the fuel efficiency (in miles per gallon) of a Beetle varies with each tank of gas according to a normal distribution with  = 34 and standard deviation  = 3.5 miles per gallon. b. What proportion of all tanks would get 40 miles per gallon or more? x –   40 – 34 3.5 6 . 3.5 Z = = = = 1.7143

P (Z > ) =  = 3.5  = 1 =0 =34 x 40 Z 1.71

P (Z > ) = 0.0436  = 3.5  = 1 =0 =34 x 40 Z 1.71

Example #3 Suppose that the fuel efficiency (in miles per gallon) of a Beetle varies with each tank of gas according to a normal distribution with  = 34 and standard deviation  = 3.5 miles per gallon. c. What proportion of all tanks would get between 27 and 42 miles per gallon? x –   27 – 34 3.5 -7 3.5 Z = = = = -2 x –   42 – 34 3.5 8 . 3.5 Z = = = = 2.285

P (-2 < Z < 2.29) = P(Z < 2.29) – P(Z < -2)  = 3.5  = 1
27  = 34 Z -2  = 0 Z 42 Z 2.29

P (-2 < Z < 2.29) = P(Z < 2.29) – P(Z < -2)
= – =  = 3.5  = 1 Z 27  = 34 Z -2  = 0 Z 42 Z 2.29

Example #3 Suppose that the fuel efficiency (in miles per gallon) of a Beetle varies with each tank of gas according to a normal distribution with  = 34 and standard deviation  = 3.5 miles per gallon. d. What proportion of all tanks would get 47 miles per gallon or more? Less than 47 miles per gallon? x –   47 – 34 3.5 13 3.5 Z = = = = 3.715

P (Z < 3.715) = P (Z > 3.715) =  = 1  = 1  = 0 =0 Z Z 3.72

P (Z < 3.715) = 1 P (Z > 3.715) =  = 1  = 1  = 0 =0 Z Z 3.72
 = 1  = 1  = 0 =0 Z 3.72 Z 3.72

Example #4: Golf courses have a wide range of difficulty
Example #4: Golf courses have a wide range of difficulty. Similarly, players differ in ability. In order to adjust for variations between players, they are often assigned a handicap score. To adjust for variations between courses, a handicapper decides to compare the golfer’s score against the data from the course. Suppose course A plays at a mean score of 76 with a standard deviation of 8 strokes with a normal distribution of scores. The mean score for course B is 80 with a standard deviation of 6 strokes and the scores are normally distributed. If a golfer regularly shoots an 80 on course A, what should be the comparable score on course B?

 = 8  = 76 x 80  = 6  = 80 x=?

x –   80 – 76 8 4 8 Z = = = = 0.5 x –   Z = x – 80 6 0.5 = 3 = x – 80 83 = x

Inverse Normal Probability Calculations: Un-standardizing
**Sometimes the proportion or percentage is given and you must find the corresponding z-score and un-standardize the value by finding the X-value. Example #5: What value(s) of Z cut off the region described? a. The lowest 11%  = 1 0.11 Z=? =0

Inverse Normal Probability Calculations: Un-standardizing
**Sometimes the proportion or percentage is given and you must find the corresponding z-score and un-standardize the value by finding the X-value. Example #5: What value(s) of Z cut off the region described? a. The lowest 11% Z = -1.23  = 1 0.12 Z=? =0

Example #5: What value(s) of Z cut off the region described?
b. The highest 30%  = 1 0.30 =0 Z = ?

b. The highest 30% Z = 0.52  = 1 0.30 =0 Z = ?

c. The highest 7%  = 1 0.07 =0 Z = ?

c. The highest 7% Z = 1.48  = 1 0.07 =0 Z = ?

d. The middle 50% 75% – 25%  = 1 Z=?  = 0 Z=?

75% – 25% Z = 0.67 and Z = -0.67 d. The middle 50%
Example #5: What value(s) of Z cut off the region described? d. The middle 50% 75% – 25% Z = 0.67 and Z = -0.67  = 1 Z=?  = 0 Z=?

Calculator Tip: Z-score given probabilities 2nd Dist – invNorm(%)

Steps for the Inverse Probability Calculation:
Draw a picture Identify the z-value from the given value of the proportion – look up the proportion in the MIDDLE of Table A. Solve for x: x –   Z = x = Z() + 

Calculator Tip: Z-score given probabilities 2nd Dist – invNorm(%, , )

x = Z() +  x = Z(32) + 150 0.10 Example # 6
A British company called Molebegon removes unwanted moles from gardens. In 1995, the European Union announced that the tiny moles are just too difficult to catch. They will not attempt to catch the smallest 10%. Molebegon’s past records indicate that weights of moles are normally distributed with a mean of 150 grams and a standard deviation of 32 grams. What is the cut off weight for the moles they will catch? x = Z() +   = 32 x = Z(32) + 150 0.10 Z=? =150

x = Z() +  x = (-1.28)(32) + 150 0.10 x = -40.96 + 150 x =
Example # 6 A British company called Molebegon removes unwanted moles from gardens. In 1995, the European Union announced that the tiny moles are just too difficult to catch. They will not attempt to catch the smallest 10%. Molebegon’s past records indicate that weights of moles are normally distributed with a mean of 150 grams and a standard deviation of 32 grams. What is the cut off weight for the moles they will catch? x = Z() +   = 32 x = (-1.28)(32) + 150 0.10 x = x = grams Z=? =150

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