The Normal Probability Distribution

Transcription

1 102 The Normal Probability Distribution C H A P T E R 7 Section 7.2 4Example 1 (pg. 71) Finding Area Under a Normal Curve In this exercise, we will calculate the area to the left of 5 inches using a normal distribution with μ = 8.72 and σ =.17. The TI-84 has two methods for calculating this area. Method 1: Normalcdf(lowerbound, upperbound, μ,σ ) computes the area between a lowerbound and an upperbound. In this example, you are computing the area from negative infinity to 5. Negative infinity is specified by (-) 1 2 nd [EE] 9 9 (Note: EE is found above the comma, ). Try entering 1 EE 99 into your calculator. Now, to find P(X < 5) we will calculate the area to the left of 5. Press 2 nd [DISTR] and select 2:normalcdf( and type in -1E99, 5, 8.72,.17 ) and press ENTER. (Note: For this example, μ = 8.72 and σ =.17)

2 Section Method 2: This method calculates the area and also displays a graph of the probability distribution.to find P(X < 5) and include a graph, you must first set up the WINDOW so that the graph will be displayed properly. You will need to set Xmin equal to ( μ - σ ) and Xmax equal to ( μ + σ ). Press WINDOW and set Xmin equal to ( μ - σ ) by typing in *.17. Press ENTER and set Xmax equal to ( μ + σ ) by typing in *.17. Set Xscl equal to σ, which is.17. Setting the Y-range is a little more difficult to do. We will start by setting Ymax. A good rule - of - thumb is to set Ymax equal to.5 /σ. For this example, set Ymax =.5/.17 by scrolling down to Ymax and typing.5/.17. Use the up arrow to highlight Ymin. A good value for Ymin is (-) Ymax / 4 so type in (-).158 / 4. Press 2 nd [DRAW] and select 1:ClrDraw and press ENTER ENTER. Press 2 nd [STATPLOT] and TURN OFF all PLOTS. Make sure that there is nothing stored in the Y-registers. Press Y= and check the Y-registers. If any of them contain a function, move the cursor to that Y-register and press CLEAR. Press 2 nd [DISTR], highlight DRAW and select 1:ShadeNorm( and type in -1E99, 5, 8.72,.17 ) and press ENTER. Conclusion: The proportion of -year-old females who are less than 5 inches tall is

3 104 Note: When using the TI-84 (or any other technology tool), the answers you obtain may vary slightly from the answers that you would obtain using the standard normal table. Consequently, your answers may not be exactly the same as the answers found in your textbook. The differences are simply due to rounding. If we want to find the proportion of -year-old females whose height is greater than 5 inches, we will calculate the area to the right of 5. Method 1: Use Normalcdf(lowerbound, upperbound, μ,σ ). In this example, you are computing the area from 5 positive infinity. Positive infinity is specified by 1 2 nd [EE] 9 9 (Note: EE is found above the comma, ). Now, to find P(X > 5) we will calculate the area to the right of 5. Press 2 nd [DISTR] and select 2:normalcdf( and type in 5, 1E99, 8.72,.17 ) and press ENTER. (Note: For this example, μ = 8.72 and σ =.17) Method 2: To find P(X >5) and include a graph, you must first set up the WINDOW so that the graph will be displayed properly. You will need to set Xmin equal to ( μ - σ ) and Xmax equal to ( μ + σ ). Press WINDOW and set Xmin equal to ( μ - σ ) by typing in *.17. Press ENTER and set Xmax equal to ( μ + σ ) by typing in *.17. Set Xscl equal to σ, which is.17. Setting the Y-range is a little more difficult to do. We will start by setting Ymax. A good rule - of - thumb is to set Ymax equal to.5 /σ. For this example, set Ymax =.5/.17 by scrolling down to Ymax and typing.5/.17. Use the up arrow to highlight Ymin. A good value for Ymin is (-) Ymax / 4 so type in (-).158 / 4. Press 2 nd [DRAW] and select 1:ClrDraw and press ENTER ENTER. Press 2 nd [STATPLOT] and TURN OFF all PLOTS. Make sure that there is nothing stored in the Y-registers. Press Y= and check the Y-registers. If any of them contain a function, move the cursor to that Y-register and press CLEAR.

4 Section Press 2 nd [DISTR], highlight DRAW and select 1:ShadeNorm( and type in 5, 1E99, 8.72,.17 ) and press ENTER.

5 106 4Example 2 (pg. 72) Finding the Probability of a Normal Random Variable In this exercise, we will calculate the area between 5 and 40 using a normal distribution with μ = 8.72 and σ =.17. Method 1:To find P(5 X 40) press 2 nd [DISTR], select 2:normalcdf( and type in 5, 40, 8.72,.17 ) and press ENTER. Method 2: To find the probability and include a graph, you must first set up the WINDOW so that the graph will be displayed properly. You will need to set Xmin equal to ( μ - σ ) and Xmax equal to ( μ + σ ). Press WINDOW and set Xmin equal to ( μ - σ ) by typing in *.17. Press ENTER and set Xmax equal to ( μ + σ ) by typing in *.17. Set Xscl equal to σ, which is.17. Setting the Y-range is a little more difficult to do. A good rule - of - thumb is to set Ymax equal to.5 /σ. For this example, set Ymax =.5/.17. Use the up arrow to highlight Ymin. A good value for Ymin is (-) Ymax / 4 so type in (-).158 / 4. Press 2 nd [DRAW] and select 1:ClrDraw and press ENTER ENTER. Press 2 nd [STATPLOT] and TURN OFF all PLOTS. Make sure that there is nothing stored in the Y-registers. Press Y= and check the Y-registers. If any of them contain a function, move the cursor to that Y-register and press CLEAR. Press 2 nd [DISTR], highlight DRAW and select 1:ShadeNorm( and type in 5, 40, 8.72,.17 ) and press ENTER.

6 Section Conclusion: The probability that a randomly selected three-year-old female is between 5 and 40 inches tall is.565 or 5.65%.

7 108 Example (pg. 74) Finding the Value of a Normal Random Variable This is an inverse normal problem and the command invnorm( area, μ,σ ) is used. In this type of problem, a percentage of the area under the normal curve is given and you are asked to find the corresponding X-value. The area value that you ENTER into the TI-84 must always be area to the left of an X-value. In this example, the percentage given is the bottom 20 %, (the 20 th percentile). This area (bottom 20%) is an area to the left, so that is the value of the area that we enter into the invnorm command. The area value must be entered in decimal form. Press 2 nd [DISTR] and select :invnorm( and type in.20, 8.72,.17 ) and press ENTER. Conclusion: The height that separates the bottom 20% of three-year-old females from the top 80% is 6.05 inches.

8 Section Example 4 (pg.75) Finding the Value of a Normal Random Variable This is an inverse normal problem and the command invnorm( area, μ,σ ) is used. In this problem the middle area is That leaves an area of 0.10 to be equally divided between the left and right tail areas. Each of these areas are, therefore, equal to The score that marks the lower edge of the middle 90% corresponds to a left area of To find the X-value corresponding to a left area of 0.05, press 2 nd [DISTR] and select :invnorm( and type in.05, 516, 116 ) and press ENTER. The score that separates the middle 90% from the top 5% is actually the score that separates the bottom 95% (the middle 90% plus the 5% in the left tail) from the top 5%. To find the X-value corresponding to a left area of 0.95, press 2 nd [DISTR] and select :invnorm( and type in.95, 516, 116 ) and press ENTER. The middle 90% of the distribution lies between the scores 25 and 707.

9 110 4Example 5 (pg. 77) Finding the value Z α This is an inverse normal problem and the command invnorm( area, μ,σ ) is used. In this type of problem, an area under the normal curve is given and you are asked to find the corresponding Z-score. In this example, the area given is the area to the right of a Z-score. The area is (The area value that you enter into the TI-84 must always be area to the left of a Z-score.). To find the Z-score corresponding to right area of 0.10, subtract 0.10 from 1 to obtain the area to the left of the Z-score. Press 2 nd [DISTR] and select :invnorm( and type in.90, 0, 1 ) and press ENTER. The Z-score that has an area to the right equal to 0.10 is Z = 1.28.

10 Section Section 7. 4Example 2 (pg. 84) A Normal Probability Plot Press STAT and select 1:Edit and press ENTER. Clear all data from L1. ENTER the data from Table 4 on pg. 8 into L1. To set up the normal probability plot, first make sure that there is nothing stored in the Y-registers. Press Y=tand check the Y-registers. If any of them contain a function, move the cursor to that Y-register and press CLEAR. Press 2 nd [STAT PLOT]. Press ENTER to select Plot 1. Highlight On and press ENTER. Set Type to the normal probability plot which is the third selection in the second row. Press ENTER. Set Data List to L1 and Data Axis to X. Next, there are three different types of Marks that you can select for the graph. The first choice, a small square, is the best one to use. Press ZOOM and select 9:ZoomStat and ENTER. The calculator draws the normal probability plot with a horizontal line at the X- axis. This plot is fairly linear, indicating that the data generally follows a normal distribution.

11 112 Section 7.4 4Example 1 (pg.90) Normal Approximation to the Binomial In this binomial experiment, the random variable, X, is the number of individuals with blood type O-negative; the probability, p, that an individual has type O- negative blood is 0.07 and the sample size, n, is 500. We will approximate the probability that fewer than 0 individuals in the sample have type O-negative blood, that is, P(X < 0) using the normal approximation to the binomial. In order to use the normal approximation to the binomial, the first step is to verify that n*p*(1-p) 10. In this example, 500 *.07 *.9 = 2.55, so the requirement is satisfied. Next, calculate the mean and standard deviation of this binomial random variable. The mean, μ, equals n*p = 500 *.07 = 5. The standard deviation, σ, equals n* p*(1 p) = 500*.07*.9 = To approximate P(X < 0) with a normal probability, we calculate P(X 29.5). (Note: This adjustment from 0 to 29.5 is called a continuity correction). Press 2 nd [DISTR], select 2:normalcdf( and type in type in -1E99, 29.5, 5, 5.71 ) and press ENTER. You can compare this probability ( ) that you obtained through a normal approximation to the actual probability obtained from the binomial distribution. Press 2 nd [DISTR], select B:binomcdf( and type in press ENTER. 500,.07, 29 ) and

12 Section The actual P(X < 0) is

13 114 4Example 2 (pg. 91) Normal Approximation to the Binomial In this binomial experiment, the random variable, X, is the number of adult Americans who favor the death penalty for individuals convicted of murder; the probability, p, that an adult American favors the death penalty 0.65 and the sample size, n, is We will approximate the probability that no more than 60 individuals in the sample favor the death penalty, that is, P(X 60) using the normal approximation to the binomial. In order to use the normal approximation to the binomial, the first step is to verify that n*p*(1-p) 10. In this example, 1000 *.65 *.5 = 227.5, so the requirement is satisfied. First, calculate the mean and standard deviation of this binomial random variable. The mean, μ, equals n*p = 1000 *.65 = 650. The standard deviation, σ, equals n * p * (1 p) = 1000 *.65 *.5 = To approximate P(X 60) with a normal probability we calculate P(X 60.5) (Note: This adjustment from 60 to 60.5 is called a continuity correction). Press 2 nd [DISTR], select 2:normalcdf( and type in -1E99, 60.5, 650, ) and press ENTER. The approximate P(X 60) is This result of.0985 indicates that there is approximately a 10% chance that a random sample will contain no more than 60 individuals who favor the death penalty if, in fact, the true proportion of adults who favor the death penalty is 65%. Since the sample result that Erica obtained is not very unusual, she concludes that her survey does not contradict Gallup s findings.

14 TI-89 Instructions 115 TI-89 Instructions: These instructions are designed to give you an overview of the normal probability distribution functions on the TI-89. The TI-89 has two methods for finding areas under the normal curve and displaying the probabilities. Method 1: The area under the normal curve is calculated and a graph of the curve is displayed. To use this method, begin by turning off all other graphs. First check the y-registers by pressing and Y= and clearing all the y-functions. In the Stats/List Editor, press F2 and select 1:Plot Setup. Press F to clear all plots. Press ESC to return to the Editor. Press F5 and select 1:Shade and select 1:Shade Normal. Fill in the entry boxes in the Shade Normal screen (lower bound, upper bound, µ and σ.) Press ENTER. Method 2: The probability is calculated without a display of the normal curve. In the Stats/List Editor, press F5 and select 4:Normal Cdf. Fill in the entry boxes in the Normal Cdf screen (lower bound, upper bound, µ and σ.) Press ENTER. To calculate an x-value that corresponds to a given left-sided area under the normal curve, press F5 and select 2:Inverse and select 1:Inverse Normal. Fill in the entry boxes (left-sided area, µ and σ). Press ENTER.

15 116 TI-Nspire Instructions: These instructions are designed to give you an overview of the normal probability distribution functions on the TI-Nspire handhelds. Press c and select 6:New Document. (Note: If you currently have a document open, the next screen will ask if you want to save the document. Press e to select No. Press.) Select 1:Add Calculator. To access the probability functions (i.e., binomial, normal, poisson, geometric, etc.), press b, select 5:Probability and select 5:Distributions. A menu of probability distributions will be displayed in a list on the left side of the screen. To calculate a probability for the Normal Distribution, select 2:Normal Cdf and enter lower bound, upper bound, µ and σ. Press. To calculate an x-value that corresponds to a given left-sided area under the normal curve, select :Inverse Normal. Fill in the entry boxes (left-sided area, µ and σ). Press.