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5.1 Normal Probability Distributions Normal distribution A continuous probability distribution for a continuous random variable, x. The most important continuous probability distribution in statistics. The graph of a normal distribution is called the normal curve. x 1 Larson/Farber

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Properties of Normal Distributions 1.The mean, median, and mode are equal. 2.The normal curve is bell-shaped and symmetric about the mean. 3.The total area under the curve is equal to one. 4.The normal curve approaches, but never touches the x-axis as it extends farther and farther away from the mean. Total area = Between μ – σ and μ + σ (in the center of the curve), the graph curves downward. The graph curves upward to the left of μ – σ and to the right of μ + σ. The points at which the curve changes from curving upward to curving downward are called the inflection points. μ 3σ μ + σ μ 2σμ σμ σ μμ + 2σμ + 3σ x Inflection points

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Means and Standard Deviations A normal distribution can have any mean and any positive standard deviation. The mean gives the location of the line of symmetry. The standard deviation describes the spread of the data. μ = 3.5 σ = 1.5 μ = 3.5 σ = 0.7 μ = 1.5 σ = Larson/Farber

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Example: Understanding Mean and Standard Deviation Which curve has the greater mean and Standard Deviation? Curve A has the greater mean (The line of symmetry of curve A occurs at x = 15. The line of symmetry of curve B occurs at x = 12.) 4 Larson/Farber Curve B has the greater standard deviation (Curve B is more spread out than curve A.)

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Interpreting Graphs The heights of fully grown white oak trees are normally distributed. The curve represents the distribution. μ = 90 (A normal curve is symmetric about the mean) σ = 3.5 (The inflection points are one standard deviation away from the mean) 5 Larson/Farber Z-Scores:

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The Standard Normal Distribution Standard normal distribution A normal distribution with a mean of 0 and a standard deviation of 1. 33 1 22 11 023 z Area = 1 Any x-value can be transformed into a z-score by using the formula 6 Larson/Farber

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The Standard Normal Distribution If each data value of a normally distributed random variable x is transformed into a z-score, the result will be the standard normal distribution. Normal Distribution x z Standard Normal Distribution Use Standard Normal Table or Calculator function: NormalCdf() to find cumulative area under the standard normal curve. 7 Larson/Farber

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Properties of the Standard Normal Distribution 1.The cumulative area is close to 0 for z-scores close to z = The cumulative area increases as the z-scores increase. 3.The cumulative area for z = 0 is The cumulative area is close to 1 for z-scores close to z = z = 3.49 Area is close to 0 z 33 1 22 1 Larson/Farber z = 3.49 Area is close to 1 Z = 0 Area = ** Note: Cumulative Area = Area under the curve to the ‘LEFT’ of the z-score.

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Using The Standard Normal Table Find the cumulative area that corresponds to a z-score of 1.15 and –0.24 The area to the left of z = 1.15 is Larson/Farber The area to the left of z = 0.24 is TI-83/84 2-NormalCdf Normalcdf(-10000, 1.15) Why use –10000 in normal cdf()?

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Finding Areas Under the Standard Normal Curve #1 To find the area to the left of z, find the area that corresponds to z in the standard normal table or Ti83/84 normalcdf(-10000, zscore) The area to the left of z = 1.23 is Larson/Farber Area to the left of z = 1.23 is #2 To find area to the right of z, find area corresponding to z in table & subtract from 1 (total area) or Ti83/84 normalcdf (zscore, 10000) Area to the right is 1 – = #3 To find the area between two z-scores, find both z-scores in table & subtract smaller from larger area, or Ti83/84 normalcdf(low-zscore, hi-zscore) The area to the left of z = 0.75 is Subtract to find the area of the region between the two zscores: =

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Practice #1: Find the area under the standard normal curve to the left of z = z Larson/Farber Answer: #2: Find the area under the standard normal curve to the right of z = 1.06 Answer: z #3: Find the area under the standard normal curve between z=-1.5 and 1.25 Answer: z 1.50

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μ = 500 σ = μ =500 x 5.2 Probability and Normal Distributions If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval. P(x < 600) = Area 12 Larson/Farber 600μ =500 P(x < 600) Normal Distribution μ = 500 σ = 100 x 1 μ = 0 Standard Normal Distribution μ = 0 σ = 1 z P(z < 1) Same Area P(x < 600) = P(z < 1) Example:

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Example1: Finding Probabilities for Normal Distributions A survey indicates that people use their computers an average of 2.4 years before upgrading to a new machine. The standard deviation is 0.5 year. A computer owner is selected at random. Find the probability that he or she will use it for fewer than 2 years before upgrading. Assume that the variable x is normally distributed. 13 Larson/Farber 22.4 P(x < 2) Normal Distribution μ = 2.4 σ = 0.5 x Standard Normal Distribution μ = 0 σ = 1 z P(z < -0.80) P(x < 2) = P(z < -0.80) = Find the probability using your Ti-84/83 Calculator___________________

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Example2: Finding Probabilities for Normal Distributions A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for between 24 and 54 minutes. P(24 < x < 54) = P(-1.75 < z < 0.75) = – = Find the probability using your Ti-84/83 Calculator_________________ 2445 P(24 < x < 54) x Normal Distribution μ = 45 σ = z Standard Normal Distribution μ = 0 σ = 1 0 P(-1.75 < z < 0.75)

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Example3: Finding Probabilities for Normal Distributions Find the probability that the shopper will be in the store more than 39 minutes. (Recall μ = 45 minutes and σ = 12 minutes) P(x > 39) = P(z > -0.50) = 1– = Find the probability using your Ti-84/83 Calculator__________________ 3945 P(x > 39) x Normal Distribution μ = 45 σ = 12 Standard Normal Distribution μ = 0 σ = P(z > -0.50) z If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 39 minutes? 200(.6915) = (approx. 138 shoppers)

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You Try this one! Assume that cholesterol levels of men in the United States are normally distributed, with a mean of 215 milligrams per deciliter and a standard deviation of 25 milligrams per deciliter. You randomly select a man from the United States. Draw pictures, use Z-scores and use your calculator to answer the questions below: 1. A lower risk of heart attack is associated with a cholesterol level below 200. What is the probability that the man’s cholesterol level is less than 200? 2. A moderate risk of heart attack is associated with a cholesterol level between 200 and 239. What is the probability that the man’s cholesterol level is between 200 and 239? 3. A higher risk of heart attack is associated with cholesterol levels above 239. What is the probability that the man’s cholesterol level is above 239? (How would you do this using a complement?) 16 Answers: 1) ) ).1685

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5.3 Finding a z-Score Given an Area Example1: Find the z-score that corresponds to a cumulative area of z 0 z Larson/Farber 4th ed The z-score is TI 83/84 3:InvNorm(.3632)

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Finding a z-Score Given an Area Example2: Find the z-score that has 10.75% of the distribution’s area to its right. z0 z – = Because the area to the right is , the cumulative area is Locate.8925 in the Standard Normal Table. Or InvNorm (.8925) Answer: Z-score = Larson/Farber 4th ed

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Finding a z-Score Given a Percentile Find the z-score that corresponds to P 5 (or the 5 th Percentile) (Recall: 5 th percentile refers to a value that is higher than 5% of the data) The z-score for P 5 is the same z-score that corresponds to an area of The areas closest to 0.05 in the table are (z = -1.65) and (z = ). Because 0.05 is halfway between the two areas in the table, use the z- score that is halfway between and The z-score is OR InvNorm (.05) = z 0 z Larson/Farber 4th ed

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Transforming a z-Score to an x-Score To transform a standard z-score to a data value x in a given population, use the formula : x = μ + zσ 20 Larson/Farber 4th ed The speeds of vehicles along a stretch of highway are normally distributed, with a mean of 67 miles per hour and a standard deviation of 4 miles per hour. Find the speeds x corresponding to z-sores of 1.96, -2.33, and 0. z = 1.96:x = (4) = miles per hour (Above the mean) z = -2.33:x = 67 + (-2.33)(4) = miles per hour (Below the mean) z = 0: x = (4) = 67 miles per hour (Equal to the mean)

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Finding a Specific Data Value Scores for a civil service exam are normally distributed, with a mean of 75 and a standard deviation of 6.5. To be eligible for civil service employment, you must score in the top 5%. What is the lowest score you can earn and still be eligible for employment? ? 0 z 5% ? 75 x 1 – 0.05 = 0.95 An exam score in the top 5% is any score above the 95 th percentile. Find the z-score that corresponds to a cumulative area of From the Standard Normal Table, the areas closest to 0.95 are (z = 1.64) and (z = 1.65). Because 0.95 is halfway between the two areas in the table, use the z-score that is halfway between 1.64 and That is, z = OR Using the equation x = μ + zσ : x = (6.5) ≈ The lowest score you can earn and still be eligible for employment is 86. invNorm (.95)

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