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Section 6.3

With a few little improvements and extras by D.R.S, University of Cordele. Section 6.3. Finding Probability Using a Normal Distribution IMPORTANT: “Area” is “_____________” IMPORTANT: “Probability” is “_________”.

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Section 6.3

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  1. With a few little improvements and extras by D.R.S, University of Cordele. Section 6.3 Finding Probability Using a Normal Distribution IMPORTANT: “Area” is “_____________” IMPORTANT: “Probability” is “_________”.

  2. Do you remember The Empirical Rule? ___%, ___%, ___% within ____, ____, ____ st.devs. 3. Applications . _____% of newborns weigh 2500-3500g 2. Where did 34% come from? _______________________ _____% of newborns weigh less than 3500g 1. TWO horizontal axes, real-world x & standard normal z. For the x axis, μ = ______, σ = _____ On the z axis, μ = ______, σ = _____ _____% of newborns weigh more than 3500g _____________ -3 ___ -2 ___ -1 ___ 0 ___ 1 ____ 2 ____ 3 ___________ z

  3. Do you remember The Empirical Rule? LIMITATION – we could only deal with z = 1, 2, 3 standard deviations away from the mean. TODAY – we expand our horizons to be able to work with any z value.

  4. z problems and x problems – which is which? It’s done by changing x problem into a z problem and bringing area answer back Lesson 6.2 was about z problems z is in TheStandard normaldistribution The horizontalaxis is a z axis We find area to left / to right/ betweenz values. • Lesson 6.3 and real-life • applications are about x problems • x is in “some” Normal Distribution • Horizontal x axis • We still find areato left / to right / between x values

  5. Kinds of Problems: “Find the probability that…” Probability that x is less than some given value Probability that x is greater than some given value Probability that x is between two given values Probability that x will be in the tails, more extreme, that is not between two given values. Probability that x will be within some given distance of some given value. Probability that x will differ from some given value by more than a certain amount. Can you draw pictures to represent all of these cases?

  6. Example 6.12: Finding the Probability that a Normally Distributed Random Variable Will Be Less Than a Given Value Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature less than 96.9 °F? Let’s do this by CONVERTING the x-PROBLEM into az-PROBLEM AND SEEING ALL THE DETAILS. Note that it’s a _________ distribution. Mean μ = ________ Standard deviation σ = ___________ Convert x = ________ to z. Formula: z =

  7. Draw a picture! Two axes. . Draw a picture – label axes. “The area to the left of x = _____ is the same as the area to the left of z = ______which is ______. And since Area is Probability, we conclude “The probability that a healthy adult has a body temperature less than _____oF is ______. z

  8. TI-84 normalcdf talks in both z and x language This is what we learned last time. In z-language:normalcdf(low z, high z)gives you the area, that is, the probability In x-language, you can give it the x values, buttell it the mean and standard deviation, too:normalcdf(low x, high x, μ, σ)This gives you the area, that is, the probability,without having to change the x into z yourself.The calculator does x-to-z conversions transparently. NEW!

  9. Example 6.12: Finding the Probability that a Normally Distributed Random Variable Will Be Less Than a Given Value (cont.) (2) DOING THIS BY using the x-language version of the TI-84’s normalcdf function: We want the area between and 96.9. The calculator does not have a button for “infinity”. We use for negative infinity. Keystrokes: (-) 1 2ND COMMA 9 9 SUPERB SHORTCUT !!! The details of changing the x problem to a z-problem are handled inside the TI-84, as long as we tell it the mean & std. deviation for this x.

  10. Example 6.12again, using Excel Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature less than 96.9 °F? =NORM.DIST(x value, mean, standard dev., TRUE)

  11. Excel NORM.S.DIST and NORM.DIST For z problems, use NORM.S.DIST • The “.S” stands for “Standard” • As in “The Standard Normal Distribution” • NORM.S.DIST(z) = area to left of that z score For x problems, use NORM.DIST • In general, for any normal distribution • NORM.DIST(x, mean, standard deviation) = area to the left of that x data value

  12. Example 6.13: Finding the Probability that a Normally Distributed Random Variable Will Be Greater Than a Given Value Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature greater than 97.6 °F? If we do the conversion to z ourselves: x = 97.6 converts to z = _______ Make the sketch and find area (it’s on the next slide) Conclusion (in a complete sentence in plain English):

  13. mean of 98.60 °F and a standard deviation of 0.73 °F. …probability … greater than 97.6 °F? On x-axis, label mean & stddev & also label 97.6. On z-axis, label 0 at mean and label the z-value. Shade under curve (which side?) Use z table to find area. . x z

  14. Example 6.13: Finding the Probability that a Normally Distributed Random Variable Will Be Greater Than a Given Value Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature greater than 97.6 °F? Same problem, but TI-84 to do it all: normalcdf ( ______, ______, ______, ______ ) left right mean stdev endpoints Conclusion (in a complete sentence):

  15. Example 6.13 with Excel Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature greater than 97.6 °F? Use 1-NORM.DIST(x, mean, stdev, TRUE)) Why is this 1– here in the front?

  16. Example 6.14: Finding the Probability that a Normally Distributed Random Variable Will Be between Two Given Values Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature between 98 °F and 99 °F? Doing it with explicit conversion to z: Convert x = 98 to z = ________. Convert x = 99 to z = ________. Sketch a picture at the right. The area __________ z = _____ and z = ______ is _________________, and that’s the probability of temp between ___ and ____.

  17. Example 6.14: Finding the Probability that a Normally Distributed Random Variable Will Be between Two Given Values Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature between 98 °F and 99 °F? normalcdf ( ______, ______, ______, ______ ) left right mean stdev endpoints Conclusion in plain English:

  18. Example 6.14 with Excel Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature between 98 °F and 99 °F? NEED TO SUBTRACT TWO NORM.DIST() VALUES

  19. Example 6.15: Finding the Probability that a Normally Distributed Random Variable Will Be in the Tails Defined by Two Given Values Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature less than 99.6 °F or greater than 100.6 °F? (Sketch is recommended!) normalcdf( _____, _____, _____, _____) + normalcdf ( _____, _____, _____, _____) Alternative: 1 – normalcdf(_____, _____, _____, _____)

  20. Example 6.16: Finding the Probability that a Normally Distributed Random Variable Will Differ from the Mean by More Than a Given Value Body temperatures of adults are normally distributed with a mean of 98.60 °F and a standard deviation of 0.73 °F. What is the probability of a healthy adult having a body temperature that differs from the population mean by more than 1 °F? x Label mean and the low and high endpoints. Shade darkly the segments that correspond to the area we want to find. TI-84 command (using the complement trick):1 - normalcdf( ______, ______, ______, ______) Conclusion:

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