“Find the probability that a randomly-selected healthy adult has a body temperature…” less than 96.9oF greater than 97.6oF between 98.0oF and 99.0oF less than 99.6oF or greater than 100.6oF differs from population mean by more than 1.0oF Method #1 – by changing the x problem into a z problem and finding the area left/right of z value Method #2 – by using TI-84 with x values directly and let the calculator handle details of the z You still need to draw pictures – you just have to.

Let’s do this by CONVERTING the x-PROBLEM 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 a z-PROBLEM AND SEEING ALL THE DETAILS. Note that it’s a _________ distribution. Mean μ = ________ Standard deviation σ = ___________ Convert x = ________ to z. Formula: 𝒛= 𝒙−𝝁 𝝈 z =

Draw a picture! Two axes. Draw a picture – label axes. . Convert x = 96.90 into a z value – use formula 𝑧= (𝑥−𝜇) 𝜎 Draw a picture – label axes. . Convert mean μ = 98.60 into a z value – easy! Mean of Standard Normal Distribution is always exactly what? TWO AXES x-axis and z-axis z

The picture leads us to the solution. The area to the left of z = _______. (Use either the printed table or normalcdf(low z, high z) to find it.) So the area to the left of x = ______ is also ________. Since Area Is Probability, the probability that a healthy adult has a body temperature less than ______o is ________, or about ______ %

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, but tell 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!

Example 6.12: Finding the Probability that a Normally Distributed Random Variable Will Be Less Than a Given Value (cont.) Previous example REPEATED but using the x-language version of the TI-84’s normalcdf : We want the area between negative infinity and 96.9. We use −1× 10 99 for −∞. (-) 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.

Example 6.12 again, 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)

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) 𝒛= 𝒙−𝝁 𝝈 Conclusion (in a complete sentence in plain English):

mean of 98. 60 °F and a standard deviation of 0. 73 °F mean of 98.60 °F and a standard deviation of 0.73 °F. …probability … greater than 97.6 °F? On x-axis, label mean & std dev & 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

Conclusion (in a complete sentence): 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 this time with TI-84 shortcut: Conclusion (in a complete sentence):

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?

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: (PICTURE IN NEXT SLIDE – WRITE ON IT) Convert x = 98 to z = ________. Convert x = 99 to z = ________. The area __________ z = _____ and z = ______ is _________________, and that’s the probability of temp between ______ and ______.

Between 98 and 99oF . x z

Same problem but with TI-84 normalcdf(low,high,mean,stdev) 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? Do directly with TI-84 and state plain conclusion:

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

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 needed, see at right →!) Recall method: 1 – area between. State conclusion plainly:

less than 99.6 °F or greater than 100.6 °F? x z

Label mean and the low and high endpoints. 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? SKETCH FIRST – on next slide → 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:

Differs from population mean by more than 1 °F? . x z