1. Normal Distributions

2. Density Curve A density curve is a smooth function meant to approximate a histogram. A density curve is a smooth function meant to approximate a histogram. The area under a density curve is one. The area under a density curve is one.

3. Density Curve

4. Density Curves: Properties

5. Density Curves Mean of density curve is point at which the curve would balance. Mean of density curve is point at which the curve would balance. For symmetric density curves, balance point (mean) and the median are the same. For symmetric density curves, balance point (mean) and the median are the same.

6. Characterization A normal distribution is bell-shaped and symmetric. A normal distribution is bell-shaped and symmetric. The distribution is determined by the mean mu,  and the standard deviation sigma, . The distribution is determined by the mean mu,  and the standard deviation sigma, . The mean mu controls the center and sigma controls the spread. The mean mu controls the center and sigma controls the spread.

7. Definitions Mean is located in center, or mode of normal curve. Mean is located in center, or mode of normal curve. The standard deviation is the distance from the mean to the inflection point of the normal curve, the place where the curve changes from concave down to concave up. The standard deviation is the distance from the mean to the inflection point of the normal curve, the place where the curve changes from concave down to concave up.

8. Construction A normal curve is drawn by first drawing a normal curve. A normal curve is drawn by first drawing a normal curve. Next, place the mean, mu on the curve. Next, place the mean, mu on the curve. Then place sigma on curve by placing the segment from the mean to the upper (or lower) inflection point on your curve. Then place sigma on curve by placing the segment from the mean to the upper (or lower) inflection point on your curve. From this information, the scale on the horizontal axis can be placed on the graph. From this information, the scale on the horizontal axis can be placed on the graph.

9. Examples Draw normal curve with mean=mu=100, and standard deviation = sigma = 10. Draw normal curve with mean=mu=100, and standard deviation = sigma = 10. Draw normal curve with mean = 20, sigma=2. Draw normal curve with mean = 20, sigma=2.

10. 68-95-99.7 Rule For any normal curve with mean mu and standard deviation sigma: For any normal curve with mean mu and standard deviation sigma: 68 percent of the observations fall within one standard deviation sigma of the mean. 68 percent of the observations fall within one standard deviation sigma of the mean. 95 percent of observation fall within 2 standard deviations. 95 percent of observation fall within 2 standard deviations. 99.7 percent of observations fall within 3 standard deviations of the mean. 99.7 percent of observations fall within 3 standard deviations of the mean.

11. Example Questions If mu=30 and sigma=4, what are the values (a, b) around 30 such that 95 percent of the observations fall between these values? If mu=30 and sigma=4, what are the values (a, b) around 30 such that 95 percent of the observations fall between these values? If mu=40 and sigma=5, what are the bounds (a, b) such that 99.7 percent of the values fall between these values? If mu=40 and sigma=5, what are the bounds (a, b) such that 99.7 percent of the values fall between these values?

12. Standard Normal Distribution The standard normal distribution has mean = 0 and standard deviation sigma=1. The standard normal distribution has mean = 0 and standard deviation sigma=1.

13. Normal Table Usage What proportion of standard normal distribution values Z are less than 1.40? That is, P(Z < 1.40) = ? What proportion of standard normal distribution values Z are less than 1.40? That is, P(Z < 1.40) = ? Ans:.9192 or 91.92 percent of values. Ans:.9192 or 91.92 percent of values.

14. Standard Normal P( 0 < Z < 1.40) = ? P( 0 < Z < 1.40) = ? Ans: P(Z < 1.40) – P(Z<0) =.9192 -.5 =.4192 Ans: P(Z < 1.40) – P(Z<0) =.9192 -.5 =.4192

15. Example P( Z < - 2.15) = ? P( Z < - 2.15) = ?

16. Normal Table Usage P(.64 < Z < 1.23) = ? P(.64 < Z < 1.23) = ? Ans: P(Z<1.23) – P(Z <.64) =.8907 -.7389 =.1518 Ans: P(Z<1.23) – P(Z <.64) =.8907 -.7389 =.1518 P(Z > 2.24) = CAREFUL !!!!! P(Z > 2.24) = CAREFUL !!!!! Ans: Either = 1 – P(Z < 2.24) = 1 -.9875 = Ans: Either = 1 – P(Z < 2.24) = 1 -.9875 = or by symmetry = P(Z 2.24) = P(Z 2.24) = P(Z< -2.24) =.0125.

17. Z-Score Formula Any normal distribution with mean=mu and standard deviation= sigma, can be converted into a standard normal Z distribution by the following transformation: Any normal distribution with mean=mu and standard deviation= sigma, can be converted into a standard normal Z distribution by the following transformation:

18. Example Consider a distribution with mean=mu=100 and standard deviation = sigma = 10. Draw density curve with number line provided. Consider a distribution with mean=mu=100 and standard deviation = sigma = 10. Draw density curve with number line provided. Now re-draw the curve and number line on horizontal axis after subtracting 100 from each value. Notice this centers the curve at zero. Now re-draw the curve and number line on horizontal axis after subtracting 100 from each value. Notice this centers the curve at zero. Then draw the resulting number line after dividing the previous number line values by 10. Then draw the resulting number line after dividing the previous number line values by 10. Voila ! We are now back to Z scale ! Voila ! We are now back to Z scale !

19. Example Example 1.26 in Page 75. Example 1.26 in Page 75. X=The SAT score of a randomly chosen student. X has N  1019,  =209). X=The SAT score of a randomly chosen student. X has N  1019,  =209). What percent of all students had SAT scores of at least 820? That is, P( X > 820) = ? What percent of all students had SAT scores of at least 820? That is, P( X > 820) = ?

20. Solution P( X > 820 ) = P( X > 820 ) = Solution =.8289 Solution =.8289

21. Example

22. Problem 1.86 (Moore&Mc) Eleanor gets 680 on SAT math exam. Mean on this exam is 500 and sd is 100. Eleanor gets 680 on SAT math exam. Mean on this exam is 500 and sd is 100. Eleanor’s standardized score is: Eleanor’s standardized score is:

23. 1.86 Continued Gerald got 27 on ACT math. Mean is 18 with sd of 6. Gerald got 27 on ACT math. Mean is 18 with sd of 6. Gerald’s Z-Score is: Gerald’s Z-Score is: Eleanor did better ! Eleanor did better !

24. Human Pregnancies What proportion of births are premature? That is, what proportion is below 240 days? P(X<240)= ? What proportion of births are premature? That is, what proportion is below 240 days? P(X<240)= ?

25. London Bus Drivers Calorie intake for drivers averages 2821 cals per day with sd=sigma=436. Calorie intake for drivers averages 2821 cals per day with sd=sigma=436. What proportion of drivers have calorie intakes, X, less than 2000 calories per day? P(X < 2000)? What proportion of drivers have calorie intakes, X, less than 2000 calories per day? P(X < 2000)?

26. London Bus Drivers What proportion of drivers consume between 2000 and 2500 cals per day? P(2000<X<2500)? What proportion of drivers consume between 2000 and 2500 cals per day? P(2000<X<2500)?

27. Finding a Percentile Backwards problem. We are now given a fraction and need to find the X-value. Backwards problem. We are now given a fraction and need to find the X-value. In past, we were provided X and found a proportion. In past, we were provided X and found a proportion. Use Formula: Use Formula:

28. London Bus Drivers Find the calorie intake at the 90 th percentile of the calorie distribution. Find the calorie intake at the 90 th percentile of the calorie distribution. Insert mean and sd into backward formula, then determine correct Z-star value. Insert mean and sd into backward formula, then determine correct Z-star value.

29. Finding a Percentile Plugging in the mean and sd are not hard. The difficulty is finding Z-star. It is simply the same percentile you are trying to find, except for the standard normal distribution. This requires you to use an inverse lookup in your z-table. Plugging in the mean and sd are not hard. The difficulty is finding Z-star. It is simply the same percentile you are trying to find, except for the standard normal distribution. This requires you to use an inverse lookup in your z-table.

30. TV Viewing Neilsen ratings service found that tv viewing for children aged 2-11 had a normal distribution with mean 23.02 hours and sigma=6.23 hours. Neilsen ratings service found that tv viewing for children aged 2-11 had a normal distribution with mean 23.02 hours and sigma=6.23 hours. What proportion of children watch more than 24 hours of tv per week? What proportion of children watch more than 24 hours of tv per week?

31. TV Viewing How many hours of tv does a child watch that is at the 95 th percentile of the tv viewing distribution? How many hours of tv does a child watch that is at the 95 th percentile of the tv viewing distribution?

32. The Central Limit Theorem (for the sample mean x) If a random sample of n observations is selected from a population (any population), then when n is sufficiently large, the sampling distribution of x will be approximately normal. If a random sample of n observations is selected from a population (any population), then when n is sufficiently large, the sampling distribution of x will be approximately normal. (The larger the sample size, the better will be the normal approximation to the sampling distribution of x.)

33. How Large Should n Be? For the purpose of applying the central limit theorem, we will consider a sample size to be large when n > 30. For the purpose of applying the central limit theorem, we will consider a sample size to be large when n > 30.