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Normal Distributions

Normal Distributions. Density Curve. A density curve is a smooth function meant to approximate a histogram. The area under a density curve is one. Density Curve. Density Curves: Properties. Density Curves. Mean of density curve is point at which the curve would balance.

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Normal Distributions

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  1. Normal Distributions

  2. Density Curve • A density curve is a smooth function meant to approximate a histogram. • 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. • For symmetric density curves, balance point (mean) and the median are the same.

  6. Characterization • A normal distribution is bell-shaped and symmetric. • The distribution is determined by the mean mu, m, and the standard deviation sigma, s. • The mean mu controls the center and sigma controls the spread.

  7. Definitions • 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.

  8. Construction • A normal curve is drawn by first drawing a normal 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. • 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 = 20, sigma=2.

  10. 68-95-99.7 Rule • 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. • 95 percent of observation fall within 2 standard deviations. • 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=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.

  13. Normal Table Usage • 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.

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

  15. Example • P( Z < - 2.15) = ?

  16. Normal Table Usage • P( .64 < Z < 1.23) = ? • Ans: P(Z<1.23) – P(Z < .64) = .8907 - .7389 = .1518 • P(Z > 2.24) = CAREFUL !!!!! • Ans: Either = 1 – P(Z < 2.24) = 1 - .9875 = • or by symmetry = P(Z < - 2.24) = .0125. In this approach you are using the fact that both tails of a standard normal are the same and so 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:

  18. Example • 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. • Then draw the resulting number line after dividing the previous number line values by 10. • Voila ! We are now back to Z scale !

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

  20. Solution • P( X > 820 ) = • 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’s standardized score is:

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

  24. Human Pregnancies • 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. • 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)?

  27. Finding a Percentile • 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. • Use Formula:

  28. London Bus Drivers • Find the calorie intake at the 90th percentile of the calorie distribution. • 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.

  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. • 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 95th 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. (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.

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