The Normal Distribution Limiting smooth bell shaped symmetric
The Normal Distribution Limiting smooth bell shaped symmetric curve is called the Normal p. d. f. curve. Is symmetric about the mean. Mean = Median 50% Mean If a random variable, X, has a Normal distribution with a mean and a standard deviation we write: X ~ Normal ( , ) parameters C 6, L 2, S 2
The Normal Distribution • The Normal distribution is important because: – it fits a lot of data reasonably well; – it can be used to approximate other distributions; – it is important in statistical inference (see later work). C 6, L 2, S 3
GPA’s of WSU Students Let X be the GPA of randomly selected WSU student. Suppose data collected from a recent survey gave the following: A sample mean GPA of 3. 065 and a sample standard deviation of. 473 These are samplebased estimates of the population mean and standard deviation s. C 6, L 2, S 6
GPA’s of WSU Students Approximately 68 % of WSU students will have a GPA within 1 standard deviation of the mean. i. e. , approximately 68 % of the WSU students have GPA’s = between 3. 065 -. 473 and 3. 065 +. 473 = between 2. 592 and 3. 538 C 6, L 2, S 7
GPA’s of WSU Students Approximately 95 % of WSU students will have a GPA within 2 standard deviation of the mean. i. e. , approximately 95 % of WSU students have GPA’s = between 3. 065 - 2 . 473 and 3. 065 + 2 . 473 = between 2. 119 and 4. 011 (4. 00) C 6, L 2, S 8
GPA’s of WSU Students Approximately 99. 73 % of WSU students will have GPA’s within 3 standard deviation of the mean. i. e. , approximately 99. 73% of WSU students have GPA’s = between 3. 065 - 3 . 473 and 3. 065 - 3 . 473 = between 1. 646 and 4. 484 (4. 00) C 6, L 2, S 9
The Normal Distribution For the Normal Distribution: A random observation has approximately: – 68% chance of falling within 1 of ; – 95% chance of falling within 2 of ; – 99. 7% chance of falling within 3 of . Or: In a Normal distribution, approximately: – 68% of observations are within 1 of ; – 95% of observations are within 2 of ; – 99. 7% of observations are within 3 s of . C 6, L 2, S 10
The Normal Distribution Probabilities and numbers of standard deviations Shaded area = 0. 683 - + 68% chance of falling between - and + Shaded area = 0. 954 - 2 + 2 95% chance of falling between - 2 and + 2 Shaded area = 0. 997 - 3 + 3 99. 7% chance of falling between - 3 and + 3 C 6, L 2, S 11
Problem: Diagnosing Spina Bifida For mothers with normal foetuses, the mean level of alpha fetoprotein is 15. 73 moles/litre with a standard deviation of 0. 72 moles/litre. For mothers carrying foetuses with spina bifida, the mean is 23. 05 and the standard deviation is 4. 08. In both groups the distribution of alpha fetoprotein appears to be approximately Normally distributed. 15. 73 23. 05 Given thiswe might like to For example we want find: information P(X > 17. 8) or to be able to find P(19 < X < 25) etc… probabilities associated with these for either group. distributions. C 6, L 2, S 12
Standard Normal Distribution FACT: If X ~ Normal( , s) then if we consider a new random variable (Z) representing the z-scores of X Z will have a standard normal distribution, i. e. Z ~ Normal (0 , 1) C 6, L 2, S 16
Obtaining Probabilities Basic method for obtaining probabilities 1. Sketch a Normal curve, marking on the mean and values of interest. 2. Shade the area under the curve corresponding to the required probability. 3. Convert all values to their z-scores 4. Obtain the desired probability using tables in C 6, L 2, S 17
Original problem: Diagnosing Spina Bifida 15. 73 23. 05 Recall: • For normal foetuses =15. 73, = 0. 72 and for foetuses with spina bifida = 23. 05 and = 4. 08. • Assume threshold for detecting spina bifida is set at 17. 8. – (A foetus would be diagnosed as not having spina bifida if the fetoprotein level is below 17. 8) C 6, L 2, S 23
Original problem: Diagnosing Spina Bifida 15. 73 23. 05 a) What is the probability that a foetus not suffering from spina bifida is correctly diagnosed? Let X be level of fetoprotein in normal foetus X ~ Normal (15. 73, 0. 72) What is P(X < 17. 8)? P(X < 17. 8) = P(Z < z-score for 17. 8) z-score = (17. 8 – 15. 73)/. 72 = 2. 07/. 72 =2. 88 P(X < 2. 88) =. 9980 15. 73 C 6, L 2, S 24 17. 8
Original problem: Diagnosing Spina Bifida 15. 73 23. 05 b) What is the probability that a foetus with spina bifida is correctly diagnosed? Let Y be the level of fetoprotein in a spina bifida foetus. Y ~ Normal (23. 05, 4. 08) P(Y > 17. 8) = P(Z > z-score for Y = 17. 8) P(Z > -1. 29) =. 4015 +. 5000 z-score = (17. 8 – 23. 05)/4. 08 = 0. 9015 = -5. 25/4. 08 = -1. 29 17. 8 -1. 29 023. 05 C 6, L 2, S 25
Original problem: Diagnosing Spina Bifida 15. 73 23. 05 If they wanted to ensure that 99% of foetuses with spina bifida were correctly diagnosed, at what level should they set T ? Find a value T so that if First theensures z-score associated with T by T = find 13. 54 From Normal Table we 99% find of foetuses finding z so that with spina bifida will be identified. Y ~ Normal (23. 05, 4. 08) P(-2. 33 < Z < 0) =. 9900 P(Z < P(Z z) =<. 0100 thus -2. 33) =. 0100 as desired. This probability is called the we will have Now convert back to original scale. . . sensitivity . 9900 –or P(Y x< 2. 33 T) = =. 0100 T = m. P(Y + s >x T) z == 23. 05 4. 08 13. 54 C 6, L 2, S 26
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