1 / 26

# Welcome to MM207 - Statistics!

Welcome to MM207 - Statistics!. Unit 5 Seminar: The Standard and Normal Distributions. Unit 4 comments. Table verses Excel follow the Excel templates much easier than equations Names in the “file name” of submitted attachments Prevents lost files Show all your work

Télécharger la présentation

## Welcome to MM207 - Statistics!

E N D

### Presentation Transcript

1. Welcome to MM207 - Statistics! Unit 5 Seminar: The Standard and Normal Distributions

2. Unit 4 comments • Table verses Excel • follow the Excel templates • much easier than equations • Names in the “file name” of submitted attachments • Prevents lost files • Show all your work • I can give partial credit • no work, no partial credit (if possible) • Read directions • 2 decimal places = 0.00 • 3 decimal places = 0.000 • Please read requirements

3. Normal Probability Distribution It is a continuous probability distribution Two values determine its shape μ = mu mean of distribution σ = sigma standard deviation of the distribution

4. Normal Probability Distribution (cont) Several important properties (page 240 text) The curve is bell-shaped with the highest point centered on µ It is symmetrical about a vertical line through µ The curve approaches the horizontal axis but never touches it The total area= 1.

5. Standard Normal Curve(a very special normal distribution) µ = 0 σ =1 z score – tells us how standard deviations away from the mean a value is: z = (x - µ)/ σ We will convert x valuesto z scores usingthe above formula [page 107 or 243] or Excel!

6. Example 1: Using the Table P(z ≤ 1.55) = ? Using the Standard Normal Table look up the area associated with z = 1.55 Note: The table only gives areas to the left of the given z score. Find the row for z = 1.5 Find the column for 0.05 The area is 0.9394 P(z ≤ 1.55) = 0.9394 This area makes sense because the area needed is more than half of the area under the curve as shown in the sketch. 1.55

7. Example 1 (cont) P(z ≤ 1.55) = 0.9394 Excel can also be used to find the probability, the function is NORMSDIST. The instructions are in Doc Sharing at Unit 5 Templates.xls

8. Excel Results

9. Example 2: Using the Table P(z ≥ -0.56) = ? Look up the area associated with -0.56 in the table. The table only gives the area to the left. The area is 0.2877. This is the area to the left of z = -0.56. We want the area to the right. P(z ≥ -0.56) =1 - 0.2877 = 0.7123 Note: P(z ≥ - 0.56) is the same as P(z ≤ 0.56) So you can also look up P(z ≤ 0.56) . Try it! This is because the curve is symmetrical. 0.2877 0.7123 -0.56

10. Example 3: Using the Table P(0 ≤ z ≤ 1) = ? This value is not given in the table. We need to get creative. P(0 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ 0) P(z ≤ 0) = 0.500 P(z ≤ 1) = 0.8413 P(0 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ 0) = 0.8413 – 0.5000 = 0.3413 Always subtract the smaller area from the larger area. The answer cannot be negative. = -

11. Example 3 (cont) P(z ≤ 0) = 0.500 P(z ≤ 1) = 0.8413 P(0 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ 0) = 0.8413 – 0.5000 = 0.3413

12. Area Between Two Z scores • =NORMSDIST(1.00)-NORMSDIST(0.00) 0.3413

13. Example 4: Finding a Probability for a given x Given: µ = 600, σ =100 P(x ≥ 750) = ? X distribution: µ = 600 σ =100 CONVERT Z distribution µ = 0 σ =1 ? 750

14. Example 4: Finding a Probability for a given x Given: µ = 600, σ =100 P(x ≥ 750) = ? Convert x to a z-score z = (x - µ)/ σ P(x ≥ 750) = P(z ≥ (750 - 600)/ 100) = P(z ≥150/100) P(z ≥ 1.5) Use the table. P(z ≥ 1.5) = 1 – P(z ≤ 1.5) = 1 – 0.9332 = 0.0668 P(x ≥ 750) = 0.0668 You can also look up the opposite, P(z ≤ - 1.5) = .0668 Z = 1.5

15. Excel solution • Call up NORMDIST from the Excel functions • Use the following values as given in the problem • X = 750 • Mean = 600 • Standard_dev = 100 • Remember we are looking for P(x>750) so we use 1-NORMDIST • 1-NORMDIST(750,600,100) = 0.0668

16. Example 5: Working backwardsFind Z0 in P(z ≤ z0 ) = 0.99 Find the areas closest to 0.9900 P(z ≤ 2.32 ) = 0.9898 P(z ≤ 2.33 ) = 0.9901 0.9901 is closer to 0.99 P(z ≤ 2.33 ) ≈ 0.99 z0 = 2.33

17. Example 6: Working Backwards An IQ test has μ = 100 and σ = 15. To get into a special program, a Student must have an IQ score in the top 10%. What is the lowest IQ score needed? Find the z score associated with 10% Since we are looking for the top 10%, we use 90% to find the z-score. 1 – 0.1000 = 0.9000 Find Z0 in P(z ≥ Z0 ) = 0.9000 Z0 = 1.28 (since 0.8997 is the area closest to 0.9000 in the table) Convert to x x = μ + zσ x = 100 + 1.28 * 15 = 118.2 Round up to 119 The lowest IQ score needed is 119.

18. Using Excel • NORMINV(0.9,100,15) = 119.223 • Notice that we used 0.9 instead of 0.1 because we were looking for the probability of greater than 10% • The difference between the manual answer and the Excel answer is the inaccuracies of the table. • In reality the score required to get into the program is 120.

19. Sampling Distributions and The Central Limit Theorem Since we are often unable to measure the population mean and standard deviation (μ and σ), we want to use the sample mean and standard deviation (xbar and s) to estimate the parameters for the population. One way we do this is by creating a sampling distribution from the population. If we take all possible samples of a certain size from a population, and compute the mean for each sample, we can create a sampling distribution of the means The mean of the sampling distribution is the “mean of the means” which is μ. μxbar = μ The standard deviation is the population standard deviation divided by the square root of n. σ xbar = σ/√n

20. Sampling Distributions and The Central Limit Theorem (con’t) The Central Limit Theorem (page 272) says that if we have samples with size n ≥ 30 from the population, then the sampling distribution is approximately normal even if the population distribution is not normal. If the population distribution is normal, then any sample size will be ok and the sampling distribution will be normal. This gives us the ability to find Confidence Intervals (Chapter 6) to estimate the population mean and to find probabilities for hypothesis testing (Chapter 7).

21. Example 7: Using Sampling Distributions Scientists determined that x (the length of a single trout) is normally distributed with µ = 10.2 and σ = 1.4 Question A:What is the probability a single trout is between 8 and 12 inches? P(8 < x < 12) = ? Question B:What is the probability that the mean length of 5 trout(xbar) will be between 8 and 12 inches? P(8 < xbar < 12) = ?

22. Example 7 Solution Given: µ = 10.2 and σ = 1.4 (normally distributed) Question A: P(8 < x < 12) = ? z = (x - µ)/σP(8 < x < 12) = P((8 – 10.2)/1.4 < z < (12 – 10.2)/1.4) = P(-1.57 < z < 1.29) = P(z < 1.29) - P(z < -1.57) = 0.9015 - 0.0582 = 0.8433 Question B: P(8 < xbar < 12) = ? z = (xbar - µxbar)/ σxbar µxbar = µ = 10.2 and σxbar = σ/√n = 1.4/√5 ≈ 0.63 P(8 < xbar < 12) = P((8 – 10.2)/0.63 < z < (12 – 10.2)/0.63) = P(-3.49 < z < 2.86) = P(z < 2.86) - P(z < -3.49) = 0.9979 - 0.0002 = 0.9977

23. Excel solution Question A NORMDIST(12,10.2,1.4,TRUE)-NORMDIST(8,10.2,1.4,TRUE) = 0.8427 Question B NORMDIST(12,10.2,0.63,TRUE)-NORMDIST(8,10.2,0.63,TRUE) = 0.9976 The main difference between the two is that Question A is asking for a particular probability of a single event and Question B is asking for a particular probability of an average of a number of events. Thus Question B is using the premise of the Central Limit Theorem and uses

24. Normal Approximations to Binomial Distributions If np and nq are greater than or equal to 5, then we can use the normal distribution to approximate a binomial distribution. Page 285 A correction must be used, called the continuity correction. Page 287 Move 0.5 units to the left and right of the midpoint to include all possible x-values in the interval. Then we can approximate a binomial probability

25. Example 8: Approximating Binomial Probabilities Try it Yourself, page 280 What is the probability that at most (≤) 85 people will say yes? n = 200 p = 0.38 np = 76 and nq = 124, so we can use the normal approximation μ = 76; σ ≈ 6.86 = Add 0.5 to 85 to get 85.5 for the continuity correction since this is x ≤ 85. P( x ≤ 85.5) = P(z ≤ (85.5 – 76)/ 6.86) ) = P(z ≤ 1.38) = 0.9162

26. Z scores using Central Limit TheoremCLT in Template

More Related