1. Chapter 5 Section 5-5

3. Chapter 5 Normal Probability Distributions Section 5-5 – Normal Approximations to Binomial Distributions 2.Correction for Continuity a.Binomial distributions only work for discrete data points. 1)When we want to calculate the exact binomial probabilities, we can find the probability of each value of x occurring and add them together. We did this in Chapter 4. b.To use a continuous normal distribution to approximate a binomial probability, you need to move.5 unit to each side of the midpoint to include all possible x-values in the interval. 1)This is called making a correction for continuity. a)We simply subtract.5 units from the lowest value and add.5 units to the highest value. 3.There is a good review chart with this information displayed on page 288 of your text book.

6. Example 1A (Page 286) 51% of adults in the US who resolved to exercise more in the new year achieved their resolution. You randomly select 65 adults in the US whose resolution was to exercise more and ask each if he or she achieved their resolution. Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If you can, find the mean and standard deviation. If you cannot, explain why. What are n, p and q? n = 65 p = 0.51q = 0.49

7. Example 1A (Page 286) 51% of adults in the US who resolved to exercise more in the new year achieved their resolution. You randomly select 65 adults in the US whose resolution was to exercise more and ask each if he or she achieved their resolution. n = 65p = 0.51q = 0.49 Are np and nq greater than or equal to 5? (65)(.51) = 33.15 and (65)(.49) = 31.85 Since both of these are greater than 5, we CAN use the normal distribution.

8. Example 1A (Page 286) 51% of adults in the US who resolved to exercise more in the new year achieved their resolution. You randomly select 65 adults in the US whose resolution was to exercise more and ask each if he or she achieved their resolution. n = 65p = 0.51q = 0.49 REMEMBER THE ROUND-OFF RULE!!!! Mean, standard deviation and variance are rounded to one decimal place more than the x-values. Since we are talking about adults, the x-values are whole numbers. Hence, we round the mean and standard deviation to the nearest tenth.

10. Example 1B (Page 286) 15% of adults in the US do not make New Year’s resolutions. You randomly select 15 adults in the US and ask each if he or she made a New Year’s resolution. Decide whether you can use the normal distribution to approximate x, the number of people who reply yes. If you can, find the mean and standard deviation. If you cannot, explain why. What are n, p and q? n = 15p = 0.15q = 0.85

11. Example 1B (Page 286) 15% of adults in the US do not make New Year’s resolutions. You randomly select 15 adults in the US and ask each if he or she made a New Year’s resolution. n = 15p = 0.15q = 0.85 Are np and nq greater than or equal to 5? (15)(.15) = 2.25 and (15)(.85) = 12.75 Since np < 5, we CANNOT use the normal distribution to approximate the distribution of x.

12. Example 2 (Page 287) Use a correction for continuity to convert each of the following binomial intervals to a normal distribution interval. 1.The probability of getting between 270 and 310 successes, inclusive. Since we are dealing with whole numbers, we subtract.5 from the low end and add.5 to the high end. 270 -.5 = 269.5 310 +.5 = 310.5 Our interval is 269.5 < x < 310.5 2.The probability of at least 158 successes. Since 158 is the low end, our interval is x > 157.5. 3.The probability of getting less than 63 successes. We want all numbers less than 63, which makes 62 the upper end. We add.5 to the upper end to get x < 62.5.

13. Example 3 (Page 288) 51% of adults in the US who resolved to exercise more in the new year achieved their resolution. You randomly select 65 adults in the US whose resolution was to exercise more and ask each if he or she achieved their resolution. What is the probability that fewer than 40 of them respond yes? We know from Example 1A that we can use the normal distribution, with a mean of 33.2 and a standard deviation of 4.0 Correcting for continuity means that we use 39.5, since 39 is the highest number less than 40, and it is at the high end of the interval. normalcdf(0,39.5,33.2,4) gives us.942. We have a 94.2% probability that fewer than 40 people will respond “Yes”.

16. Example 5 (Page 290) 1.A survey reports that 86% of internet users use Windows Internet Explorer as their browser. You randomly select 200 internet users and ask each whether he or she uses Internet Explorer as his or her browser. What is the probability that exactly 176 say yes? 2 nd VARS binompdf(176,172,4.9) =.0583 Alternatively, we can use the normalcdf to find the area between 175.5 and 176.5 (176 corrected for continuity). 2 nd VARS normalcdf(175.5,176.5,172,4.9) =.0583 We have about a 5.83% chance of getting exactly 176 out of 200 people to say that they use Internet Explorer as their browser.

17. YOUR ASSIGNMENTS TODAY ARE: Classwork: Page 291 #1-16 All Homework: Page 292-294 #17-26 All