Chapter 5 Section 5-5

Chapter 5 Normal Probability Distributions Section 5-5 – Normal Approximations to Binomial Distributions A. Properties of a Normal Approximation to a Binomial Distribution 1. If np ≥ 5, and nq ≥ 5, then the binomial random variable x is approximately normally distributed, with a mean that equals np and a standard deviation that equals 𝑛𝑝𝑞 . a. Again, if np ≥ 5, and nq ≥ 5, then μ = np and σ = 𝑛𝑝𝑞 b. We need to remember from Section 4-2 what the properties of a binomial experiment are: 1) n independent trials (we know before we start how many trials there are going to be). 2) Only two possible outcomes (success or failure). 3) Probability of success is p. 4) Probability of failure is 1 – p, which we call q. 5) p is constant for each trial (the trials have nothing to do with each other).

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.

Chapter 5 Normal Probability Distributions Section 5-5 – Normal Approximations to Binomial Distributions a. The steps to using the Normal Distribution to Approximate Binomial Probabilities are: 1) Verify that the binomial distribution applies. a) Specify n, p, and q. 2) Determine if you can use the normal distribution to approximate x, the binomial variable. a) Are np and nq both greater than or equal to 5? 3) Find the mean and standard deviation for the distribution. a) μ = np and σ = 𝑛𝑝𝑞 . 4) Apply the approximate continuity correction. Shade the corresponding area under the normal curve. a) Subtract .5 unit from lowest value, add .5 unit to highest value.

Chapter 5 Normal Probability Distributions Section 5-5 – Normal Approximations to Binomial Distributions 5) Find the corresponding z-score(s). a) 𝑧= 𝑥−𝜇 𝜎 6) Find the probability. a) Use the calculator.

Example 1A (Page 286) 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. 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. 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.

Example 1A (Page 286) 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. 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 are n, p and q? n = 65 p = 0.51 q = 0.49

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 = 65 p = 0.51 q = 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.

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 = 65 p = 0.51 q = 0.49 𝜇=𝑛𝑝=33.15 (we already knew this) 𝜎= 𝑛𝑝𝑞 = 65 (.51)(.49) =4.030 So, we CAN use the normal distribution, with a mean of 33.15 and a standard deviation of 4.03

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. What are n, p and q? n = 15 p = 0.15 q = 0.85

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 = 15 p = 0.15 q = 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.

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. 2. The probability of at least 158 successes. 3. The probability of getting less than 63 successes.

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

Example 2 (Page 287) Use a correction for continuity to convert each of the following binomial intervals to a normal distribution interval. 2. The probability of at least 158 successes. Since 158 is the low end, our interval is x > 157.5.

Example 2 (Page 287) Use a correction for continuity to convert each of the following binomial intervals to a normal distribution interval. 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.

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 33.15 and a standard deviation of 4.03 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.

What is the probability that fewer than 40 of them respond yes? 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? Our answer will be between .841 and .977. 33.15 37.18 41.21 45.24 21.06 25.09 29.12 0.341 0.136 0.0215 0.0015

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? 2nd VARS 2 (-1E99, 39.5, 33.15, 4.03) gives us .942. This fits between .841 and .977. Also, 2nd VARS 3 (.942, 33.2, 4.03) = 39.534, so it checks. We have a 94.2% probability that fewer than 40 people will respond “Yes”.

Example 4 (Page 289) 38% of people in the US admit that they snoop in other people’s medicine cabinets. You randomly select 200 people in the United States and ask each if they snoop in other people’s medicine cabinets. What is the probability that at least 70 will say yes? Can we use the normal distribution? 𝑛𝑝= 200 .38 =76 and 𝑛𝑞= 200 .62 =124 Since both of these are ≥ 5, we CAN use the normal distribution with a mean of 76 (np) The standard deviation = 200 .38 .62 ≈6. 864

What is the probability that at least 70 will say yes? Example 4 (Page 289) 38% of people in the US admit that they snoop in other people’s medicine cabinets. You randomly select 200 people in the United States and ask each if they snoop in other people’s medicine cabinets. What is the probability that at least 70 will say yes? 𝜇=76 and 𝜎=6.864 This is going to be just less than .841 76 82.864 89.728 96.592 55.408 62.272 69.136 0.341 0.136 0.0215 0.0015

Example 4 (Page 289) 38% of people in the US admit that they snoop in other people’s medicine cabinets. You randomly select 200 people in the United States and ask each if they snoop in other people’s medicine cabinets. What is the probability that at least 70 will say yes? Correcting for continuity means that we subtract .5 from 70 (the low end of the interval) to get 69.5. 2nd VARS 2 (69.5, 1E99, 76, 6.864) = .8282 This matches our estimate of just less than .841 2nd VARS 3 (1 - .8282, 76, 6.864) = 69.499 (it checks) We have an 82.8% probability that at least 70 people will respond “Yes”.

Example 5 (Page 290) 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? SOLUTION: Can we use the normal distribution? np = 200(.86) = 172 nq = 200(.14) = 28 This means that we can use the normal approximation. The mean is 𝑛𝑝=172 The standard deviation is 200 .86 (.14) ≈4.907 Since we want exactly 176, we use normalpdf (2nd VARS 1)!! 2nd VARS 1 (176, 172, 4.907) = .0583

Example 5 (Page 290) 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? Alternatively, we can use the normalcdf to find the area between 175.5 and 176.5 (176 corrected for continuity). 2nd VARS 2 (175.5, 176.5, 172, 4.907) = .0583 Either way, 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.

YOUR ASSIGNMENTS ARE: Classwork: Page 291 #1-16 All Homework: Pages 292-294 #19-26 All