Normal Distribution Revisited

1. Normal Distribution Revisited The area under the normal distribution is 1 (100% of the data points lie somewhere)

2. To estimate binomial experiment probabilities • If np>5 and n(1-p)>5 • The normal distribution with • x = np and σ = np(1-p) • Can approximate the binomial distribution of the random variable X

3. E.G. 100 coins are tossed • Find P(60 coins are heads) • P(60 heads)=100C60(1/2)60(1/2)40 • = 1.08 % • What if I asked the probability that the coin comes up heads 60 times or less?

4. Without the normal distribution • We could calculate 61 different probabilities and add them all together. • (possible with a computer but impractical otherwise)

5. Applying the Normal Distribution • x = np and σ = np(1-p) • x=np=100(1/2)=50 • σ = np(1-p) = 100(1/2)(1/2) =5 • Find z score for x=60.5 • Z =60.5 - 50 =2.1 • 5 • P(x < 60)=P(Z<2.1)=98.21% (from z score table page 606-607)

6. Why did we use 60.5? • I am 32 years old. • My daughter weighs 32 pounds. • Is the 32 in the first sentence the same as the 32 in the second?

7. In the first 32 means at least 32 but not yet 33 years old. • In the second it means closer to 32 than 31 or 33.

8. 32 years old means in the range • 32<x<33 • 32 pounds means in the range • 31.5<x<32.5

9. Less than 32 years old would mean less than 32.0 years old. • Less than 32 pounds would mean less than 29.5 pounds. • We use the latter interpretation of whole numbers when using the continuous normal distribution to approximate binomial experiment results.

10. When we specify that a random variable is less than or equal 60 • We mean in the range from 0 to 60.5 • So we say x = 60.5 and find the z score • to use the associated percentage as the probability that x is less than or equal to 60. • For p(X<60) we would use 59.5 • The 0.5 is called a continuity correction.

11. To Find the Probability for a Range • P(20<x<30) = P(X<30) - P(X<20)

12. Relating Back to StatisticsImagine that you want to predict an elections outcome. • You pole 20 people selected through a simple random process to determine who supports TPP (The Political Party) • If in truth, 60% of the population support TPP and the remaining 40% support TOPP (the other political party) what is the probability that your survey predicts the winner correctly?

13. Probability that the survey is right: Let R.V. X denote the number of sampled people that support TPP. If X>10 then you will accurately predict the outcome of the election. Re-read the question to see why this is so.

14. P(X>10) • P(X>10)=1-P(X≤10) • Lets find P(X≤10) • µ = np=20(0.6)=14 • σ = sqrt(20(.6)(.4))=4.8 • Z=(10.5-14)/4.8= -0.73 • P(X≤10)=P(Z<-0.73)= 23.27% from Z score table

15. P(X>10)=1-P(X≤10) • P(X>10)=1- 23.27% = 76.73 %

16. What if we doubled our sample size? • Everything else remains but n= 40 and we want P(X>20) • P(X>20)=1-P(X≤20) • Lets find P(X≤10) • µ = np=40(0.6)=24 • σ = sqrt(40(.6)(.4))=3.1 • Z=(20.5-24)/3.1= -1.13 • P(X≤10)=P(Z<-1.13)= 12.92% from Z score table • P(X>20)=1-12.92%=87.08% • As you can see, as we increase the sample size, we increase the probability that the conclusion accurately represents the sample.

17. Limitations: • The above calculations require knowing what the underlying percentage of the population supports TPP ahead of time. • In a real study this is not realistic as there would be no reason to conduct a poll if you already know what the underlying trend is. • You can verify for your self that the probability of predicting accurately depends on the underlying population as well as the sample size (just try changing the 60% in the above example to any other number.

18. Practice Page 448 • Questions 1 to 11