1. 7.4 and 7.5 Obj: Assess normality of a distribution and find the normal approximation to a binomial distribution

2. 7.4 Assessing Normality A Normal Probability Plot is used to decide if data is normally distributed. If sample data are taken from a population that is normally distributed, a normal probability plot of the observed values versus the expected Z-scores will be approximately linear.

3. Using Technology 1.Enter the data into L1. 2.STAT PLOTS 3.Plot 1 – turn it on, press ENTER. 4.Choose the normal probability plot icon. 5.Data list should be set at L1. 6.ZOOM STAT Decide if the data is linear. If so, the data is normally distributed.

4. Example Following is a random sample of the hours of tv watched in one week by college students. Decide if the data is normally distributed. 36.130.52.917.5 21.023.525.616.0 28.929.67.820.4 33.836.80.09.9 25.819.519.118.5 22.99.739.219.0 8.6

5. 7.5 Normal Approximation to the Binomial Probability Distribution Review Binomial Probability Distributions Properties Expected Value (mean) Standard Deviation Histograms

6. Rule of Thumb As the number of trials increases, the probability distribution becomes more nearly symmetric and bell shaped. If np(1 – p) > 10, the probability distribution will be approximately symmetric and bell shaped. Then we can use the normal distribution with μ = np and σ =

7. Correction for Continuity Remember that binomial distributions are for discrete data while normal distributions are for continuous data. Therefore, we must make some allowances in our approximations. Approximate: P(X = a) P(X < a) P(X > a) P(a < X < b)

8. But what about…? P(X a)P(a < X < b) Remember that the data is discrete, so replace each inequality with an inequality/equal.

9. Example According to a study, 6% of all humans have blood type O-negative. What is the probability that, in a simple random sample of 500, fewer than 25 have blood type O-negative? 1)Check that the experiment is binomial. 2)Can the data be approximated with the normal distribution? 3)(Convert to Z-scores.) 4)Use the table or the normalcdf.

10. Continued… 1. Yes, the experiment is binomial. The trials are independent, the probability is the same for each, and there are only 2 possible outcomes. 2. 500(.06)(.94) = 28.2 > 10. 3.Correct for continuity: < 25 = < 24 = < 24.5. μ= 500(.06) = 30 σ = √(500(.06)(.94)) = 5.31 4. normalcdf(-1E99, -1.04, 0, 1) =.1492

11. Exact Probability Remember that this is an APPROXIMATION! The exact probability would be found by binomcdf(500,.06, 24) =.1494 Assignmentpage 3993 – 9 page 4065 - 20