Normal Curve with Standard Deviation |  + or - one s.d.  |

Figure 6-5 (p. 170) The distribution for Example 6.2. Example 6.2 What is the probability of selecting someone from this population with an SAT score greater then 700? Or p(X>700) = ? Shaded area in figure 6.5 is the proportion of SAT score above 700. Translate to z scores, z = (X – μ) ∕ σ z = ( )/100 = 2 So we are looking for p(z>2.00) = ? Go back and look at normal curve, what proportion is greater then z of 2.00? Only 2.28% of the population so p(z>2.00) = 2.28%.

Figure 6-13 (p. 181) Determining probabilities or proportions for a normal distribution is shown as a two-step process with z-scores as an intermediate stop along the way. Note that you cannot move directly along the dashed line between X values and probabilities or proportions. Instead, you must follow the solid lines around the corner.

Figure 6-6 (p. 171) A portion of the unit normal table, see Appendix B page 699. This table lists proportions of the normal distribution corresponding to each z-score value. Column A of the table lists z-scores. Column B lists the proportion in the body of the normal distribution up to the z-score value. Column C lists the proportion of the normal distribution that is located in the tail of the distribution beyond the z-score value. Column D lists the proportion between the mean and the z-score value.

Figure 6.7 Proportions of a normal distribution corresponding to z = and z = Unit Normal table divided into Body and Tail. +

Figure 6-8 (p. 175) The distributions for Example 6.3A–6.3C. Finding proportions for specific Z-scores. Example 6.3A p(z > 1.00) = 15.87% Look up z score in column A then find proportion in column C Example 6.3B p(z < 1.50) = 93.32% Look up z score in column A then find proportion in column B Example 6.3C p(z < -0.50) = 30.85% Look up z score in column A then find proportion in column C

Figure 6-9 (p. 176) The distributions for Examples 6.4A and 6.4B. Finding Z-scores from proportions Example 6.4A Find z score for top ten percent of distribution, locate in column C then use the closest z score from column A, z = 1.28 Example 6.4b Find z scores for the middle 60 % of the distribution. Locate in column D then use the closest z score from column A, z = 0.84 and z =

Figure 6-10 (p. 178) The distribution of IQ scores. The problem is to find the probability of proportion of the distribution corresponding to scores less than 120. Example 6.5 What is the probability of selecting someone with an IQ of less then 120? p(X<120) = ? Convert to z-scores, z = (X – μ) ∕ σ z = ( )/15 = look up z of in unit normal table We want the area below 120 so find Proportion in column B p(X<120) = p(z<1.33) = or 90.82%

Figure 6-11 (p. 180) The distribution for Example 6.6. What is the proportion of cars traveling between 55 and 65 miles per hour? p(55 65) = ? Covert scores to z-scores, find z of and 0.70 in Column A, find proportion in column D for each z score, because D gives proportion above and below mean add proportions = % Z-score of -.30 is Z-score of 0.70 is

Figure 6-12 (p. 181) The distribution for Example 6.7. Example 6.7 What proportion of cars are traveling between 65 and 75? p(65< X <75) = ? OR p(0.70< z < 1.70) p(0.70< z) = from column D p(z < 1.70) = from column D Subtract proportion for 65 from proportion for = %

Converting proportions to z-score Example 6.8, how much commuting time in the top 10% ? Look up 0.10 proportion in column C (or closest value) find corresponding z score in column A z = Then convert z-sore to X value X = μ + zσ X = (10) so X = 37.1 which is the cutoff for top 10%

Example 6.9 Find the range of commuting times for the middle 90% Middle 90% is 45% above and below the mean. Look up in column D (or closest score) find z score of 1.65 in column A convert z score to a X value X = μ + zσ X = (10) X = 40.8 But Wait!! You Are Not Done, this is only the upper score X = (10) X = 7.8 Because we are looking for the range of scores for middle 90% 90 % is between 7.8 and 40.8

Probability and the Binomial Distribution Binomial distributions are formed by a series of observations (for example, 100 coin tosses) for which there are exactly two possible outcomes (heads and tails). The two outcomes are identified as A and B, with probabilities of p(A) = p and p(B) = q. For example the probability of getting heads or tails when flipping a coin. –Probability of getting heads: p = p(heads) = ½ –Probability of getting tails: q = p(tails) = ½ –p + q is always equal to 1.00

Figure 6-16 (p. 185) The binomial distribution showing the probability for the number of heads in 2 tosses of a balanced coin.

Figure 6-17 (p. 186) Binomial distributions showing probabilities for the number of heads (a) in 4 tosses of a balanced coin and (b) in 6 tosses of a balanced coin.

Figure 6-18 (p. 187) The relationship between the binomial distribution and the normal distribution. The binomial distribution is always a discrete histogram, and the normal distribution is a continuous smooth curve. Each X value is represented by a bar in the histogram or a section of the normal distribution.

Probability and the Binomial Distribution Binomial distributions with a large number of outcomes closely approximated a normal distribution. When pn and qn are both greater than 10 –mean of μ = pn –standard deviation of σ =  npq. In this situation, a z-score can be computed for each value of X and the unit normal table can be used to determine probabilities for specific outcomes.

Figure 6-19 page 188 The binomial distribution (normal approximation) for example 6.11, predicting the suit of a card The probability of guessing the correct suit is p = ¼. The number of correct predictions from a series of 48 trials. The shaded area corresponds to the probability of guessing correctly more than 15 times. Note that the score X = 15 is defined by its real limits, so the upper real limit is n = 48, p = ¼ q = ¾ pn = ¼(48) =12 qn = ¾(48) =36 pn =  = 12 pnq =  = 3 z = (X – μ) ∕ σ Z = (15.5 – 12)/3 = 1.17 Look up Z = 1.17 p = So probability of predicting the correct suit more than 15 times in 48 trials is 12.10%.

Figure 6-20 (p. 190) A diagram of a research study. One individual is selected from the population and receives a treatment. The goal is to determine whether or not the treatment has an effect.

Figure 6-21 (p. 191) Using probability to evaluate a treatment effect. Values that are extremely unlikely to be obtained from the original population are viewed as evidence of a treatment effect.

Probability and Inferential Statistics Probability is important because it establishes a link between samples and populations. For any known population it is possible to determine the probability of obtaining any specific sample. In the Statistics course you will use this link as the foundation for inferential statistics.