Understanding Normal Distribution and Probabilities

1. Section 5.3 Normal Distributions Finding Probabilities

2. Probabilities and Normal Distributions If a random variable, x is normally distributed, the probability that x will fall within an interval is equal to the area under the curve in the interval. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability that a person selected at random will have an IQ score less than 115. 100 115 To find the area in this interval, first find the standard score equivalent to x = 115.

3. Probabilities and Normal Distributions Normal Distribution Find P(x < 115). 100 115 Standard Normal Distribution SAME SAME Find P(z < 1). 1 0 P(z < 1) = 0.8413, so P(x <115) = 0.8413

4. Application Normal Distribution Monthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. A utility bill is randomly selected. Find the probability it is between $80 and $115. P(80 < x < 115) P(–1.67 < z < 1.25) 0.8944 – 0.0475 = 0.8469 The probability a utility bill is between $80 and $115 is 0.8469.

5. From Areas to z-Scores Find the z-score corresponding to a cumulative area of 0.9803. z = 2.06 corresponds roughly to the 98th percentile. 0.9803 –4 –3 –2 –1 0 1 2 3 4 z Locate 0.9803 in the area portion of the table. Read the values at the beginning of the corresponding row and at the top of the column. The z-score is 2.06.

6. Finding z-Scores from Areas Find the z-score corresponding to the 90th percentile. .90 z 0 The closest table area is .8997. The row heading is 1.2 and column heading is .08. This corresponds to z = 1.28. A z-score of 1.28 corresponds to the 90th percentile.

7. Finding z-Scores from Areas Find the z-score with an area of .60 falling to its right. .40 .60 z 0 z With .60 to the right, cumulative area is .40. The closest area is .4013. The row heading is 0.2 and column heading is .05. The z-score is 0.25. A z-score of 0.25 has an area of .60 to its right. It also corresponds to the 40th percentile

8. 0 Finding z-Scores from Areas Find the z-score such that 45% of the area under the curve falls between –z and z. .275 .275 .45 –z z The area remaining in the tails is .55. Half this area is in each tail, so since .55/2 = .275 is the cumulative area for the negative z value and .275 + .45 = .725 is the cumulative area for the positive z. The closest table area is .2743 and the z-score is 0.60. The positive z score is 0.60.

9. From z-Scores to Raw Scores To find the data value, x when given a standard score, z: The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard score of: (a) 2.33 (b) –1.75 (c) 0 (a) x = 152 + (2.33)(7) = 168.31 (b) x = 152 + (–1.75)(7) = 139.75 (c) x = 152 + (0)(7) = 152

10. Finding Percentiles or Cut-off Values Monthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. What is the smallest utility bill that can be in the top 10% of the bills? $115.36 is the smallest value for the top 10%. 90% 10% z Find the cumulative area in the table that is closest to 0.9000 (the 90th percentile.) The area 0.8997 corresponds to a z-score of 1.28. To find the corresponding x-value, use x = 100 + 1.28(12) = 115.36.

11. Homework : 1-37 (odd) pgs. 242-243 42-46 even pgs.243-244. Day 2:Homework : 2-36 (even) pgs. 242-243 41-45 odd pgs.243-244.