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The Standard Normal Distribution Section 5.2. The Standard Score The standard score, or z-score, represents the number of standard deviations a random.

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Presentation on theme: "The Standard Normal Distribution Section 5.2. The Standard Score The standard score, or z-score, represents the number of standard deviations a random."— Presentation transcript:

1 The Standard Normal Distribution Section 5.2

2 The Standard Score The standard score, or z-score, represents the number of standard deviations a random variable x falls from the mean. The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the standard z-score for a person with a score of: (a) 161 (b) 148 (c) 152 (a)(b)(c)

3 The Standard Normal Distribution The standard normal distribution has a mean of 0 and a standard deviation of 1. Using z-scores any normal distribution can be transformed into the standard normal distribution. –4–3–2–101234 z

4 Cumulative Areas The cumulative area is close to 1 for z-scores close to 3.49. 0123–1–2–3 z The total area under the curve is one. The cumulative area is close to 0 for z-scores close to –3.49. The cumulative area for z = 0 is 0.5000.

5 Find the cumulative area for a z-score of –1.25. 0123–1–2–3 z Cumulative Areas 0.1056 Read down the z column on the left to z = –1.25 and across to the column under.05. The value in the cell is 0.1056, the cumulative area. The probability that z is at most –1.25 is 0.1056.

6 Finding Probabilities To find the probability that z is less than a given value, read the cumulative area in the table corresponding to that z-score. 0123–1–2–3 z Read down the z-column to –1.4 and across to.05. The cumulative area is 0.0735. Find P(z < –1.45). P (z < –1.45) = 0.0735

7 Finding Probabilities To find the probability that z is greater than a given value, subtract the cumulative area in the table from 1. 0123–1–2–3 z P(z > –1.24) = 0.8925 Find P(z > –1.24). The cumulative area (area to the left) is 0.1075. So the area to the right is 1 – 0.1075 = 0.8925. 0.1075 0.8925

8 Finding Probabilities To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger. Find P(–1.25 < z < 1.17). 1. P(z < 1.17) = 0.8790 2. P(z < –1.25) = 0.1056 3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734 0123–1–2–3 z

9 0 123-2-3 z Summary 0123-2-3 z To find the probability is greater than a given value, subtract the cumulative area in the table from 1. 0123-2-3 z To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger. To find the probability that z is less than a given value, read the corresponding cumulative area.

10 Homework :1-15 all pgs. 232-233 Day 2: 16-30 all pgs. 233-235


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