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Section 5.2. The Standard Normal Distribution. The Standard Score. The standard score , or z-score , represents the number of standard deviations a random variable x falls from the mean.
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Section 5.2 The Standard Normal Distribution
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 standard deviation of 7. Find the standard z-score for a person with a score of: (a) 161 (b) 148 (c) 152
4 3 2 1 0 1 2 3 4 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. z
z -3 -2 -1 0 1 2 3 Cumulative Areas 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 • The cumulative area is close to 1 for z scores close to 3.49.
0.1056 z -3 -2 -1 0 1 2 3 Cumulative Areas Find the cumulative area for a z-score of -1.25. Read down the z column on the left to z = -1.2 and across to the column under .05. The value in the cell is0.1056, the cumulative area. The probability that z is at most -1.25 is 0.1056. P ( z -1.25) = 0.1056
z -3 -2 -1 0 1 2 3 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. Find P( z < -1.45) Read down the z-column to -1.4 and across to .05. The cumulative area is 0.0735. P ( z < -1.45) = 0.0735
0.1075 Required area 0.8925 z -3 -2 -1 0 1 2 3 Finding Probabilities To find the probability that z is greater than a given value, subtract the cumulative area in the table from 1. 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. P( z > -1.24) = 0.8925
z -3 -2 -1 0 1 2 3 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) 2. P(z < -1.25) =0.1056 1. P(z < 1.17) = 0.8790 3. P( -1.25 < z < 1.17) = 0.8790 - 0.1056 = 0.7734
To find the probability that z is less than a given value, read the corresponding cumulative area. To find the probability is greater than a given value, subtract the cumulative area in the table from 1. z z 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. -3 -1 0 -1 0 1 2 3 -2 -3 -2 -1 0 1 2 3 -3 -2 1 2 3 Summary
Guided Practice Find the probabilities for each using the standard normal distribution. 1. P(z < -1.25) 2. P(z > -0.5) 3. P(z > 2.5) 4. P(-2.05 < z < 1.85)
Classwork: Practice • Find the probabilities for each using the standard normal • distribution. • 1. P(z < 2.46) 2. P(z > -3.3) • P(z > 1.27) 4. P(-3.05 < z < 2.65) • 5. P(-2.01 < z < 0.84) 6. P(z< -0.9)