5 4 Finding Probabilities for a Normal Distribution

2 Continuous Probability Distribution • The histogram and polygon of the distribution in Table 1 is shown to the right. • As stated in Chapter 2, as the class width gets smaller and the number of classes increases, the polygon becomes a smooth curve. • The smooth curve is an approximation of the probability distribution curve of a continuous random variable.

3 Continuous Probability Distribution A probability distribution of a continuous random variable must satisfy the following two conditions: 1. The probability that x assumes a value in any of the intervals (class) ranges from 0 to 1. In other words, the area under a probability distribution curve of a continuous random variable between any two points is between 0 and 1. 2. The total probability of all intervals within which x can assume a value is 1. 0. That is, the total area under the probability distribution curve of a continuous random variable is always 1. 0.

Properties of a Normal Curve A normal curve is unimodal and symmetric. The mean

Area-Probability Equality Property The area under a normal curve for an interval is equal to the probability of randomly selecting an observation that lies in the interval. The total area under a normal curve is equal to 1. Copyright © 2016 Pearson Education, Inc. Slide 5

Example: Sketching Normal Curves Suppose that a prestatistics class takes a first test and students’ scores are described well by the normal curve shown in the Figure. Copyright © 2016 Pearson Education, Inc. Slide 6

Example: Sketching Normal Curves 1. What is the probability that a randomly selected student scored more than 80 points? 2. On Test 2, the mean score is 20 points less than the mean score on Test 1, but the standard deviation is the same. Assuming the scores are still normally distributed, sketch the curve that describes the scores. 3. On Test 3, the mean score is the same as on Test 2, but the standard deviation is larger. Assuming the scores are still normally distributed, sketch a curve that describes the scores. Copyright © 2016 Pearson Education, Inc. Slide 7

Solution 1. The area to the right of 80, the median, is half of

Solution 2. Because the standard deviation hasn’t changed, the shape and spread of the Test 1 and Test 2 distributions are exactly the same. But because the mean is 20 points lower, we move the curve for Test 1 20 points to the left to get the curve for Test 2 (see Figure). Copyright © 2016 Pearson Education, Inc. Slide 9

Solution 3. Since the means for Test 2 and Test 3 are equal, we do not move the curve for Test 2 left or right (see Fig. ). But because the standard deviation is larger for Test 3, we sketch the curve wider. By drawing the curve wider, we must also make it flatter, because the area under the curve must still equal 1. As a check, we could calculate that each little square in the grids of both Test 2 and Test 3 scores and 50 is 0. 1 (try it). So, 10 little squares should cover the region below each curve, which is true. Copyright © 2016 Pearson Education, Inc. Slide 10

Comparing Normal Curves with Different Standard Deviations A normal curve for a distribution with larger standard deviation will be wider and flatter than a normal curve for a distribution with smaller standard deviation. Copyright © 2016 Pearson Education, Inc. Slide 11

Empirical Rule If a distribution is normal, then the probability that a randomly selected observation lies within • one standard deviation of the mean is approximately 0. 68. So, about 68% of the data lie between • two standard deviations of the mean is approximately 0. 95. So, about 95% of the data lie between • three standard deviations of the mean is approximately 0. 997. So, about 99. 7% of the data lie between Copyright © 2016 Pearson Education, Inc. Slide 12

Empirical Rule Copyright © 2016 Pearson Education, Inc. Slide 13

z-Score The z-score of an observation is the number of standard deviations that the observation is from the mean. If the observation lies to the left of the mean, then its z-score is negative. If the observation lies to the right of the mean, then its z-score is positive. z-Score Formula The z-score of an observation x is given by Copyright © 2016 Pearson Education, Inc. Slide 14

Example: Compute and Interpret a z-Score The 2014 draft picks for NBA basketball teams have heights that are approximately normally distributed with mean 79. 1 inches and standard deviation 3. 0 inches (Source: nbadraft. net). Shabazz Napier was the shortest 2014 draft pick with a height of 72 inches. Find the z-score for 72 inches. What does it mean? Copyright © 2016 Pearson Education, Inc. Slide 15

Solution We substitute 72 for x, 79. 1 for , and 3. 0 for s in the formula: The z-score is – 2. 37, which means that Napier’s height is 2. 37 standard deviations less than the mean. Copyright © 2016 Pearson Education, Inc. Slide 16

z-Scores Because the z-score of an observation is the number of standard deviations that the observation is from the mean, we can restate the Empirical Rule in terms of zscores. So, if a distribution is normally distributed, then 68% of its z-scores lie between – 1 and 1 95% of its z-scores lie between – 2 and 2 99. 7% of its z-scores lie between – 3 and 3 Copyright © 2016 Pearson Education, Inc. Slide 17

Standard Normal Distribution If a distribution is normally distributed, then the distribution of the observations’ z-scores is also normally distributed with mean 0 and standard deviation 1. We call the distribution of z-scores the standard normal distribution. Copyright © 2016 Pearson Education, Inc. Slide 18

Example: Find a Probability for the Standard Normal Distribution Find the probability that a z-score randomly selected from the standard normal distribution is less than – 1. 35. Solution First, we sketch the standard normal curve, which has mean 0 and standard deviation 1. Next, we mark the z -score – 1. 35 on the horizontal axis and use orange to shade the region to the left of – 1. 35. Copyright © 2016 Pearson Education, Inc. Slide 19

Solution First, we sketch the standard normal curve, which has mean 0 and standard deviation 1. Next, we mark the z -score – 1. 35 on the horizontal axis and use orange to shade the region to the left of – 1. 35. Copyright © 2016 Pearson Education, Inc. Slide 20

Solution Then we find the area by using Table 1 in Appendix C. To do this, we write the absolute value of – 1. 35 as 1. 35 = 1. 3 + 0. 05 and identify the row with – 1. 3 in the far left, identify the column with. 05 at the top, and find the number that’s in both that row and column, which is 0. 0885. The number 0. 0885 describes the area to the left of the z-score – 1. 35, which is what we are interested in. Copyright © 2016 Pearson Education, Inc. Slide 21

Example: Using z-Scores and Tables to Find Probabilities Earlier in this section, we worked with the Wechsler IQ test, which measures a person’s intelligence. Recall that people’s IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Let X be the IQ (in points) of a randomly selected person. Find the probability that a randomly selected person has an IQ 1. less than 78 points. 2. greater than 117 points. Copyright © 2016 Pearson Education, Inc. Slide 22

Solution 1. We first sketch a normal curve with mean 100, mark the number 78 on the horizontal axis, and use orange to shade the region to the left of 78. Next calculate the z-score. Copyright © 2016 Pearson Education, Inc. Slide 23

Solution Then we find the area by using Table 1 in Appendix C. To do this, we write the absolute value of – 1. 47 as 1. 47 = 1. 4 + 0. 07 and identify the row with – 1. 4 in the far left, identify the column with. 07 at the top, and find the number that’s in both that row and column, which is 0. 0708. The number 0. 0708 describes the area of the orange region to the left of 78 points, which is what we are interested in. So, the probability is 0. 0708. The probability is less than 0. 5, which checks with the orange region having area less than half of the area under the entire curve. Copyright © 2016 Pearson Education, Inc. Slide 24

Solution 2. We first sketch a normal curve with mean 100, mark the number 117 on the horizontal axis, and use blue to shade the region to the right of 117. Copyright © 2016 Pearson Education, Inc. Slide 25

Solution Then we use Table 1 in Appendix C to find an area. To do this, we write 1. 13 = 1. 1 + 0. 03 and identify the row with 1. 1 in the far left, identify the column with. 03 at the top, and find the number that’s in both that row and column, which is 0. 8708. The number 0. 8708 describes the area (in orange) to the left of 117 points, which is not what we are interested in. Because the area under the entire curve is 1, we can find the area (in blue) to the right of 117 by finding 1 minus 0. 8708: 1 – 0. 8708 = 0. 1292 Copyright © 2016 Pearson Education, Inc. Slide 26

Solution So, the probability is 0. 1292, which we indicate in the figure. The probability is less than 0. 5, which checks with the blue region having area less than half of the area under the entire curve. Copyright © 2016 Pearson Education, Inc. Slide 27

Interpreting the Area under a Normal Curve for an Interval We can interpret the area A under the normal curve for an interval as • the probability that a randomly selected observation will lie in the interval. • the proportion of observations that lie in the interval. Copyright © 2016 Pearson Education, Inc. Slide 28