Section 7.1 The Standard Normal Curve

Section 7.1 The Standard Normal Curve Objectives: 1. Use a probability density curve to describe a population 2. Use a normal curve to describe a normal population 3. Find areas under the standard normal curve 4. Find z-scores corresponding to areas under the normal curve The curve used to describe the distribution of a continuous random variable is called the probability density curve of the random variable. The probability density curve tells us what proportion of the population falls within any given interval. The area under a probability density curve between two values is equal to the proportion of the population that falls between the two values.  A probability density curve represents the probability distribution of a continuous variable.  The area under the entire curve is equal to 1.  The area under the curve between two values a and b is: a) the proportion of the population whose values are between a and b. b) the probability that a randomly selected individual will have a value between a and b. EXAMPLE 1: Following is a probability density curve for a population. a. What proportion of the population is between 4 and 6? b. If a value is chosen at random from this population, what is the probability that it will be between 4 and 6? c. What proportion of the population is not between 4 and 6? d. If a value is chosen at random from this population, what is the probability that it is not between 4 and 6? Solution a. The proportion of the population between 4 and 6 is equal to the area under the curve between 4 and 6, which is 0.16. b. The probability that a randomly chosen value is between 4 and 6 is equal to the area under the curve between 4 and 6, which is 0.16. c. The area under the entire curve is equal to 1. Therefore, the proportion that is not between 4 and 6 is equal to 1 − 0.16 = 0.84. d. The probability that a randomly chosen value is not between 4 and 6 is equal to the area under the curve that is not between 4 and 6, which is 0.84. Another way to answer part (d) is to use the Rule of Complements: P(Not between 4 and 6) = 1 − P(Between 4 and 6) = 1 − 0.16 = 0.84 Example #2 1. Following is a probability density curve with the area between 0 and 1 and the area between 1 and 2 indicated. What proportion of the population is between 0 and 1? a. b. What is the probability that a randomly selected value will be between 1 and 2? c. What proportion of the population is between 0 and 2? d. What is the probability that a randomly selected value will be greater than 2? Solution: A) 0.63 B) 0.23 C) 0.86 D) 0.14 Properties of Normal Distributions 1. Normal distributions have one mode. 2. Normal distributions are symmetric around the mode. 3. The mean and median of a normal distribution are both equal to the mode. In other words, the mean, median, and mode of a normal distribution are all the same. Finding Areas under the Standard Normal Curve Objective 3 Find areas under the standard normal curve The standard normal distribution= has mean 0 and standard deviation 1. SUMMARY    Points on the horizontal axis to the left of the mode have negative z-scores. Points on the horizontal axis to the right of the mode have positive z-scores. The mode itself has a z-score of 0. To find an area under the standard normal curve by using Table A.2: Step 1 Graph a normal curve, label the z-score or scores, and shade in the area to be found. Step 2: Look up the area in Table A.2 corresponding to each z-score. Step3  To find the area to the left of a z-score, use the area in Table A.2.  To find the area to the right of a z-score, find the area to the left  and subtract from 1. To find the area between two z-scores, find the area to the left of each, then subtract the smaller area from the larger area Example #1 Finding an area to the left of a z-score Use Table A.2 to find the area to the left of z = 1.26. Solution: Step 1: Sketch a normal curve, label the point z = 1.26, and shade in the area to the left of it. Note that z = 1.26 is located to the right of the mode, since it is positive. Step 2: To look up z = 1.26, find the row containing 1.2 and the column containing 0.06. The value in the intersection of the row and column is 0.8962. This is the area to the left of z = 1.26 Example #2 Find the area to the right of z = −0.58. Solution Step 1: Graph a normal curve, label the point z = −0.58, and shade in the area to the right of it. Note that z = −0.58 is located to the left of the mode, since it is negative. Step 2: To look up z = −0.58, find the row containing −0.5 and the column containing 0.08. The value in the intersection of the row and column is 0.2810. This is the area to the left of z = −0.58. To find the area to the right, we subtract from 1 Finding an area between two z-scores Example #3 Find the area between z = −1.45 and z = 0.42. Solution Step 1: Sketch a normal curve, label the points z = −1.45 and z = 0.42, and shade in the area between them. See Figure 7.9. Step 2: Use Table A.2 to find the areas to the left of z = −1.45 and to the left of z = 0.42. The area to the left of z = −1.45 is 0.0735, and the area to the left of z = 0.42 is 0.6628. Step 3: Subtract the smaller area from the larger area to find the area between the two z-scores: Solution We present results from the TI-84 Plus calculator. The area is found by using the normalcdf command. We enter the left endpoint of the interval (−1.45), the right endpoint (0.42), the mean (0), and the standard deviation (1). Step-by-step instructions are given in the Using Technology section on page 303. The TI-84 Plus gives the area as .5892279372. Rounding to four decimal places gives 0.5892. example 7.6 Find the area to the right of a z-score by using technology In Example 7.3, we found the area to the right of z = −0.58. Find this area by using technology. Solution We present output from the TI-84 Plus calculator. We use the normalcdf command. We enter the left endpoint of the interval (−0.58). Since there is no right endpoint, we enter 1E99, which represents the very large number that is written with a 1 followed by 99 zeros. Then we enter the mean (0) and the standard deviation (1). Step-by-step instructions are given in the Using Technology section on page 303. The TI-84 Plus gives the area as .7190427366. Rounding to four decimal places gives 0.7190. Finding the z-score corresponding to an area to the left Example #4: Use Table A.2 to find the z-score that has an area of 0.26 to its left. Solution In Table A.2, look through the body of the table to find the area closest to 0.26. This value is 0.2611. The values in the left-hand column and top row corresponding to 0.2611 are −0.6 and 0.04. Therefore, the z-score is z = −0.64. . Example #5: Find the z-score that has an area of 0.2170 Extra credit**** Finding a z-score corresponding to an area: In Table A.2, the numbers in the body of the table represent areas, and numbers down the left-hand column and across the top row represent zscores. When given an area, we find that area in the body of the table, and look at the left-hand column and top row to obtain the z-score. Example 7.8 Using technology to find the z-score corresponding to an area to the left In Example 7.7, we found the z-score that has an area of 0.26 to its left. Use technology to find this z-score. Solution We present results from the TI-84 Plus calculator. The z-score is found by using the invNorm command. We enter the area to the left (.26), the mean (0), and the standard deviation (1). Stepby-step instructions are given in the Using Technology section on page 303. The TI-84 Plus gives z = −.6433454021. Rounding to two decimal places gives z = −0.64. Section 7.2 Applications of the Normal Distribution Objectives 1. Convert values from a normal distribution to z-scores 2. Find areas under a normal curve 3. Find the value from a normal distribution corresponding to a given proportion DEFINITION Let x be a value from a normal distribution with mean μ and standard deviation σ. The z-score of x is Properties of the z-Score 1. The z-score follows a standard normal distribution. 2. Values below the mean have negative z-scores, and values above the mean have positive z-scores. 3. The z-score tells how many standard deviations the original value is above or below the mean. Finding and interpreting a z-score Example Heights in a certain population of women follow a normal distribution with mean μ = 64 inches and standard deviation σ = 3 inches. a. A randomly selected woman has a height of x = 67 inches. Find and interpret the z-score of this value. b. Another randomly selected woman has a height of x = 63 inches. Find and interpret the z-score of this value. Solution a. The z-score for x = 67 is We interpret this by saying that a height of 67 inches is 1 standard deviation above the mean height of 64 inches. b. The z-score for x = 63 is We interpret this by saying that a height of 63 inches is 0.33 standard deviation below the mean height of 64 inches. Example #2 A normal distribution has mean μ = 15 and standard deviation σ = 4. Find and interpret the z-score for x = 11. Answer: z= -1 A value of 11 is one standard deviation below the mean