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Statistics Chapter 12

Statistics Chapter 12. The Normal Distribution: A Problem-Solving Tool Section 12.4. Properties of a Normal Curve. The normal curve is a bell-shaped curve with the following properties: It is smooth and symmetric. Its highest point occurs over the mean μ of the entire population.

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Statistics Chapter 12

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  1. StatisticsChapter 12

  2. The Normal Distribution: A Problem-Solving ToolSection 12.4

  3. Properties of a Normal Curve The normal curve is a bell-shapedcurve with the following properties: It is smooth and symmetric. Its highest point occurs over themean μ of the entire population. It never touches the x axis. The total area under the curve is 1.

  4. z-Score (Standardized Score) • If x is a given score and μ and σ are themean and the standard deviation of the entire set of scores, then the correspondingz-score is The z-score gives the number of standard deviations that x is from the mean.

  5. The following slide is a z-score table with four place accuracy. The Text book supplies a three place accuracy table at the back of the textbook on the inside of the hard cover. Some of the problems however give answers with four decimal places. So this table may be helpful to you.

  6. Example Farmer Brown has planted a field of experimental corn, and it is estimated that there are 20,000 plants. The mean height of farmer Brown’s corn is 100 inches and the standard deviation is 10 inches. a. What percent of the cornstalks is between 90 and 110 inches tall? b. What percent of the cornstalks is more than 120 inches tall? c. What percent of the cornstalks is less than 90 inches tall? d. About how many stalks are less than 90 inches tall?

  7. Solution 0.341 0.341 a. When z=-1, the area is 0.341. When z=1, the area is also 0.341. Therefore, the percent between 90 and 110 inches is 0.682 or 68.2%. 0.023 b. When z=2, the area is 0.477. 0.477 The desired area is 0.500 – 0.477 = 0.023. The percent that is more than 120 inches is 0.023 or 2.3%.

  8. Solution Continued 0.341 c. When z = -1, the area is 0.341. 0.159 The desired area is 0.500 – 0.341 = 0.159. The percent that is less than 90 inches is 0.159 or 15.9%. d. If the percent of cornstalks that is less than 90 inches is 15.9%, then 15.9% of 20,000 is 3180 cornstalks.

  9. Example For a certain standardized placement test, it was found that the scores were normally distributed, with a mean of 200 and a standard deviation of 30. Suppose that this test is given to 1000 students. a.How many are expected to make scoresbetween 170 and 230?b. How many are expected to score above 260?

  10. Solution 0.341 0.341 a. When z=-1, the area is 0.341. When z=1, the area is also 0.341. The percent between 170 and 230 is 0.682 or 68.2%. Therefore, 68.2% of 1000 is 682 students. b. When z=2, the area is 0.477. 0.477 The desired area is 0.500 – 0.477 = 0.023. The percent that is more than 260 is 0.023 or 2.3%. Therefore, 2.3% of 1000 is 23 students. 0.023

  11. Example An express workout at the gym is normally distributed with a mean of 30 minutes and a standard deviation of 5 minutes. What is the probability that Latasha completes the workout in 30 to 35 minutes? more than 25 minutes? less than 20 minutes? Between 33 and 37 minutes?

  12. Solutions 0.341 When z = 1, the area is 0.341. The probability Latasha will complete her workout in 30 to 35 minutes is 0.341. 0.500 0.341 When z = -1, the area is 0.341. The probability Latasha will complete her workout in more than 25 minutes is 0.341 + 0.500 = 0.841. 0.841 0.477 When z = -2, the area is 0.477. The probability Latasha will complete her workout in less than 20 minutes is0.500 – 0.477 = 0.023. 0.023 0.500

  13. Solution Continued When z = 0.60, the area is 0.226. 0.226 0.419 When z = 1.40, the area is 0.419. The desired area is 0.419 – 0.226 = 0.193. 0.193 The probability that Latasha takes between 33 to 37 minutes to finish her work out is 0.193.

  14. Example Suppose you are the manager of a cereal packing company, and you know that the weights of the boxes have a normal distribution with a mean of 16.15 ounces and a standard deviation of 0.05 ounces. What is the weight of a box with a z-score of 0? 1?

  15. Solution

  16. Percentile Kim took a test in a class of 52 students, and only 4 students scored better than she did. What was her percentile? Joanne took a test in a class of 43 students, and 28 of the students scored less than she did. What was her percentile? The scores in a class were as follows: 76, 48, 53, 77, 72, 64, 95, 78, 84, 86. What percentile corresponds to a score of 72? END

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