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IS 310 Business Statistics CSU Long Beach

IS 310 Business Statistics CSU Long Beach. Continuous Probability Distributions. A continuous random variable can assume any value in an interval on the real line or in a collection of intervals.

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IS 310 Business Statistics CSU Long Beach

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  1. IS 310 Business Statistics CSU Long Beach

  2. Continuous Probability Distributions • A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. • It is not possible to talk about the probability of the random variable assuming a particular value. • Instead, we talk about the probability of the random variable assuming a value within a given interval.

  3. Exponential f (x) Uniform f (x) Normal f (x) x x1 x2 x1 x2 x x1 x2 x x1 x2 Continuous Probability Distributions • The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the graph of the probability density function between x1and x2.

  4. Normal Probability Distribution • The normal probability distribution is the most important distribution for describing a continuous random variable. • It is widely used in statistical inference.

  5. Normal Probability Distribution • It has been used in a wide variety of applications: Heights of people Scientific measurements

  6. Normal Probability Distribution • It has been used in a wide variety of applications: Test scores • Amounts • of rainfall

  7.  = mean  = standard deviation  = 3.14159 e = 2.71828 Normal Probability Distribution • Normal Probability Density Function where:

  8. Normal Probability Distribution • Characteristics The distribution is symmetric; its skewness measure is zero. x

  9. Normal Probability Distribution • Characteristics The entire family of normal probability distributions is defined by itsmeanm and its standard deviations . Standard Deviation s x Mean m

  10. Normal Probability Distribution • Characteristics The highest point on the normal curve is at the mean, which is also the median and mode. x

  11. Normal Probability Distribution • Characteristics The mean can be any numerical value: negative, zero, or positive. x -10 0 20

  12. Normal Probability Distribution • Characteristics The standard deviation determines the width of the curve: larger values result in wider, flatter curves. s = 15 s = 25 x

  13. Normal Probability Distribution • Characteristics Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and .5 to the right). .5 .5 x

  14. Normal Probability Distribution Since the area under the curve represents probability, the probability of a normal random variable at one specific value is zero . With a single value, one can’t find the area since the area must be bound by two values. Thus, P(x = 10) = 0 P(x = 3) = 0 P(x = 7.5) = 0 However, one can find the following probabilities: P( 1 < x < 3) P(2.2 < x < 3.7) P(x > 3)

  15. of values of a normal random variable are within of its mean. 68.26% +/- 1 standard deviation of values of a normal random variable are within of its mean. 95.44% +/- 2 standard deviations of values of a normal random variable are within of its mean. 99.72% +/- 3 standard deviations Normal Probability Distribution • Characteristics

  16. 99.72% 95.44% 68.26% Normal Probability Distribution • Characteristics x m m + 3s m – 3s m – 1s m + 1s m – 2s m + 2s

  17. Normal Probability Distribution There may be thousands of normal distribution curves, each with a different mean and a different standard deviation. Since the shapes are different, the areas under the curves between any two points are also different. To make life easier, all normal distributions can be converted to a standard normal distribution. A standard normal distribution has a mean of 0 and a standard deviation of 1.

  18. Standard Normal Probability Distribution The letter z is used to designate the standard normal random variable. s = 1 z 0

  19. Standard Normal Probability Distribution • Converting to the Standard Normal Distribution requires the use • of this formula

  20. Applications of Standard Normal Distribution Example Problem Test scores of a special examination administered to all potential employees of a firm are normally distributed with a mean of 500 points and a standard deviation of 100 points. What is the probability that a score selected at random will be higher than 700? P(x > 700) = ? If we convert this normal variable, x, to a standard normal variable, z, z = (x - µ) / σ = (700 – 500) / 100 = 2 --------------500----------700 x-scale P(x > 700) = P(z > 2) ----------------0-----------2 z-scale

  21. Standard Normal Probability Distribution The area under the normal curve between any two points represents the probability. Refer to pages 235 through 238 (10th edition) or pages 242-245 (11th edition) of your text book and get a clear understanding of how to calculate probabilities . Be proficient in using Table 1 of Appendix B (10-Pages 918 and 919; 11-Pages 978-979)

  22. Sample Problems Problem # 10 (10-Page 241; 11-Page 248) Using Table 1 (10-Page 919; 11-Page 979) • P(z ≤ 1.5) = 0.9332 • P(z ≤ 1.0) = 0.8413 • P(1 ≤ z ≤ 1.5) = 0.9332 – 0.8413 = 0.09 • P(0 < z < 2.5) = 0.9938 – 0.5000 = 0.4938

  23. More Sample Problems Problem # 11 (10-Page 241; 11-Page 248) • P(z ≤ - 1.0) = 0.1587 • P(z ≥ - 1) = 1 – P(z ≤ - 1) = 1 – 0.1587 = 0.8413 • P(z ≥ - 1.5) = 1 – P(z ≤ - 1.5) = 1 – 0.0668 = 0.9332 • Do it yourself • P(- 3 < z ≤ 0) = 0.5 – 0.0013 = 0. 4987

  24. More Sample Problem • Problem # 15 (10-Page 242; 11-Page 249) • Let’s do this problem in class!

  25. More Sample Problems Problem # 20 (10-Page 242; 11-Page 249) Given: µ = 77 σ = 20 • P(x < 50) = ? Convert to z: z = (x - µ) / σ = (50 – 77) / 20 = - 1.35 P(x < 50) = P(z < - 1.35) = 0.0885 • P(x > 100) = ? z = (100 – 77) / 20 = 1.15 P(x > 100) = P(z > 1.15) = 1 – P(z ≤ 1.15) = 1 – 0.8749 = 0.1251 or 12.51 %

  26. Continuation of Sample Problem • x = ? to be considered a heavy user Upper 20% of the area is in the right tail of the normal curve. 80% of the area is to the left. Go to Table 1 and locate 0.8 (or 80%) as the table entry. The closest entry is 0.7995. That point represents a z-value of 0.84. Use this value of z in the following equation: z = (x - µ) / σ 0.84 = (x – 77)/ 20 x = 93.8 hours

  27. Sample Problems The service life of a certain brand of automobile battery is normally distributed with a mean of 1000 days and a standard deviation of 100 days. The manufacturer of the battery wants to offer a guarantee, but does not know the length of the warranty. It does not want to replace more than 10 percent of the batteries sold. What should be the length of the warranty? z = (x - µ) / σ - 1.2817 = (x – 1000) / 100 x = 871.83 or 872 days

  28. More Sample Problems A statistics instructor grades on a curve. He does not want to give more than 15 percent A in his class. If test scores of students in statistics are normally distributed with a mean of 75 and a standard deviation of 10, what should be the cut-off point for an A? z = (x - µ) / σ 1.04 = (x – 75) / 10 x = 85.4 or 85

  29. End of Chapter 6

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