1. 1 Chapter 6 Continuous Probability Distributions

2. 2 Chapter Outline  Uniform Probability Distribution  Normal Probability Distribution f ( x ) x x Uniform x Normal

3. 3 Continuous Probability Distributions  A continuous random variable assume values that have no gap or jump between them.  Since between any two values, a continuous random variable assumes infinite number of values, the probability that any particular value occurs is zero.  Therefore, we study the probability of the random variable assuming a value within a given interval.

4. 4 Continuous Probability Distributions  The probability of the random variable assuming a value within some given interval from x 1 to x 2 is defined to be the area under the line of the probability density function between x 1 and x 2. f ( x ) x x Uniform x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2 x Normal x1 x1x1 x1 x1 x1x1 x1 x2 x2x2 x2 x2 x2x2 x2

5. 5 Uniform Probability Distribution  A random variable is uniformly distributed whenever the probability is proportional to the interval’s length.  The uniform probability density function is: f ( x ) = 1/( b – a ) for a < x < b f ( x ) = 1/( b – a ) for a < x < b = 0 elsewhere = 0 elsewhere f ( x ) = 1/( b – a ) for a < x < b f ( x ) = 1/( b – a ) for a < x < b = 0 elsewhere = 0 elsewhere where: a = smallest value the variable can assume b = largest value the variable can assume b = largest value the variable can assume

6. 6 Uniform Probability Distribution Var( x ) = ( b - a ) 2 /12 E( x ) = ( a + b )/2 Expected Value of x Variance of x

7. 7 Uniform Probability Distribution  Example: Slater's Buffet Slater customers are charged for the amount of Slater customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces.

8. 8 Uniform Probability Distribution  Example: Slater's Buffet  Uniform Probability Density Function f ( x ) = 1/10 for 5 < x < 15 f ( x ) = 1/10 for 5 < x < 15 = 0 elsewhere = 0 elsewhere f ( x ) = 1/10 for 5 < x < 15 f ( x ) = 1/10 for 5 < x < 15 = 0 elsewhere = 0 elsewhere where: x = salad plate filling weight x = salad plate filling weight

9. 9 Uniform Probability Distribution  Expected Value of x E( x ) = ( a + b )/2 E( x ) = ( a + b )/2 = (5 + 15)/2 = (5 + 15)/2 = 10 = 10 E( x ) = ( a + b )/2 E( x ) = ( a + b )/2 = (5 + 15)/2 = (5 + 15)/2 = 10 = 10  Variance of x Var( x ) = ( b - a ) 2 /12 Var( x ) = ( b - a ) 2 /12 = (15 – 5) 2 /12 = (15 – 5) 2 /12 = 8.33 = 8.33 Var( x ) = ( b - a ) 2 /12 Var( x ) = ( b - a ) 2 /12 = (15 – 5) 2 /12 = (15 – 5) 2 /12 = 8.33 = 8.33

10. 10 Uniform Probability Distribution  Uniform Probability Distribution for Salad Plate Filling Weight x x 1/10 Salad Weight (oz.) 5 5 10 15 0 0 f(x)f(x) f(x)f(x)

11. 11 Uniform Probability Distribution What is the probability that a customer will take between 12 and 15 ounces of salad? x x 1/10 Salad Weight (oz.) 5 5 10 15 0 0 f(x)f(x) f(x)f(x) P(12 < x < 15) = (1/10)(3) =.3 12

12. Area as A Measure of Probability  The area under the graph of f ( x ) is simply the measure of probability.  This is valid for all continuous random variables.  The probability that x takes on a value between some lower value x 1 and some higher value x 2 can be found by computing the area under the graph of f ( x ) over the interval from x 1 to x 2.  The overall area under the graph of f ( x ) over the value range of x is 1 or 100%.

13. 13 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.  It has been used in describing a wide variety of real-life applications including: Heights of people Heights of people Rainfall amounts Rainfall amounts Test scores Test scores Scientific measurements Scientific measurements

14. 14 Normal Probability Distribution  Normal Probability Density Function  = mean  = standard deviation  = 3.14159 e = 2.71828 where:

15. 15 Normal Probability Distribution  Characteristics The distribution is symmetric; its skewness The distribution is symmetric; its skewness measure is zero. measure is zero. The distribution is symmetric; its skewness The distribution is symmetric; its skewness measure is zero. measure is zero. x

16. 16 Normal Probability Distribution  Characteristics The entire family of normal probability The entire family of normal probability distributions is defined by its mean  and its standard deviation . standard deviation . The entire family of normal probability The entire family of normal probability distributions is defined by its mean  and its standard deviation . standard deviation . x  Mean  Standard Deviation 

17. 17 Normal Probability Distribution  Characteristics The highest point on the normal curve is at the The highest point on the normal curve is at the mean, which is also the median and mode. mean, which is also the median and mode. The highest point on the normal curve is at the The highest point on the normal curve is at the mean, which is also the median and mode. mean, which is also the median and mode. x  Mean 

18. 18 Normal Probability Distribution  Characteristics The mean can be any numerical value. When only The mean can be any numerical value. When only the mean (the central location) changes, the whole the mean (the central location) changes, the whole normal curve simply shifts horizontally. normal curve simply shifts horizontally. The mean can be any numerical value. When only The mean can be any numerical value. When only the mean (the central location) changes, the whole the mean (the central location) changes, the whole normal curve simply shifts horizontally. normal curve simply shifts horizontally. x -8024

19. 19 Normal Probability Distribution  Characteristics The standard deviation determines the width of the curve. A smaller  results in a narrower, taller curve. The standard deviation determines the width of the curve. A smaller  results in a narrower, taller curve. x  = 5  = 12

20. 20 Normal Probability Distribution  Characteristics Probabilities for the normal random variable are given by areas under the curve. The total area under curve is 1 (.5 to the left of the mean and.5 to the right..5 to the right. Probabilities for the normal random variable are given by areas under the curve. The total area under curve is 1 (.5 to the left of the mean and.5 to the right..5 to the right. x  Mean .5.5

21. 21 Normal Probability Distribution  Characteristics (basis for the empirical rule) Expected number of correct answers of values of a normal random variable of values of a normal random variable are between  -  and  + . are between  -  and  + . of values of a normal random variable of values of a normal random variable are between  -  and  + . are between  -  and  + . 68.26%68.26% of values of a normal random variable of values of a normal random variable are between  - 2  and  + 2 . are between  - 2  and  + 2 . of values of a normal random variable of values of a normal random variable are between  - 2  and  + 2 . are between  - 2  and  + 2 . 95.44%95.44% of values of a normal random variable of values of a normal random variable are between  - 3  and  + 3 . are between  - 3  and  + 3 . of values of a normal random variable of values of a normal random variable are between  - 3  and  + 3 . are between  - 3  and  + 3 . 99.72%99.72%

22. 22 Normal Probability Distribution  Characteristics (basis for the empirical rule) Expected number of correct answers x  – 3   – 1   – 2   + 1   + 2   + 3   68.26% 95.44% 99.72%

23. 23 Standard Normal Probability Distribution  Characteristics (basis for the empirical rule) Expected number of correct answers A normally distributed random variable A normally distributed random variable with a mean of 0 and a standard deviation of 1 is with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability said to have a standard normal probability distribution. Any other normal distribution can be distribution. Any other normal distribution can be converted to the standard normal distribution. converted to the standard normal distribution. A normally distributed random variable A normally distributed random variable with a mean of 0 and a standard deviation of 1 is with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability said to have a standard normal probability distribution. Any other normal distribution can be distribution. Any other normal distribution can be converted to the standard normal distribution. converted to the standard normal distribution.

24. 24 Standard Normal Probability Distribution  Characteristics Expected number of correct answers The letter z is used to designate the stand normal The letter z is used to designate the stand normal random variable. random variable. The letter z is used to designate the stand normal The letter z is used to designate the stand normal random variable. random variable. z 0  

25. 25 Standard Normal Probability Distribution  Converting to the Standard Normal Distribution 0 We can think of z as a measure of the number of standard deviations x is from .

26. 26 Standard Normal Probability Distribution  Example: What is the probability that z  1.58? 01.58 z Area of z  1.58 is the probability.

27. 27 Standard Normal Probability Distribution  Example: What is the probability that z  1.58?  Cumulative Probability Table for the Standard Normal Distribution: it provides the areas (probabilities) to the LEFT of the z values. the z values. z.00.01.02.03.04.05.06.07.08.09........... 1.5.9332.9345.9357.9370.9382.9394.9406.9418.9429.9441 1.6.9452.9463.9474.9484.9495.9505.9515.9525.9535.9545 1.7.9554.9564.9573.9582.9591.9599.9608.9616.9625.9633 1.8.9641.9649.9656.9664.9671.9678.9686.9693.9699.9706 1.9.9713.9719.9726.9732.9738.9744.9750.9756.9761.9767........... P (z  1.58) =.9429

28. 28 Standard Normal Probability Distribution  Example: What is the probability that z  1.58? 01.58 z Area =.9429 Area = 1-.9429 =.0571

29. 29 Standard Normal Probability Distribution  Example: Computer Prices The average price of personal computers manufactured by Company APO is $1,000 with a standard deviation of $150. Furthermore, it is known that the computer prices manufactured by APO are normally distributed. The average price of personal computers manufactured by Company APO is $1,000 with a standard deviation of $150. Furthermore, it is known that the computer prices manufactured by APO are normally distributed. a. What is the probability that a randomly selected computer will have a price of at least $1125? a. What is the probability that a randomly selected computer will have a price of at least $1125? P ( x  1125) = ? P ( x  1125) = ?

30. 30 Standard Normal Probability Distribution  Example: Computer Prices Step 1: Convert x to the standard normal distribution. z = ( x -  )/  z = ( x -  )/  = (1125 – 1000)/150 = (1125 – 1000)/150 =.83 =.83 z = ( x -  )/  z = ( x -  )/  = (1125 – 1000)/150 = (1125 – 1000)/150 =.83 =.83 Step 2: Find the area under the standard normal curve to the left of z =.83. curve to the left of z =.83. Step 2: Find the area under the standard normal curve to the left of z =.83. curve to the left of z =.83. see next slide see next slide

31. 31 Standard Normal Probability Distribution  Example: Computer Prices  Cumulative Probability Table for the Standard Normal Distribution: P ( z <.83)

32. 32 Standard Normal Probability Distribution  Example: Computer Prices Step 3: Compute the area under the standard normal curve to the right of z =.83. curve to the right of z =.83. Step 3: Compute the area under the standard normal curve to the right of z =.83. curve to the right of z =.83. P ( z .83) = 1 – P ( z <.83) P ( z .83) = 1 – P ( z <.83) = 1-.7967 = 1-.7967 =.2033 =.2033 P ( z .83) = 1 – P ( z <.83) P ( z .83) = 1 – P ( z <.83) = 1-.7967 = 1-.7967 =.2033 =.2033 P ( x  1125)

33. 33 Standard Normal Probability Distribution  Example: Computer Prices 0.83 Area =.7967 Area = 1 -.7967 =.2033 =.2033 z

34. 34 Standard Normal Probability Distribution  Example: Computer Prices b. To attract buyers like college students, Company APO decides to provide promotions on the cheaper computers. What is the maximum price such that only 15% of all the computers are eligible for a promotion sale? ---------------------------------------------------------------------- (Hint: Given a probability (area), we can use the standard normal table in an inverse fashion to find the corresponding z value.)

35. 35 Standard Normal Probability Distribution  Example: Computer Prices z Area =.15 0 z.85 z.15

36. 36 Standard Normal Probability Distribution  Example: Computer Prices Step 1: Determine whether the z value is negative or positive. The question is asking for the maximum computer price such that the cheapest computers (15% of all the computers) can enjoy a promotion sale. It is translated as: <price max ) =.15 P (x <price max ) =.15 Therefore, the z value should be negative and it cuts off an area of.15 in the left tail of the standard normal distribution. Step 1: Determine whether the z value is negative or positive. The question is asking for the maximum computer price such that the cheapest computers (15% of all the computers) can enjoy a promotion sale. It is translated as: <price max ) =.15 P (x <price max ) =.15 Therefore, the z value should be negative and it cuts off an area of.15 in the left tail of the standard normal distribution.

37. 37 Standard Normal Probability Distribution  Example: Computer Prices Step 2: Find the z-value. The following table shows the positive z-values. To find the negative z-values, we can utilize the symmetric feature of the Normal Distributions. Step 2: Find the z-value. The following table shows the positive z-values. To find the negative z-values, we can utilize the symmetric feature of the Normal Distributions. The closest z-value that cuts off the left area of.85 is 1.04. Therefore, the z-value we look for is –1.04.

38. 38 Standard Normal Probability Distribution  Example: Computer Prices Step 3: Convert z.15 to the corresponding value of x. x =  + z.15  x =  + z.15   = 1000 + (-1.04)(150) = 844 = 844 x =  + z.15  x =  + z.15   = 1000 + (-1.04)(150) = 844 = 844 So, the maximum price at which a computer can enjoy So, the maximum price at which a computer can enjoy a promotion sale is $844, and there are about 15% of all the computers manufactured by APO are eligible for the promotion sale.

39. 39 Standard Normal Probability Distribution  Example: Computer Prices z Area =.15 1000 844  = 150