4 2 continuous probability distributionn
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Here are the steps to solve this problem: a) Find the z-score corresponding to 115 mm Hg: (115 - 85)/13 = 2.31 The proportion that is NOT severely hypertensive is 1 - P(Z >= 2.31) = 1 - 0.0103 = 0.9897 b) Find the z-score corresponding to 90 mm Hg: (90 - 85)/13 = 0.3846 The proportion that will be asked to consult a physician is P(Z >= 0.3846) = 0.6507 c) Find the z-scores corresponding to the mildly hypertensive range: (90 - 85)/13 = 0.3846 (
4 2 continuous probability distributionn
- 2. Recall Continuous probability distribution A continuous probability distribution can assume an infinite number of values within a given range – for variables that take continuous values. The distance students travel to class. The time it takes an executive to drive to work. The length of an afternoon nap. The length of time of a particular phone call. The amount of money spent on your last haircut.
- 4. Normal Probability Distribution The normal probability distribution is the most important distribution for describing a continuous random variable. It has been used in a wide variety of applications: Heights and weights of people Test scores Scientific measurements Amounts of rainfall It is widely used in statistical inference
- 5. Normal Probability Distribution Normal Probability Density Function where: µ = mean σ = standard deviation π = 3.14159 e = 2.71828 22 2/)( 2 1 )( σµ σπ −− = x exf This formula generates the density curve which gives the shape of the normal distribution.
- 6. According to the Germans: “That’s right on the money” 2 2 1 2 1 − − σ µ πσ X e Herr Gauss The Normal distribution is also known as the Gaussian distribution
- 7. “Bell shaped” and Symmetrical about the mean µ. Unimodal and its mode occurs at x = µ. Mean, median and mode are equal Interquartile range equals 1.33 σ Random variable has infinite range from –∞ to ∞. The total area under the curve and above the horizontal axis is equal to 1. Properties of the Normal Distribution µ x f(x) Mean Median Mode
- 8. Most observations in the distribution are close to the mean, with gradually fewer observations further away Properties of the Normal Distribution µ x f(x) Mean Median Mode
- 9. Properties of the Normal Distribution Empirical Rule 68.3 % of all observations lie within 1 standard deviation of the mean 95.4 % of all observations lie within 2 standard deviations of the mean 99.7 % of all observations lie within 3 standard deviations of the mean µ-3σ µ-2σ µ-1σ µ µ+1σ µ+2σ µ+3σ 68.26% 95.45% 99.74% 55 70 14585 100 115 130 µ-3σ µ-2σ µ-1σ µ µ+1σ µ+2σ µ+3σ 55 70 14585 100 115 130 For a normal distribution, the following rules hold true; These rules are commonly known as the Empirical Rule
- 10. Women participating in a three-day experimental diet regime have been demonstrated to have normally distributed weight loss with mean 600 g and a standard deviation 200 g. a) What percentage of these women will have a weight loss between 400 and 800 g? b) What percentage of women will lose weight too quickly on the diet (where too much weight is defined as >1000g)? Application of the Empirical Rule
- 11. X : (600,200) 600 800 1000 12004002000 ~ 68% a)
- 12. X : (600,200) 600 800 1000 12004002000 b) 2.3%
- 13. σ2 2 2 µ x σ1 2 1 µ µ1 < µ2 a. The normal distribution depends on the values of the parameters µ, the population mean and σ2, the population variance. a. Two normal curves, which have the same standard deviation but different means. Properties of the Normal Distribution
- 14. σ2 2 xµ1 = µ2 σ1 2 σ2 2 σ1 2 < b. Two normal curves with the same mean but different standard deviations Properties of the Normal Distribution
- 15. σ2 2 xµ1 < µ2 σ1 2 σ2 2 σ1 2 < µ1 µ2 c. Two normal curves that have different means and different standard deviations. Properties of the Normal Distribution
- 16. Areas under the Normal Curve The area under the curve bounded by the two ordinates x = x1 and x = x2 equals the probability that the random variable x assumes a value between x = x1 and x = x2. Thus, for the normal curve in the Figure below, the P(x1 < x < x2) is represented by the area of the shaded region. µx1 x2
- 17. The area under the curve between any two ordinates depend upon the values of µ and σ and consequently, the probability associated with distributions differ in mean and standard deviation will be different for the two given values of x. Areas under the Normal Curve I II x1 x2 P(x1 < x < x2) for different Normal curves
- 18. Many Normal Distributions By varying the parameters σ and µ, we obtain different normal distributions There are an infinite number of normal distributions
- 19. It would be hopeless task to set up separate tables of normal curve areas for every conceivable value of µ and σ Areas under the Normal Curve
- 20. Which Table to Use? An infinite number of normal distributions means an infinite number of tables to look up! Yet we must use tables if we hope to avoid the use of integral calculus
- 21. P(x1 < x < x2) = ? Finding Probabilities Probability is the area under the curve! x1 x2 X
- 22. Fortunately, all the observations of any normal random variable x could be transformed to a new set of observations of a normal random variable z with mean zero and variance 1, by using the transformation namely, σ µ− = X Z The Standard Normal Distribution
- 23. σ µ z1 σ =1 0x1 z2x2 Any Normal Distribution Standard Normal Distribution Areas will be equal. zx The Standard Normal Distribution The new distribution is called Standard Normal Distribution, with mean equal to 0 and it’s standard deviation equal to 1.
- 24. By standardising any normally distributed random variable, we can use just the table namely, Areas Under the Normal Curve Or Areas of a Standard Normal Distribution Such tables are usually found in the Appendix of any statistics book. Standard Normal Distribution Tables
- 25. Learn how to use the different table
- 27. Entries give the probability that a standard normally distributed random variable will assume a value to the left of a given z value. This means we have to sometimes perform minor calculations to determine probabilities. Use of the Normal Probability Table
- 28. Example 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 Find the probability that z is less than 1.74. 1. Locate a value of z equal to 1.7 in the left column. 2. Move across the row to the column under 0.04, where we read 0.9591. Therefore, P(z < 1.74) = 0.9591. z
- 29. 1. State the problem. 2. What is the appropriate probability statement? 3. Draw a picture and shade required area 4. Convert to a standard normal distribution 5. Find the probability in the standard normal table Computing Normal Probabilities
- 30. Example: Suppose the number of a particular type of bacteria in samples of 1-ml of drinking water tend to be approximately normally distributed with µ =85 and σ2 = 81. What is the probability that a given 1-ml sample will contain more than 100 bacteria?
- 31. σ 85 100 Find this area σ = 9 We are to find the P(x> 100) Sketch a curve To find the P(x> 100), we need to evaluate the area under the normal curve to the right of x = 100. x
- 32. transform x = 100 to the corresponding z value σ µ− = 1 1 x z Since µ = 85 and σ = 981 = 9 85100 − =z = 1.67
- 33. 0 1.67 σ = 1 Find this area z Hence P(x> 100) = P(z > 1.67) Since the area to the right and the table only gives the probability to the left of the distribution we use P(z > 1.67) = 1 − P(z < 1.67)
- 34. Z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 1.5 0.9332 0.9345 0.9357 0.937 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.975 0.9756 0.9761 0.9767
- 35. 0 1.67 σ = 1 Find this area z Hence P(x> 100) = P(z > 1.67) Since the area to the right and the table only gives the probability to the left of the distribution we use P(z > 1.67) = 1 − P(z < 1.67) = 1 − 0.9525 = 0.0475. So 1-ml sample will has 0.0475 probability to contain more than 100 bacteria
- 36. Exercise If the total cholesterol values for a certain target population are approximately normally distributed with a mean of 200 (mg/100 mL) and a standard deviation of 20 (mg/100 mL), What is the probability that a person picked at random from this population will have a cholesterol value greater than 240 (mg/100 mL)? Answer = 0.0228
- 37. Example 2 IQ scores (IQ is Intelligence Quotient) on the Wechsler Adult Intelligence Scale are approximately normally distributed with mean µ = 100 and σ = 15. a. What is the proportion of persons having IQs between 80 and 120?
- 38. Solution Let x1 = 80 and x2 = 120 This question means that we are to find the probability P(x1<IQ<x2) = P(80< IQ <120) Sketch a normal curve like the one in the figure below. Shade in the area desired. σ =15 10080 120 x Example 2 cont.
- 39. Find the z values corresponding to x1 = 80 and x2 = 120 Example 2 cont. Since, P(x1<x<x2) =P (z1<z<z2) Therefore, P(80<IQ<120) = P(−1.33<z<1.33)
- 40. The P(−1.33<z<1.33) is given by the area of the shaded region in the figure. This area may be found by subtracting the area to the left of the ordinate z = −1.33 from the entire area to the left of z = 1.33. Example 2 cont. σ = 1 0-1.33 1.33 z
- 41. P(80<IQ<120) = P(−1.33<z<1.33) = P(z<1.33) − P(z<−1.33) = 0.9082 – 0.0918 = 0.8164. Therefore the proportion of persons having IQs between 80 and 120 is 0.8164, about 82%. σ =1 0-1.33 1.33 z
- 42. Example 3 Suppose that the hemoglobin levels for healthy adult males are approximately normally distributed with a mean of 16 and variance of 0.81. Find the probability that a randomly chosen healthy adult male has a hemoglobin level less than 14. Solution We are to find P(x < 14) Since µ = 16 and σ = = 0.9 We find Z = − 2.22 P(x < 14) = P(Z < − 2.22) = 0.0132 Therefore, only about 1.3% of healthy adult males have σ=0.81 σ= 1 16 0 14 -2.22 x z
- 44. . Inverse Normal Distribution Finding x value when probabilities (areas) are given
- 45. 1. State the problem 2. Draw a picture 3. Use table to find the probability closest to the one you need 4. Read off the z-value 5. Unstandardise i.e. x = µ + zσ Finding a Value (X) given a Probability
- 46. Example Given a normal distribution with µ = 40 and σ = 6, find the value of x that has a. 45% of the area below it b. 14% of the area above it. Solution In this problem we reverse the process and begin with a known area or probability, find the z value, and then determine x by rearranging the formula to give σ µ− = x z µσ += zx
- 48. a. We require a z value that leaves an area of 0.45 to the left. From the table we find P(z < -0.13) = 0.45 so that the desired z value is -0.13. σ=6 40 x 0.45 µσ += zx = (-0.13 X 6 )+ 40 = 39.22 x =? Example cont.
- 49. b. 14% of the area above it. This time we require a z value that leaves 0.14 of the area to the right and hence an area of 0.86 to the left. From the table we find P(z < 1.08) = 0.86 so that the desired z value is 1.08 and x = (6)(1.08) + 40 = 46.48 σ =6 x =? 0.86 40 µσ += zx Example cont. 0.14
- 51. Classification of arterial diastolic blood pressure (mm Hg) in adults, 18 years and older Consider that blood pressure readings are obtained from nearly 200,000 participants in a a large-scale community blood pressure screening program, and that these measurements follow a normal distribution. The mean is 85 mm Hg, with a standard deviation of 13 mm Hg. Diastolic blood pressure (mm Hg) Blood Pressure Classification Less than 85 Normal 85-89 High normal 90-104 Mild hypertension 105-114 Moderate hypertension Greater or equal to 115 Severe hypertension Exercise
- 52. a) What proportion of our sample will NOT be categorised as severely hypertensive? b) Suppose that we recommend that a physician be consulted if an individual has an arterial diastolic blood pressure equal or greater than 90 mm Hg. What proportion of individuals in our screening program will be asked to consult a physician? c) What proportion of individuals have diastolic blood pressure in the mildly hypertensive range? d) What diastolic blood pressure will 75% of the population be above? Exercise
- 53. EXERCISE (a) The distribution of serum levels of alpha tocopherol (serum vitamin E) is approximately normal with mean 860 mg/dL and standard deviation 340mg/dL. What serum level will 85% of the population be below? (b) Suppose a person is identified as having toxic levels of alpha tocopherol if his or her serum level is greater than 2000mg/dL. What percentage of people will be so identified?
- 54. Under certain conditions, a binomial random variable has a distribution that is approximately normal. Normal approximation of the Binomial Distribution
- 55. If n, p, and q are such that: np and nq are both greater than 5. We use a normal distribution with qpnandpn =σ=µ Normal approximation of the Binomial Distribution
- 56. Application of Normal Distribution If 22% of all patients with high blood pressure have side effects from a certain medication, and 100 patients are treated, find the probability that at least 30 of them will have side effects. Using the Binomial Probability Formula we would need to compute: P(30) + P(31) + ... + P(100) or 1 − P( x < 29)
- 57. Using the Normal Approximation to the Binomial Distribution Check if Is it appropriate to use the normal distribution? n p = 22 n q = 78 Both are greater than five.
- 58. Find the mean and standard deviation µ = n p = 100(.22) = 22 and σ = 14.416.17 )78)(.22(.100 = = qpnandpn =σ=µ Using the Normal Approximation to the Binomial Distribution
- 59. σ 22 30 Find this area To find the probability that at least 30 of them will have side effects, find P( x ≥ 30) σ = 4.14 Using the Normal Approximation to the Binomial Distribution
- 60. Applying the Normal Distribution z = 30 – 22 = 1.81 4.14 Find P( z ≥ 1.81) = 1 – P (z < 1.81) = 0.0351 0 1.81 The probability that at least 30 of the patients will have side effects is 0.0351. σσ = 1