C. The Normal Distribution

Transcription

1 C The Normal Distribution Probability distributions of continuous random variables Probability distributions of discrete random variables consist of a function that matches all possible outcomes of a random phenomenon with their associated probabilities Continuous random variables have probability distributions as well, but because there are an infinite number of outcomes, a definite probability cannot be matched with any particular outcome Probability distributions for continuous random variables are usually defined with a mathematical formula or density curve Probabilities are determined for ranges of outcomes by finding the area under the curve Probability distributions defined for continuous random variables using density curves must meet the same criteria as discrete random variables The key is that the sum of all probabilities must be 1 With continuous random variables the total area under the curve is 1 The probability that the continuous random variable X lies between any two values is equal the area under the curve between the two values In the two continuous probability density functions below, P 5 X 10 ( ) equals 05 and 075, respectively X X Properties of the normal distribution The Normal Distribution is a continuous probability distribution that appears in many situations, both natural and man-made It is also, as will later be described, the basis for many of the statistical inference procedures covered by the statistics curriculum The Normal distribution has several important properties: (1) It is perfectly symmetric, unimodal, and bell-shaped In fact, it is where the term bell curve comes from (2) The curve continues infinitely in both directions, and is asymptotic to the horizontal axis as it approaches ± (3) It can be defined by only two parameters: mean µ and standard deviation (4) The distribution is centered at the mean µ (5) The points of inflection, where the curve changes from curving downward to curving upward, occur at exactly ±1 Normal Curves Page 1 of 9

2 (6) The total area under the curve equals 1 Continues to infinity Total Area = 1 Points of Inflection Continues to infinity µ Probabilities of outcomes from random phenomena that have a Normal distribution are computed by finding the area under the curve These areas are determined by first determining how many standard deviations a point lies from the mean Regardless of the value of the mean and standard deviation, all Normal distributions have the same area between given points The Rule (sometimes called the Empirical Rule) gives some benchmarks for understanding how probability is distributed under a normal curve If a continuous random variable has a Normal distribution, approximately 68% of all outcomes will be within 1 standard deviation of the mean About 95% of all outcomes are within 2 standard deviations of the mean and 997% are within 3 standard deviations of the mean Therefore, only 03% about one outcome in 330 will be more than 3 standard deviations from the mean 68% of area 95% of area 997% of area µ Many phenomena are normally distributed, or at least are very close to being so For instance, American adult males have heights that are approximately distributed with a mean of µ = 173 cm and a standard deviation of = 75 cm Normal Curves Page 2 of 9

3 The probability distribution for males would look like this: µ height (cm) If an adult male were randomly selected from the population, it would not be surprising to find one of 173 cm or even 180 cm However, finding a male with a height of 188 cm or more looks unlikely Example: American adult males have heights that are approximately distributed with a mean of µ = 173 cm and a standard deviation of = 75 cm What percent of males have heights between 1655 cm and 1805 cm? Answer: Using the Rule, about 68% of all outcomes lie within one standard deviation of the mean Since 1655 cm is 1 standard deviation below the mean of 173 cm and 1805 cm is 1 standard deviation above, about 68% of all males have heights between 1655 cm and 1805 cm Example: American adult males have heights that are approximately distributed with a mean of µ = 173 cm and a standard deviation of = 75 cm What percent of males have heights over 188 cm? Answer Using the Rule, about 95% of all outcomes lie within two standard deviations of the mean This means that about 5% fall below 2 standard deviations and above 2 standard deviations combined Thus, about 25% would be above 2 standard deviations from the mean The height 188 cm is 2 standard deviations above the mean of 173 cm, so roughly 25% of all males have heights above 188 cm Example: A particular college entrance exam has two parts: math and verbal The distribution of math scores is Normal with a mean of 500 and a standard deviation of 100 The middle 997% of all test scores will be between roughly what two values? Normal Curves Page 3 of 9

4 Answer Using the Rule, about 997% of all outcomes lie within three standard deviations of the mean The standard deviation of the distribution is 100, so three standard deviations is 300 The middle 997% of all scores will lie within 300 of the mean, or between 200 and 800 The Rule only gives benchmarks values for the area under the Normal distribution at whole numbers of standard deviations from the mean If locations other than whole standard deviations are desired, then Table A: Standard Normal Probabilities is required Determining probabilities from the Table A requires one to compute a standard score or z-score The z-score is simply the number of standard deviations an observation lies from the mean The z-score is computed by z = x µ, where x is the value of the observed outcome Example: American adult males have heights that are approximately distributed with a mean of µ = 173 cm and a standard deviation of = 75 cm What is the z-score for a male of height 183 cm? Answer: The z-score for a male of height 183 cm is z = x µ = = The table of Standard Normal probabilities gives the probability of observing an outcome below the given z-score The table is read by cross-referencing the row of the table containing the whole number and tenths place of the z-score with the column holding the hundredths place Normal Curves Page 4 of 9

5 Example: What is the value of the Standard Normal probability table (Table A) for a z-score of 133? Answer: Cross-reference the 13 row of the table with the 03 column z The value of the Standard Normal probability table (Table A) for a z-score of 133 is Example: American adult males have heights that are approximately distributed with a mean of µ = 173 cm and a standard deviation of = 75 cm What percent of adult males have heights less than 183 cm? Answer: Since Table A gives the gives the probability of observing an outcome below the given z-score, and the z-score for 183 cm is z = 133, about 9082% of adult males have heights of less than 183 cm z 133 x = 183 cm z-score height x Note: Using probability notation, the above solution would be written: P( x < 183cm) P( z < 133) Normal Curves Page 5 of 9

6 Note: Because a definite probability cannot be matched with any particular outcome, only a range of outcomes, the chance of observing a male with height of less than 183 cm is the same as the chance of observing a male with a height of 183 cm or less That is, P( x < 183cm)= P( x 183cm) Example: A particular college entrance exam has two parts: math and verbal The distribution of math scores is Normal with a mean of 500 and a standard deviation of 100 An advanced summer math program requires a score of at least 625 to participate What proportion of students is eligible to participate in the program? Answer: First, the z-score must be computed z = x µ = = 125 Cross indexing 100 row 12 with column 05 on Table A, the result is 8944 The table gives the proportion below a score of 625, so the proportion of students with scores of at least 625 is = Example: American adult females have heights that are normally distributed with a mean of 161 cm and a standard deviation of 65 cm A company is seeking women of a certain height to operate a particular vehicle Women cannot be shorter than 146 cm (because they will not be able to reach the controls) or taller than 173 cm (because they hit their heads on the roof) What percent of women could hold this job? Answer: The z-scores for 146 cm and 173 cm are z = 231 and z = 185, respectively Table A shows that of females are below 146 cm and of females are below 173 cm Therefore, the proportion of females between 146 cm and 173 cm is = Almost 96% of women could operate this vehicle z 231 x = 146 cm z 133 x = 173 cm z-score height x Normal Curves Page 6 of 9

7 It is possible to determine outcomes (or mean or standard deviation) given a probability To do this, one first has to read Table A in reverse That is, find the probability in the center of the table and read horizontally and vertically to the margins Example: What z-score corresponds to a Table A value of 08907? z Answer: Since the probability is greater than 05, the z-score must be positive Locating and reading backward gives a z-value of 123 Once a z-score is determined, unknown values in the equation z = x µ can be determined Example: A particular college entrance exam has two parts: math and verbal The distribution of verbal scores is Normal with a mean of 500 and a standard deviation of 100 What is the verbal score of a student who scores in the 89 th percentile? Answer The 89 th percentile is the point where 89% of outcomes are at or below that point The probability in the table that is closest to 089 is 08907, which corresponds to a z-score of 123 Solving z = x µ for the unknown x, x = µ + z = ( 100)= 623 Thus, a score of 623 corresponds to the 89 th percentile Normal Curves Page 7 of 9

8 Example: American adult males have heights that are approximately distributed with a mean of µ = 173 cm and a standard deviation of = 75 cm Between what two heights is the middle 90% of all males? Answer: Considering the middle 90% of males leaves 10% of the males in the two tails of the distribution, or 5% in each distribution Finding the area in Table A closest to and reading its corresponding z-score gives z = 1645 This is the z-score that cuts off the lowest 5% of the distribution By symmetry, the top 5% will be cut off by z = 1645 Solving Š1645 = x 173 and 1645 = x 173 for x, gives values of x = 1607 cm and x = 1853 cm respectively The middle 90% of male heights are between those two values Trial Run 1 The diameters of a certain type of ball bearing are approximately normally distributed with a mean of 220 cm and a standard deviation of 002 cm The largest 1% of all ball bearings will have diameters greater than (A) 215 cm (B) 220 cm (C) 222 cm (D) 225 cm (E) 229 cm 2 A standardized test has scores that normally distributed with a mean of 680 and a standard deviation of 25 Approximately what proportion of scores is between 650 and 720? (A) 011 (B) 017 (C) 038 (D) 083 (E) The distribution of heights of students at 20 high schools in a Midwestern city follows an approximate normal distribution Twenty percent of the students are less than 55 inches and 10% of the students are more than 62 inches What are the mean and standard deviation of the distribution of heights of high school students in this Midwestern city? (A) mean = 578, standard deviation = 330 (B) mean = 580, standard deviation = 357 (C) mean = 580, standard deviation = 476 (D) mean = 590, standard deviation = 234 (E) mean = 590, standard deviation = 313 Normal Curves Page 8 of 9

9 Trial Run Solutions 1 D The z-value that cuts off an area of 001 in the right tail of a Normal distribution, using Table A, is 233 Given z = x µ x 220, solve 233 = for x; x = D The z-scores for 650 and 720 are z = = 120 and z = = 160, respectively The area of the Normal distribution to the left of z = 120 is and the area of the Normal distribution to the left of z = 120 is The area between the two z-values is = A The lowest 20% of the Normal distribution is cut off by z = 084, thus 084 = 55 µ The highest 10% is cut off by z = 128, thus 128 = 62 µ Solving each for µ gives µ = and µ = Setting them equal to each other = and solving for gives 212 = Normal Curves Page 9 of 9