Normal Probability Distribution

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1 Normal Probability Distribution The Normal Distribution functions: #1: normalpdf pdf = Probability Density Function This function returns the probability of a single value of the random variable x. Use this to graph a normal curve. Using this function returns the y-coordinates of the normal curve. Syntax: normalpdf (x, mean, standard deviation) #2: normalcdf cdf = Cumulative Distribution Function This function returns the cumulative probability from zero up to some input value of the random variable x. Technically, it returns the percentage of area under a continuous distribution curve from negative infinity to the x. You can, however, set the lower bound. Syntax: normalcdf (lower bound, upper bound, mean, standard deviation) #3: invnorm( inv = Inverse Normal Probability Distribution Function This function returns the x-value given the probability region to the left of the x-value. (0 < area < 1 must be true.) The inverse normal probability distribution function will find the precise value at a given percent based upon the mean and standard deviation. Syntax: invnorm (probability, mean, standard deviation) Example 1: Given a normal distribution of values for which the mean is 70 and the standard deviation is 4.5. Find: a) the probability that a value is between 65 and 80, inclusive. b) the probability that a value is greater than or equal to 75. c) the probability that a value is less than 62. d) the 90 th percentile for this distribution. (answers will be rounded to the nearest thousandth) 1a: Find the probability that a value is between 65 and 80, inclusive. (This is accomplished by finding the probability of the cumulative interval from 65 to 80.) Syntax:normalcdf(lower bound, upper bound, mean, standard deviation) : The probability is 1 P a g e

2 1b: Find the probability that a value is greater than or equal to 75. (The upper boundary in this problem will be positive infinity. The largest value the calculator can handle is 1 x Type 1 EE 99. Enter the EE by pressing 2nd, comma -- only one E will show on the screen.) : The probability is 1c: Find the probability that a value is less than 62. (The lower boundary in this problem will be negative infinity. The smallest value the calculator can handle is -1 x Type -1 EE 99. Enter the EE by pressing 2nd, comma -- only one E will show on the screen.) : The probability is 1d: Find the 90 th percentile for this distribution. (Given a probability region to the left of a value (i.e., a percentile), determine the value using invnorm.) : The x-value is Example 3: Graph and examine a situation where the mean score is 46 and the standard deviation is 8.5 for a normally distributed set of data. Go to Y=. Adjust the window. GRAPH. 2 P a g e

3 1. Practice and Homework (Independent # odd and HW even #) The amount of mustard dispensed from a machine at The Hotdog Emporium is normally distributedwith a mean of 0.9 ounce and a standard deviation of 0.1 ounce. If the machine is used 500 times, approximately how many times will it be expected to dispense 1 or more ounces of mustard. Choose: Professor Halen has 184 students in his college mathematics lecture class. The scores on the midterm exam are normally distributed with a mean of 72.3 and a standard deviation of 8.9. How many students in the class can be expected to receive a score between 82 and 90? Express answer to the nearest student. 3. A machine is used to fill soda bottles. The amount of soda dispensed into each bottle varies slightly. Suppose the amount of soda dispensed into the bottles is normally distributed. If at least 99% of the bottles must have between 585 and 595 milliliters of soda, find the greatest standard deviation, to the nearest hundredth, that can be allowed. 4. Residents of upstate New York are accustomed to large amounts of snow with snowfalls often exceeding 6 inches in one day. In one city, such snowfalls were recorded for two seasons and are as follows (in inches): 8.6, 9.5, 14.1, 11.5, 7.0, 8.4, 9.0, 6.7, 21.5, 7.7, 6.8, 6.1, 8.5, 14.4, 6.1, 8.0, 9.2, P a g e

4 What are the mean and the population standard deviation for this data, to the nearest hundredth? 5. Neesha's scores in Chemistry this semester were rather inconsistent: 100, 85, 55, 95, 75, 100. For this population, how many scores are within one standard deviation of the mean? 6. Battery lifetime is normally distributed for large samples. The mean lifetime is 500 days and the standard deviation is 61 days. To the nearest percent, what percent of batteries have lifetimes longer than 561 days? 7. The number of children of each of the first 41 United States presidents is given in the accompanying table. For this population, determine the mean and the standard deviation to the nearest tenth. How many of these presidents fall within one standard deviation of the mean? 8. 4 P a g e From 1984 to 1995, the winning scores for a golf tournament were 276, 279, 279, 277, 278, 278, 280, 282, 285, 272, 279, and 278. Using the standard deviation for

5 this sample, Sx, find the percent of these winning scores that fall within one standard deviation of the mean. 9. A shoe manufacturer collected data regarding men's shoe sizes and found that the distribution of sizes exactly fits the normal curve. If the mean shoe size is 11 and the standard deviation is 1.5, find: a. the probability that a man's shoe size is greater than or equal to 11. b. the probability that a man's shoe size is greater than or equal to c. 10. Five hundred values are normally distributed with a mean of 125 and a standard deviation of 10. a. What percent of the values lies in the interval , to the nearest percent? b. What percent of the values is in the interval , to the nearest percent? c. What interval about the mean includes 95% of the data? d. What interval about the mean includes 50% of the data? 5 P a g e

6 11. A group of 625 students has a mean age of 15.8 years with a standard deviation of 0.6 years. The ages are normally distributed. How many students are younger than 16.2 years? Express answer to the nearest student? Binomial Probability "Exactly", "At Most", "At Least" Example 4 Problem used for demonstration: A fair coin is tossed 100 times. What is the probability that: a. heads will appear exactly 52 times? b. there will be at most 52 heads? c. there will be at least48 heads? Dealing with "Exactly": A fair coin is tossed 100 times. What is the probability that: part a. heads will appear exactly 52 times? We have seen that the formula used with Bernoulli trials (binomial probability) computes 6 P a g e

7 the probability of obtaining exactly "r" events in "n" trials: where n = number of trials, r = number of specific events you wish to obtain, p = probability that the event will occur, and q = probability that the event will not occur (q = 1 - p, the complement of the event). We have also seen that the built-in command binompdf(binomial probability density function) can also be used to quickly determine "exactly". (Remember, the function binompdf is found under DISTR (2nd VARS), arrow down to #0 binompdf and the parameters are: binompdf (number of trials, probability of occurrence, number of specific events) binompdf (n, p, r) Here is our answer to part a. Dealing with"at Most": A fair coin is tossed 100 times. What is the probability that: part b. there will be at most 52 heads? The formula needed for answering part b is : There is a built-in command binomcdf(binomial cumulative density function) that can be used to quickly determine "at most". Because this is a "cumulative" function, it will find the sum of all of the probabilities up to, and including, the given value of P a g e (The function binomcdf is found under DISTR (2nd VARS), arrow down to #A binomcdf

8 and the parameters are: binomcdf (number of trials, probability of occurrence, number of specific events) binomcdf (n, p, r) Here is our answer to part b. Dealing with "At Least": A fair coin is tossed 100 times. What is the probability that: part c. there will be at least48 heads? The formula needed for answering part b is : Keep in mind that "at least" 48 is the complement of "at most" 47. In a binomial distribution,. While there is no built-in command for "at least", you can quickly find the result by creating this complement situation by subtracting from 1. Just remember to adjust the value to 47. The adjusted formula for "at least" is 1 - binomcdf (n, p, r - 1).Here is our answer to part c. 8 P a g e

9 The fact that this answer is the same as the "at most" answer for the number 52, is due to the symmetric nature of the distribution about its mean of 50. Practice (independent odd # HW even #) 1. The probability that Kyla will score above a 90 on a mathematics test is 4/5. What is the probability that she will score above a 90 on exactly three of the four tests this quarter? Choose: 2. Which fraction represents the probability of obtaining exactly eight heads in ten tosses of a fair coin? Choose: 45/ / / / Experience has shown that 1/200 of all CDs produced by a certain machine are defective. If a 9 P a g e

10 quality control technician randomly tests twenty CDs, compute each of the following probabilities: P(exactly one is defective) P(half are defective) P( no more than two are defective) 4. A fair coin is tossed 5 times. What is the probability that it lands tails up exactly 3 times? Choose: 5. After studying a couple's family history, a doctor determines that the probability of any child born to this couple having a gene for disease X is 1 out of 4. If the couple has three children, what is the probability that exactly two of the children have the gene for disease X? 6. If a binomial experiment has seven trials in which the probability of success is p and the probability of failure is q, write an expression that could be used to compute each of the following probabilities: P(exactly five successes) P(at least five successes) 10 P a g e P(at most five successes) 7. On any given day, the probability that the entire Watson family eats dinner together is 2/5. Find the probability that, during any 7-day period, the Watson's each dinner together at least six times.

11 8. When Joe bowls, he can get a strike (knock down all of the pins) 60% of the time. How many times more likely is it for Joe to bowl at least three strikes out of four times as it is for him to bowl zero strikcs out of four tries? Round answer to the nearest whole number. 9. A board game has a spinner on a circle that has five equal sectors, numbered 1, 2, 3, 4, and 5, respectively. If a player has four spins, find the probability that the player spins an even number no more than two times on those four spins. 10. Give an example of an experiment where it is appropriate to use a normal distribution as an approximation for a binomial probability. Explain why in this example an approximation of the probability is a better approach than finding the exact probability. Key 11 P a g e

12 3. The 99% implies a distribution within 3 standard deviations of the mean. The difference from 585 milliliters to 595 milliliters is 10 milliliters. Symmetrically divided, there are 5 milliliters used to create 3 standard deviations on one side of the mean. Dividing 5 by 3, we get the standard deviation to be 1.67 milliliters, to the nearest hundredth. 4. Snowfall: 8.6, 9.5, 14.1, 11.5, 7.0, 8.4, 9.0, 6.7, 21.5, 7.7, 6.8, 6.1, 8.5, 14.4, 6.1, 8.0, 9.2, 7.1 The mean = 9.46 inches. Standard deviation = Scores: 100, 85, 55, 95, 75, 100. From the graphing calculator we have the mean = 85 and the population standard deviation = = = 101 and = 69 All but one of the scores falls in the range from 69 to 101. : 5 scores 12 P a g e

13 8. Scores: 276, 279, 279, 277, 278, 278, 280, 282, 285, 272, 279, and 278. Mean = Sample Standard Deviation = One standard deviation above and below the mean creates a range from to (275 < score < 281). Of the 12 scores, 9 fall into this range, which is 75% of the scores. 7. Use your graphing calculator. Be sure to remember that you are working with grouped data (frequency table). Mean = 3.6 Standard deviation = 2.9 One standard deviation from the mean creates a range from 0.7 to 6.5 which includes presidents with from 1 to 6 children. There are a total of 31 such presidents. 13 P a g e

14 9. Mean = 11 and standard deviation = 1.5 a. 50% In a normal distribution, the mean divides the data into two equal areas. Since 11 is the mean, 50% of the data is above 11 and 50% is below 11. b is exactly one standard deviation above the mean. Examining the normal distribution chart shows that 15.9%will fall above one standard deviation. Probability is.159. c and 8 are exactly one standard deviation above the mean and 2 standard deviations below the mean respectively. Using the chart we know: scores, mean 125, standard deviation 10 a. What percent of the values is in the interval ? mean + one standard deviation = 135 mean - one standard deviation = % % % + 15% = 68.2% (from chart) Percent within one standard deviation of the mean = 68.2% = 68% b. What percent of the values is in the interval ? mean standard deviations = 150 mean standard deviations = 100 2(1.7% + 4.4% + 9.2% + 15% %) = 98.8% (from chart) Percent with 2.5 standard deviations of the mean = 98.8% = 99% c. What interval about the mean includes approximately 95% of the data? 2 standard deviations about the mean for a total interval size of 40, with the mean in the center. mean + 2 standard deviations = 145 mean - 2 standard deviations = 105 Interval: [105,145] d. What interval about the mean includes 50% of the data?.50% of the distribution lies within standard deviations about the mean for a total interquartile range (size) of , with the mean in the center. mean standard deviations = mean standard deviations = Interval: [ , ] 14 P a g e

15 Binomial 3. a.) P(exactly one is defective) b.) P(half are defective) = P(exactly 10 are defective) c.) P( no more than two are defective) = P( at most two are defective). None, one, or two could be defective. Sum = which is approximately 1 or 100% P a g e

16 6. a.) b.) c.) 7. At least means 6 times or 7 times. 8. At least three strikes: Exactly zero strikes: 19 times more likely. 16 P a g e

17 9. There are 2 even numbers out of the 5 numbers. No more than = at most. 10. One possible answer: Find the probability of getting at most 250 heads when flipping a fair coin 300 times. When dealing with very large samples, it can become very tedious to compute certain probabilities. In such cases, the normal distribution can be used to more quickly approximate the probabilities that otherwise would have been obtained through laborious computations. 17 P a g e

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