Complement: 0.4 x 0.8 = =.6

Transcription

1 Homework The Normal Distribution Name: 1. Use the graph below 1 a) Why is the total area under this curve equal to 1? Rectangle; A = LW A = 1(1) = 1 b) What percent of the observations lie above 0.8? =.2; A = LW A = 1(.2) =.2; 20% c) What percent of the observations lie between 0.25 and 0.75? =.5; A = LW A = 1(.5) =.5; 50% d) What is the mean of this distribution? μ =.5 2. Use the graph below 3 2 a) Verify that the graph is a valid density curve. All on or above horizontal axis; Rectangle + Triangle1(. 8) + 1 (1). 4) = b) The median of this density curve is a point between X = 0.2 and X = 0.4. Explain why. 0 x.2 =.35 and. 4 x.8 =.4;.5 has to be between.2 and.4 c) Find the proportion of observations within each interval i 0.6 < x < 0.8 ii 0 < x < 0.4 iii 0 < x < =.2 1(. 2) =.2 Complement: 0.4 x 0.8 = =.6 Trapezoid: A = 1 (. 2)( ) = ) The sketch below contains three normal curves; think of them as approximating the distribution of exam scores for three different classes. One (call it A) has a mean of 70 and a standard deviation of 5; another (call it B) has a mean of 70 and a standard deviation of 10; the third (call it C) has a mean of 50 and a standard deviation of 10. Identify which is which by labeling each curve with its appropriate letter. A C B

2 4) For each of the following normal curves, identify (as accurately as you can from the graph) the mean and standard deviation of the distribution. μ = 50 σ = 5 μ = 1100 σ = 300 μ = 10 σ = 40 μ = 225 σ = 75 5) Suppose the average height of women collegiate volleyball players is 5 9, with a standard deviation of 2.1. Assume that heights among these players follow a mound-shaped distribution. a) Draw the curve and label it b) According to the empirical rule, about 95% of women collegiate volleyball players have heights between what two values? 64.8 to 73.2 c) What does the empirical rule say about the proportion of players who are between 62.7 inches and 75.3 inches? 99.7% of players are between 62.7 and 75.3 inches tall d) Reasoning from the empirical rule, what is the tallest we would expect a woman collegiate volleyball player to be? 75.3 inches would cover all but.15% e) About what percent of women collegiate volleyball players are taller than 71.1 inches? 16% f) About what percent of women collegiate volleyball players are shorter than 64.8 inches? 2.5% g) Find the percentiles for the following heights: th 50 th 97.5 th th

3 6) Given an approximately normal distribution, N(175, 37). a. Draw a normal curve and label 1, 2, and 3 standard deviations on both sides of the mean. b. What percent of values are within the interval (138, 212)? 68% c. What percent of values are within the interval (101, 249)? 95% d. What percent of values are within the interval (64, 286)? 99.7% e. What percent of values are outside the interval (138, 212)? 32% f. What percent of values are outside the interval (64, 286)? 0.3% ) The incubation time for Rhode Island Red chicks is normally distributed with a mean 21 days and standard deviation approximately 1 day. a. Draw a normal curve and label 1, 2, and 3 standard deviations on both sides of the mean If 1000 eggs are being incubated, how many chicks do we expect will hatch in each of the following situations? b. In 19 to 23 days c. In 18 days or more d. In 22 days or less 95% of 1000 = % of 1000 = % of 1000 = 840 8) Use the table of standard normal probabilities to determine the proportion of the normal curve that falls within: a. one standard deviation of its mean (in other words, between z-scores of 1 and 1) =.6826 b. two standard deviations from the mean =.9544 c. three standard deviations from the mean =.9974 d. Compare these values to the values obtained from the empirical rule. They are all very close to the proportions described by the Empirical Rule.

4 9) Use the table of standard normal probabilities to determine the proportion of observations from a standard normal distribution that satisfies each of the following statements. For each, sketch a standard normal curve and shade the area under the curve that is the answer. a. z < 2.85 b. z > 2.85 c. z > d < z < 2.85 P(z < 2.85) =.9978 P(z > 2.85) = =.0022 P(z > 1.66) = =.9515 P( 1.66 < z < 2.85) = = ) Use the table of standard normal probabilities to find the value z of a standard normal variable that satisfies each of the following statements. For each sketch a standard normal curve and mark your value of z on the axis. a. 25% of observations fall below it b. 40% of observations fall above it z =.67 z =.25 c. proportion less than it is 0.8 d. 90% of observations are greater than it z =.84 z = ) What are the quartiles (all three) of a standard normal distribution? Q 1 : z =.67 Q 2 : z = 0 Q 3 : z =.67 12) The deciles of any distribution are the points that mark off the lowest 10% and the highest 10%. What are the deciles of the standard normal distribution? z = 1.28 and z = 1.28

5 13) Which of the following are true statements? a. The area under a normal curve is always equal to 1, no matter what the mean and standard deviation are. b. The smaller the standard deviation of a normal curve, the higher and narrower the graph. c. Normal curves with different means are centered around different numbers. i. I and II ii. I and III iii. II and III iv. I, II, and III v. None of the above gives the complete set of true responses. 14) Which of the following are true statements? a. The area under the standard normal curve between 0 and 2 is twice the area between 0 and 1. b. The area under the standard normal curve between 0 and 2 is half the area between 2 and 2. c. For the standard normal curve, the interquartile range is approximately 3. i. I and II ii. I and III iii. II only iv. I, II and III v. None of the above gives the complete set of true responses. 15) Populations P1 and P2 are normally distributed and have identical means. However, the standard deviation of P1 is twice the standard deviation of P2. What can be said about the percentage of observations falling within two standard deviations of the mean for each population? i. The percentage for P1 is twice the percentage for P2. ii. The percentage for P1 is greater, but not twice as great, as the percentage for P2. iii. The percentage for P2 is twice the percentage for P1. iv. The percentage for P2 is greater, but not twice as great, as the percentage for P1. v. The percentages are identical.

6 16) Suppose that a college admissions office needs to compare scores of students who take the SAT with those who take the ACT. Suppose that among the college s applicants who take the SAT, scores have a mean of 896 and a standard deviation of 174. Further suppose that among the college s applicants who take the ACT, scores have a mean of 20.6 and a standard deviation of 5.2. a) If applicant Bobby scored 1080 on the SAT, how many points above the SAT mean did he score? ; 184 points above the SAT mean. b) If applicant Kathy scored 28 on the ACT, how many points above the ACT mean did she score? ; 7.4 points above the ACT mean. c) Is it sensible to conclude that since your answer to (a) is greater than your answer to b), Bobby outperformed Kathy on the admissions test? Explain. No, they are different tests with different scoring scales. d) Determine how many standard deviations above the mean Bobby scored by dividing your answer to (a) by the standard deviation of the SAT scores. Do the same for Kathy s score. Bobby: z = ; z = 1.06 Kathy: z = ; z = e) Which applicant has the higher z-score for his or her admissions test score? Kathy had the higher z-score, 1.42 > 1.06 f) Explain in your own words which applicant performed better on his or her admissions test. Kathy did better. Since she is more standard deviations above the mean she would be in a higher percentile than Bobby. g) Calculate the z-score for applicant Peter, who scored 740 on the SAT, and for applicant Kelly, who scored 19 on the ACT. Peter: z = ; z =.90 Kelly: z = ; z = h) Which of Peter and Kelly has the higher z-score? Kelly has the higher z-score, -.31 > -.90: i) Under what conditions does a z-score turn out to be negative? A z-score will be negative when a value is below the mean.

7 17) Data from the National Vital Statistics Report reveal that the distribution of the duration of human pregnancies (i.e., the number of days between conception and birth) is approximately normal with mean = 270 and standard deviation = 15. Use this normal model to determine the probability that a given pregnancy comes to term in: a. less than 244 days (which is about 8 months) b. more than 275 days (which is about 9 months) c. over 300 days d. between 260 and 280 days e. Data from the National Vital Statistics Report reveal that of 3,880,894 births in the US in 1997, the number of pregnancies that resulted in a preterm delivery, defined as 36 or fewer weeks since conception, was 436,600. Compare this to the prediction that would be obtained from the model vs ; pretty close 18) Suppose that the IQ scores of students at a certain college follow a normal distribution with mean 115 and standard deviation 12. a. Use the normal model to determine the proportion of students with an IQ score below b. Find the proportion of these undergraduates having IQs greater than c. Find the proportion of these undergraduates having IQs between 110 and d. With his IQ of 75, what would the percentile of Forrest Gump s IQ be?.04 th percentile f. Determine how high one s IQ must be in order to be in the top 1% of all IQs at this college

8 19) Suppose that Professors Wells and Zeddes have final exam scores that are approximately normally distributed with mean 75. The standard deviation of Wells scores is 10, and that of Zeddes scores is 5. a. With which professor is a score of 90 more impressive? Support your answer with appropriate probability calculations and with a sketch. Zeddes; z-score is 3 which is farther above the mean than Wells where the z-score is.33 b. With which professor is a score of 60 more discouraging? Again support your answer with appropriate probability calculations and with a sketch. Zeddes; z-score is -3 which is farther below the mean than Wells where the z-score is ) Suppose that the wrapper of a certain candy bar lists its weight as 2.13 ounces. Naturally, the weights of individual bars vary somewhat. Suppose that the weights of these candy bars vary according to a normal distribution with mean = 2.2 ounces and standard deviation = 0.04 ounces. a. What proportion of candy bars weigh less than the advertised weight?.0401 b. What proportion of candy bars weigh more than 2.25 ounces?.1056 c. What proportion of candy bars weigh between 2.2 and 2.3 ounces?.4938 d. If the manufacturer wants to adjust the production process so that only 1 candy bar in 1000 weighs less than the advertised weight, what should the mean of the actual weights be (assuming that the standard deviation of the weights remains 0.04 ounces)? μ =

9 21) A trucking firm determines that its fleet of trucks averages a mean of 12.4 miles per gallon with a standard deviation of 1.2 miles per gallon on cross-country hauls. What is the probability that one of the trucks averages fewer than 10 miles per gallon? With a z-score of -2, the probability that the trucks will average fewer than 10 mpg is ) A factory dumps an average of 2.43 tons of pollutants into a river every week. If the standard deviation is 0.88 tons, what is the probability that in a week more than 3 tons are dumped? With a z-score of 0.65, the probability that in a week more than 3 tons of pollutants are dumped is ) An electronic product takes an average of 3.4 hours to move through an assembly line. If the standard deviation is 0.5 hour, what is the probability that an item will take between 3 and 4 hours? With z-scores of -0.8 and 1.2, the probability that an item will take between 3 and 4 hours to move through an assembly line is ) The mean score on a college placement exam is 500 with a standard deviation of 100. Ninety-five percent of the test takers score above what? 95% of the test takers score above ) The average noise level in a restaurant is 30 decibels with a standard deviation of 4 decibels. Ninety-nine percent of the time it is below what value? 99% of the time the average noise level in a restaurant is less than decibels. 26) The mean income per household in a certain state is $9500 with a standard deviation of $1750. The middle 95% of incomes are between what two values? With a z-score of -2, the probability that the trucks will average fewer than 10 mpg is ) Jay Olshansky from the University of Chicago was quoted in Chance News as arguing that for the average life expectancy to reach 100, 18% of people would have to live to 120. What standard deviation is he assuming for this statement to make sense (assuming life expectancies are normally distributed)? Jay Olshansky is assuming a standard deviation of approximately 21.7.

10 28) Cucumbers grown on a certain farm have weights with a standard deviation of 2 ounces. What is the mean weight if 85% of the cucumbers weigh less than 16 ounces? The mean weight is ounces if 85% of cucumbers weigh less than 16 ounces. 29) A coffee machine can be adjusted to deliver any fixed number of ounces of coffee. If the machine has a standard deviation in delivery equal to 0.4 ounce, what should be the mean setting so that an 8-ounce cup will overflow only 0.5% of the time? For an 8 oz cup to overflow 0.5% of the time, the mean setting should be 6.97 oz. 30) If 75% of all families spend more than $75 weekly for food, while 15% spend more than $150, what is the mean weekly expenditure and what is the standard deviation? The mean weekly food expenditure is $ with a standard deviation of $ ) Three landmarks of baseball achievement are Ty Cobb s batting average of.420 in 1911, Ted Williams s.406 in 1941, and George Brett s.390 in These batting averages cannot be compared directly because the distribution of major league batting averages has changed over the decades. The distributions are quite symmetric and reasonably normal. While the mean batting average has been held roughly constant by rule changes and the balance between hitting and pitching, the standard deviation has dropped over time. Here are the facts Decade Mean Std. Dev. 1910s 1940s 1970s Comparatively, who ranked highest amongst his peers? Justify your answer. Ted Williams batting average of.406 is 1941 would be ranked highest amongst his peers. Ted was 4.26 standard deviations above the mean compared to 4.15 for Ty Cobb and 4.06 for George Brett.

11 32) The following graph and data output are from the 108 years of rainfall in Austin. Let the calculated mean and standard deviation from the data represent µ and σ for the calculations. Annual Rainfall in Austin Rainfall in in S1 = mean Rainfall S2 = stddev Rainfall a) What proportion of rainfall was between 30 and 40 inches? Since the z-score for 30 inches is and the z-score for 40 inches is 0.66, the approximate proportion of rainfall between 30 and 40 inches is b) Comment on the accuracy of your calculations. The calculations should be fairly accurate. The histogram of the distribution of rainfall for Austin appears to be approximately Normal with no obvious outliers. Therefore, it is appropriate to use standard Normal probabilities. 33) A person with too much time on his hands collected 1000 pennies that came into his possession in 1999 and calculated the age (as of 1999) of each. The distribution has mean years and standard deviation years. Knowing these summary statistics but without seeing the distribution, can you comment on whether the normal distribution is likely to provide a reasonable model for these penny ages? Explain. It is probably not normally distributed (9.613) is well below 0 (a new penny). This distribution is most likely skewed to the right as there are more newer pennies in circulation. 34) Use the following data set Draw a stemplot, boxplot and normal probability plot to assess normality. Data appears to be fairly normally distributed

12 35) Use the following data set: Is this data set best modeled by using a normal distribution? Create and draw a histogram, boxplot and normal probability plot to decide. Explain your results. The histogram and box-plot both show that the data is skewed to the left. The normal probability plot has too much curvature to be considered normal.