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Jack 1) weighs 160 pounds and his sister weighs 110 pounds. If the mean weight for men his age is 175 with a standard deviation of 14 pounds and the mean weight for women is 145 with a standard deviation of 10 pounds, determine whose weight is closer to "average." Write your answer in terms of z-scores and areas under the normal curve. for Jack z = x 160 175 14 110 145 10 15 14 25 10 1.07 , when z = -1.07, area = .1423 for his sister x 2.50 , when z = -2.50, area = .0062 Jack is closer to average, but he is still in the lower 14%, while his sister is in the bottom less than 1% of the population. 2) Explain the difference between a population and a sample. In which if these is it important to distinguish between the two in order to use the correct formula? mean; median; mode; range; quartiles; variance; standard deviation. A sample is a subset of a population. A population consists of every member of a particular group of interest. The variance and the standard deviation require that we know whether we have a sample or a population. 3) The following numbers represent the weights in pounds of six 7-year old children in Mrs. Jones' 2nd grade class. {25, 60, 51, 47, 49, 45} Find the mean; median; mode; range; quartiles; variance; standard deviation. mean = 46.166.... median = 48 mode does not exist range = 35 Q1 = 45 Q2 = median = 48 Q3 = 51 variance = 112.1396 standard deviation =10.59 4) The Student Services office did a survey of 500 students in which they asked if the student is part-time or full-time. Another question asked whether the student was a transfer student. The results follow. Transfer Non-Transfer Row Totals Part-Time Full-Time 100 170 110 120 230 210 290 500 Column Totals 270 a) If a student is selected at random (from this group of 500 students), find the probability that the student is a transfer student. P (Transfer) The total number of transfer students is 270. The total number of students in the survey is 500 P(Transfer) = 270/500 = .54 b) If a student is selected at random (from this group of 500 students), find the probability that the student is a part time student. P (Part Time) The total number of part time students is 210. The total number of students in the survey is 500. P(Part Time) = 210/500 = .42 c) If a student is selected at random (from this group of 500 students), find the probability that the student is a transfer student and a part time student. P(transfer part time). From the table we see that there are 100 students which are both transfer and part time. This is out of 500 students in the sample. P(transfer part time) = 100/500 = .20 d) If a student is selected at random (from this group of 500 students), find the probability that the student is a transfer student if we know he is a part time student. P(transfer | part time). This is conditional probability and so we must change the denominator to the total of what has already happened. There are 100 students which are both transfer and part time. There are 210 part time students. P(transfer | part time) = 100/210 .4762 e) If a student is selected at random (from this group of 500 students), find the probability that the student is a part time given he is a transfer student. P(part time | transfer) P(part time | transfer) = 100/270 .3704 f) Are the events part time and transfer independent? Explain mathematically. The definition of independent is P(A|B) = P(A). To test we ask if P(part time | transfer) = P(part time)? Is .3704 = .42? No, there for the events are not independent. We could also test P(transfer | part time) = P(transfer) Is .4762 = .54? Again, the answer is no. g) Are the events part time and transfer mutually exclusive. Explain mathematically. For events to be mutually exclusive their intersection must be 0. In part c we found that P(transfer part time) = 100/500 = .20. Therefore the events are not mutually exclusive. 5) How do you recognize a binomial experiment? There are exactly 2 outcomes: success and failure trials The are independent. The probability of success is the same in each trial. There are a fixed number of trials. 6) How do you recognize a Poisson experiment? You are measuring things in an interval and you know the average from past experience. 7) How do you recognize a normal distribution? It is symmetric about the mean. We can use it when n 30. 8) How do you recognize a discrete distribution? The outcomes are integers. 9) If a normal curve is continuous how are we able to use it for countable random variables as well? Because of the normal approximations to the binomial. See section 5.6. 10) Which of the following represent continuous distributions? a) b) c) d) The lengths of fish in a certain lake. -- continuous The number of fish in a certain lake. -- discrete The diameter of 15 trees in a forest. -- continuous How many trees are on a farmer's acre. . -- discrete 11) On a dry surface, the braking distance (in meters) of a certain car is a normal distribution with mu = 45.1 m and sigma = 0.5 m. a) Find the braking distance that corresponds to z = 1.8 z b) x mu ----> 1.8 sigma x 45.1 ----> .9 .5 x 45.1 ----> x = 46 Find the braking distance that represents the 91st percentile. We need to look in the table for the z-score which corresponds to an area of .9100. The closest 1.34. Using the same technique as part a we have 1.34 x 45.1 ----> .67 5 x 45.1 ----> x = 45.77 c) Find the z-score for a braking distance of 46.1 m Using the same formula a little differently z x mu sigma 46.1 45.1 .5 1 .5 2 d) Find the probability that the braking distance is less than or equal to 45 m The z score corresponding to x = 45 is z P(x 45) = P(z 45 45.1 .5 .1 .5 .2 -.2) = area to the left of x = -.2 which is .4207 e) Find the probability that the braking distance is greater than 46.8 m The z score corresponding to x = 46.8 is z The area to the left of z = 3.4 is .9997 P(x > 46.8) = P(z > 3.4) = 1 P(z < 3.4) = 1 - .9997 = .0003 46.8 .5 45.1 1.7 .5 3.4 f) Find the probability that the braking distance is between 45 m and 46.8 m. P(45 < x < 46.8) = P( -.2 < z < 3.4) = P(z < 3.4) P(z < -.2) = .9997 - .4207 = .579 12) To predict the annual rice yield in pounds we use the equation ^ y 859 5.76 x1 3.82x2 where x1 represents the number of acres planted (in thousands) and where x2 represents the number of acres harvested (in thousands) and where r2 = .94. a) Predict the annual yield when 3200 acres are planted and 3000 are harvested. ^ y = 859 + 5.76*3200 + 3.82*3000 = 859 + 18432 + 11460 = 30751 which is 30,751,000 pounds of rice b) Interpret the results of this r2 value. 94% of the variation in the annual rice yield can be explained by the number of acres planted and harvested. The remaining 6% is unexplained and is due to other factors or to chance. c) What do we call the r2 value? It is the coefficient of determination. 13) What type of relationship is shown by this scatter plot? 45 40 35 30 25 20 15 10 5 0 0 5 10 15 20 weak positive linear correlation 14) What values can r take in linear regression? Select 4 values in this interval and describe how they would be interpreted. the values are between 1 and +1 inclusive. -1 means strong negative correlation +1 means strong positive correlation 0 means no correlation .5 means moderate positive correlation etc. 15) Does correlation imply causation? No. 16) What do we call the r value. The correlation coefficient. 17) If we have data with min = 2; Q1 = 10; median = 12; Q3 = 15; and max = 21; draw a box and whisker. ______ |----------|--|----|---------| ------5-----10-----15-----20 18) If we have the following data 34, 38, 22, 21, 29, 37, 40, 41, 22, 20, 49, 47, 20, 31, 34, 66 draw a stem and leaf 2 3 4 5 6 | | | | | 219200 48714 0197 6 19) Be prepared to do a hypothesis test particularly to state the claim, the null, the alternative, and to write a conclusion. 20) Be prepared to write a confidence interval. ... View Full Document