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 Subject: General Statistics & Probabilities Category: Reference, Education and News > Homework Help Asked by: sunshine2-ga List Price: \$40.00 Posted: 19 Jun 2003 09:05 PDT Expires: 19 Jul 2003 09:05 PDT Question ID: 219231
 ```Review: Looking for simple step-by-step explanations and solutions (preferably by Sunday afternoon) --1. Suppose the manager of a large farm wanted to study the impact of the fruit flies on the orange crops on a daily basis over a 6-week period. On each day a random sample of orange trees were selected from within a random sample of acres. The daily average number of damaged oranges per tree and the proportion of trees having damaged oranges were calculated. The two main measures calculated each day (i.e., average number of damaged oranges per tree and proportion of trees having damaged oranges) are called? --2. This table contains the opinions of a sample of 200 people broken down by gender about the latest congressional plan to eliminate anti-trust exemptions for professional baseball. For Neutral Against Totals Female 38 54 12 104 Male 12 36 48 96 Totals 50 90 60 200 (a), construct a table of total percentages. (b) of those FOR the plan, ________ percent were females. (c) of the males in the sample, ________ percent were FOR the plan. --3. Health care issues are receiving much attention in both academic and political arenas. A sociologist recently conducted a survey of citizens over 60 years of age whose net worth is too high to qualify for Medicaid by who have no private health insurance. The ages of 25 uninsured senior citizens were as follows: 60 61 62 63 64 65 66 68 68 69 70 73 73 74 75 76 76 81 81 82 86 87 89 90 92 *Identify the third quartile of the ages of the uninsured senior citizens. --4. The stem-and-leaf display represents the number of vitamin supplements sold by a health food store in a sample of 16 days. Stem Leaves 1H 99 2L 0023 2H 567 3L 034 3H 568 4L 1 For this sample, the sum of the observations is 448, while the sum of the squares of the observations is 13,356. Note: 1H means “high tens”: 15, 16, 17, 18, or 19; 2L means “low twenties”: 20, 21, 22, 23, or 24; 2H means “high twenties”: 25, 26, 27, 28, or 29; etc. *What is the standard deviation of the number of vitamin supplements sold in this sample? --5. The probability that a new advertising campaign will increase sales is assessed as being 0.80. The probability that the cost of developing the new ad campaign can be kept within the original budget allocation is 0.40. Assuming that the two events are independent, the probability that the cost is kept within budget and the campaign sales increases is? --6. Suppose A and B are mutually exclusive events where P(A) = 0.4 and P(B) = 0.5. Then P(A and B) = ? --7. A survey is taken among customers of a fast-food restaurant to determine preference for hamburger or chicken. Of 200 respondents selected, 75 were children and 125 were adults. 120 preferred hamburger and 80 preferred chicken. 55 of the children preferred hamburger. What is the probability that a randomly selected individual is a child or prefers hamburger? --8. Suppose there are three events A, B, and C with probabilities: P(A) = 0.50, P(B) = 0.40, P(C) = 0.30. Also, A and B are independent; A and C are mutually exclusive; P(B and C) = .05. (a) P(C') = (b) P(A and C) = --9. For a potential investment of \$5000, a portfolio has an EMV of \$1000 and a standard deviation of \$100. What is the coefficient of variation? --10. If we know that the length of time it takes a college student to find a parking spot in the library parking lot follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute, find the point in the distribution in which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot. --11. A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with a mean of 10.5 ounces and a standard deviation of 0.3 ounce. Find the proportion of all jars packaged by this process that have weights that fall below 10.875 ounces. --12. Suppose Z has a standard normal distribution with a mean of 0 and standard deviation of 1. The probability that Z is between -2.89 and -1.03 is? --13. The amount of time necessary for assembly line workers to complete a product is a normal random variable with a mean of 15 minutes and a standard deviation of 2 minutes. So, 70% of the products would be assembled within how many minutes?```
 ```Hello sunshine2, Glad to see you again! Here are the answers to your questions. 1. The average number of damaged oranges per tree is called, by definition, the "sample mean". It's calculated by taking the number of observed damaged oranges and dividing it by the number of trees in the sample. Sample Mean http://www.animatedsoftware.com/statglos/sgxbar.htm The proportion of trees having damaged oranges, is called "sample proportion". It's calculated in the same way as the sample mean, as you take the number of observed trees that contain damaged oranges, and divide thsi number by the number of trees in the sample. 2. In the following table, I take the total sample size. Thus, the percentage you see in the square (Female, For), would mean "what percentage of all people in the mean is BOTH a female and for the anti-trust plan?". For Neutral Against Totals Female 19% 27% 6% 52% Male 6% 18% 24% 48% Totals 25% 45% 30% 100% These percentages are calculated in the following way. For example, the percentage in (Female, For) is calculated by taking the number of females that are for the plan (38), dividing it by 200 (total number of people in the sample) and multiplying it by 100. The same goes for the other elements in the table. 3. The third quartile is "the value of a set of data at the third quarter, 75%, when that data has been arranged in ascending order. It is the median value of the upper half of all the values in the data set" Glossary O-U http://www.tta.gov.uk/training/skillstests/numeracy/glossary/o-u.htm In your case, we first have: 60 61 62 63 64 65 66 68 68 69 70 73 73 74 75 76 76 81 81 82 86 87 89 90 92 The upper half of your data is: 73 74 75 76 76 81 81 82 86 87 89 90 92 And the median of this subset is 81. Thus, the third quartile is 81. For more information on what the median is, check Glossary H-N http://www.tta.gov.uk/training/skillstests/numeracy/glossary/h-n.htm 5. First let's see what "independent" means. Events A and B are said to be independent if Prob(A and B) = P(A)*P(B). Independent Random Variables http://www.uwm.edu/~stockbri/14indep.pdf Therefore, A = The campaign will increase sales B = The campaign can be kept within the original budget We know that P(A)=0.8 and P(B)=0.4 and that A and B are independent. Thus, P(A and B) = 0.8*0.4 = 0.24 = 24% There is a 24% probability that both events will happen. 6. First, let's see what the definition of mutually exclusive events is. "Two events are mutually exclusive if it is not possible for both of them to occur..." Mutually exclusive events http://davidmlane.com/hyperstat/A132677.html Thus, events A and B cannot happen at the same time; so, P(A and B)=0 7. In order to solve this one, we must first build a table like the on in question 2. From the problem, we know the following data: Chicken Hamburguer Totals Children 55 75 Adult 125 Totals 80 120 200 Now, it's easy to fill the blank spaces, since we know that the sum of lines and columns whould add up to the values in "Totals". For example, we know that there are 75 children, 55 of which prefer burguers; therefore 75-55=20 children prefer chicken. Now, we know that a total of 80 people prefer chicken, 20 of which are children, so 80-20=60 adults prefer chicken. Lastly, since there are 125 adults, and 60 prefer chicken, we know that 125-60=65 adults prefer hamburguers. Thus we get the following table Chicken Hamburguer Totals Children 20 55 75 Adult 60 65 125 Totals 80 120 200 From this table, we make one with percentages, just like in question 3. We get Chicken Hamburguer Totals Children 10% 27.5% 37.5% Adult 30% 32.5% 62.5% Totals 40% 60% 100% Finally, let's get to the question. What is the probability that a randomly selected individual is a child or prefers hamburger? Let's see which events fulfill the condition (Children or Hamburguer). Children that like chicken fulfill it, so do children that like hamburguers, and so do adults that like hamburguers. Adults that like chickens do not fulfill the condition. So now we just have to add the probability of the events that fulfill the conditions in order to get the solution. So, the probability of randomly sleecting someone who is a children or likes hamburguers is 10% + 27.5% + 32.5% = 70% 8. First I'll solve P(C'). C' is the notation for "the complement of C", which is equivalent to saying "not C". Thus P(C')=P(not C). Check the following definition: "Definition: The complement of an event A is the set of all outcomes in the sample space that are not included in the outcomes of event A. The complement of event A is represented by A'. Rule: Given the probability of an event, the probability of its complement can be found by subtracting the given probability from 1. P(A') = 1 - P(A)" Therefore, P(C') = 1-P(C) = 1-0.3 = 0.7 Regarding P(A and C), this is just the same as in question 6. We know from the question that A and C are mutually exclusive, so P(A and C) = 0 9. We know that EMV is the mean of the value of the portfolio, and the formula for the coefficient of variation is given in the following page Coefficient of Variation http://engineering.uow.edu.au/Courses/Stats/File1586.html Therefore, the coefficient of variation is: 100/1000=0.1 10. Let X be the number of minutes it takes a student to find a parking spot. We know that X hasa normal distribution with mean 3.5 and standard deviation 1. The question here is find the number a that solves P(X > a) = 0.758 In order to solve this we must use a table for the normal distribution. The tables you will find in books and on the web are for standard normal distribution (that is, with mean 0 and standard deviation 1). So we have to convert our random variable X into a standard normal distributed variable. This is easily done: we have to substract the mean and divide by the standard deviation. Thus, we have that P(X > a) =P(X - 3.5 > a-3.5) But X - 3.5 has a standard normal distribution, so I'll rename it as Z. We are looking for: P(Z > a-3.5) = 0.758 In the following page, you will find an applet that calculates P(Z < a) Z table http://davidmlane.com/hyperstat/z_table.html Thus, we need to rewrite your problem P(Z > a-3.5)= 1-P(Z < a-3.5) = 0.758 So, P(Z < a-3.5) = 1-0.758= 0.242 So, now we write 0.242 in the applet box called "Area below Z", and we find that the z-value that solves this is -0.6996. Thus, a-3.5=-0.6996 a=3.5-0.6996=2.8004 Therefore, 75.8% of the students take more than 2.8004 minutes to find a parking spot. 11. This problem is similar to the previous one. Here we know that X is distributed normally with mean 10.5 and standard deviation 0.3. We have to find P(X<10.875). Again, we have to convert X to standard normal, so we have that: P(X < 10.875) =P(X-10.5 < 10.875-10.5) =P( (X-10.5)/0.3 < (10.875-10.5)/0.3) =P( Z < 1.25) We now have to plug 1.25 in the applet box called Z to find that 89.44% of the observations fall below this value; and thus 89.44% of the jars weigh less than 10.875 ounces. 12. We will use the same applet here. We have to find out: P(-2.89 < Z < -1.03) But this is equal to say P( Z < -1.03) - P( Z < -2.89) So now we just have to plug these numbers in the applet box called Z, finding that: P(Z<-1.03)=0.1515 P(Z<-2.89)=0.0019 So, 0.1515-0.0019=0.1496. Thus, the answer is 14.96% I hope these answers are clear enough. If you have any doubts, please don't hesitate to request a clarification, I'll be glad to explain further. Best regards, elmarto``` Clarification of Answer by elmarto-ga on 19 Jun 2003 11:52 PDT ```I'm sorry, I neglected questions 4, 13, part of question 2 and the search terms I used. Here they are. -------------------------------------------- 2. I'll repost the whole answer to question 2 here and then add the missing answers: In the following table, I take the total sample size. Thus, the percentage you see in the square (Female, For), would mean "what percentage of all people in the mean is BOTH a female and for the anti-trust plan?". For Neutral Against Totals Female 19% 27% 6% 52% Male 6% 18% 24% 48% Totals 25% 45% 30% 100% These percentages are calculated in the following way. For example, the percentage in (Female, For) is calculated by taking the number of females that are for the plan (38), dividing it by 200 (total number of people in the sample) and multiplying it by 100. The same goes for the other elements in the table. (b) of those FOR the plan, ________ percent were females. From the table in the question, we have that there are 50 people for, 38 of which are women. Thus the percentage of those FOR the plan who were females would be (38/50)*100 = 76% (c) of the males in the sample, ________ percent were FOR the plan We know from the table that there are 96 males, 12 of which are FOR the plan. Thus the percentages of males who are FOR the plan is (12/96)*100 = 12.5% 4. In order to find the standard deviation (SD), we must first find the variance. Then we will take the square root of the variance and that will be the SD. The formula for the sample variance is: N ___ (1/N-1)* \ (xi-x)^2 / --- i=1 where N is the sample size (16), xi is each observation and x is the sample mean. The sample mean in this case would be 448/16= 16. This formula was taken from Variance http://mathworld.wolfram.com/Variance.html Let's "play" with the term: (xi-16)^2 We can expand it to get: (xi-16)^2 =xi^2 - 2*xi*16 + 16^2 =xi^2 - 32*xi + 256 So, N ___ \ (xi-x)^2 / --- i=1 is equal to N ___ \ xi^2-32*xi+256 / --- i=1 which is equal to N ___ ___ ___ \ xi^2 - \ 32*xi + \ 256 / / / --- --- --- i=1 (All sums are between i=1 and N) We know what the first term is: it's the sum of the squares of the observations, which we know is 13,356. The third term is simply 256 summed N times, so it's 256*16=4096. The second term can be rewritten as N ___ 32*\ xi / --- i=1 But we know that the sum of xi is 448, so the second term would be 32*448=14,336 Therefore, we have that N ___ ___ ___ \ xi^2 - \ 32*xi + \ 256 / / / --- --- --- i=1 equals 13,356 - 14,336 + 4096 = 3,116 So, the variance is (1/N-1)*3,116 = 3,116/15 = 207.73... Finally, the standard deviation is the square root of the variance, which would be 14.412... 13. In this question, we are looking for "a" such that P( 15-a < X < 15 +a ) = 0.70 Again, we have to convert X to a standard normal in order to use tables. Thus P( 15-a < X < 15 +a ) =P(-a < X-15 < a) =P(-a/2 < (X-15)/2 < a/2) =P(-a/2 < Z < a/2) Now, P(-a/2 < Z < a/2) = 0.70 =P(Z-a/2) And we also know that P(Z<-a/2)=1-P(Z>-a/2) So, by replacing this in (*), we get that P(Z-a/2)) =P(Z
 sunshine2-ga rated this answer: and gave an additional tip of: \$2.00 `Another great help session. Thank you.`