Q & A for Stats
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Specific weekly problems will be updated the Friday AFTER the Wednesday due date. <br/><br/>If you would like to receive help on how to do problems BEFORE the due date, feel free to come to my TA office hours prior to turning in your homework.<br/><br/>Wed - 6:00-8:00pm<br/>Thur - 4:30-6:30pm<br/>Lokey Bldg - Rm 123iWeb 3.0.4http://www.stats4girls.com/Stats4girls/Q_%26_A/Q_%26_A_files/QA.jpgQ & A for Stats
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Homework 7 #10
http://www.stats4girls.com/Stats4girls/Q_%26_A/Entries/2012/3/14_Homework_7_10.html
8466a369-13f8-4dba-98b0-125808f6d1adWed, 14 Mar 2012 01:37:18 -0700Question 10<br/><br/>A machine is programmed to put 737 grams of salt in a container. Due to uncontrolled variation in the process, there is variation in content from container to container. To estimate the mean amount of salt per container, a sample of 50 boxes is selected and x-bar = 739.5 grams. From experience with the machine, it is known that the std dev = 7.5 grams. Find a 90% confidence interval for mew.<br/><br/>Lower confidence limit (LCL) = (to 1 decimal place)<br/><br/>Upper confidence limit (UCL) = (to 1 decimal place)<br/><br/><br/><br/><br/><br/><br/><br/><br/><br/>Homework 6 #5
http://www.stats4girls.com/Stats4girls/Q_%26_A/Entries/2011/10/27_Homework_6_5.html
08a3e5dd-2076-42b3-b165-20ea0021cac3Thu, 27 Oct 2011 01:07:13 -0700Question 5<br/><br/>In order to see what happens when the normal approximation is used improperly, consider the binomial distribution with n = 15 and p = 0.05. Since np = 0.75, the rule of thumb (np > 5 and nq > 5) is not satisfied. Using the binomial tables, find the probability of one or fewer successes and compare this with the normal approximation. (Give both answers to the 3rd decimal place.)<br/><br/>Using binomial table: P(X<2) =<br/><br/><br/>Using normal approximation: P(X<2) =<br/><br/><br/>Homework 5 #10
http://www.stats4girls.com/Stats4girls/Q_%26_A/Entries/2011/10/11_Homework_5_10.html
f1756298-e3f0-4313-bf45-25220d1d0304Tue, 11 Oct 2011 00:55:36 -0700Question 10<br/><br/> 1. If boys and girls are equally likely to be born, what is the probability that in a randomly selected family of six children, there will be boys? (Find the answer using a formula, not a branching diagram)<br/><br/>P(x = 1, 2, 3, 4, 5, 6 | B (n = 6, p = 0.5)) =<br/><br/><br/>Step 1:<br/>This problem can be solved using the binomial probability function. Start by writing down the given variables.<br/><br/>n=6 x=1,2,3,4,5,6 p=.5 q=.5 (*hint: q=1-p)<br/><br/>Step 2:<br/>X is dynamic, which means it can be represented as 1, 2, 3, 4, or 5. Plug the variables into the binomial probability function, using 1 to represent x. <br/>Exam 1 Review #20
http://www.stats4girls.com/Stats4girls/Q_%26_A/Entries/2011/9/25_Exam_1_Review_20.html
8e79da90-96ff-4430-9ed0-1e468321976cSun, 25 Sep 2011 21:47:19 -0700Question 20<br/><br/>You are the CEO of a large company. At today’s board meeting, the last item on the agenda is your proposal to merge with a competitor. You’ve heard through the grapevine that only 5 of your 15 board members support your proposal. At lunch, you’ll be seated randomly with 3 board members. What is the probability that all three of your tablemates are supporters? Diagram the sample space to figure out the answer. (You don’t need to calculate the answer. It is enough to show the probability for each branch of your diagram).<br/><br/><br/>Step 1:<br/>To solve this problem, we will need to draw a tree diagram to find out the probability that you will be seated with 3 out of the 5 board members who support your proposal.<br/>Start by drawing the first two branches: supp (with a 5/15 prob - the numerator is the number of supporters and 15 is the number of board members) and non-supp (with a 10/15 prob).<br/><br/><br/><br/>Step 2:<br/>Now draw another stage of branches and notice the two adjacent supp branches that are highlighted in pink.<br/>Hint: Remember that each time a board member is picked to sit at your table,one of the numerators will decrease by one (either supp or non-supp), but the denominator will decrease by one for both the supp and non-supp.<br/><br/><br/><br/>Step 3:<br/>Now draw a third stage of branches and you will now see three adjacent supp branches highlighted in pink, which shows the probability that 3 supporters will be randomly seated at your table.<br/>Hint: Remember that each time a board member is picked to sit at your table,one of the numerators will decrease by one (either supp or non-supp), but the denominator will decrease by one for both the supp and non-supp.<br/><br/><br/><br/>Nice job - Another problem solved!<br/><br/>Homework 4 #14
http://www.stats4girls.com/Stats4girls/Q_%26_A/Entries/2011/9/25_Homework_4_14.html
8064ba86-4399-426b-97a6-45954993ff3fSun, 25 Sep 2011 21:14:09 -0700Question 14<br/><br/>P(G) = 0.5, P(H) = 0.4, and P(G and H) = 0.1 (see diagram).<br/><br/><br/><br/><br/><br/><br/><br/><br/><br/> 1. P(G|H) = .25<br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/> 1. P(H|G) = .2<br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/> 1. P(H-bar) = .6<br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/><br/> 1. P(G or H) = .8<br/><br/><br/><br/> 1. P(G or H-bar) = .7<br/><br/><br/> 1. YES or NO: Are events G and H mutually exclusive? NO<br/>Mutually exclusive events cannot occur at the same time. Because circle G and circle H overlap, they are not mutually exclusive; which means G and H could occur simultaneously where they intersect)<br/><br/><br/> 1. YES or NO: Are events G and H independent? NO<br/><br/><br/>