Understanding Midterm Solutions for Statistics 401: Key Concepts
School
Rutgers University**We aren't endorsed by this school
Course
STATS 401
Subject
Statistics
Date
Dec 10, 2024
Pages
3
Uploaded by MinisterThunder15795
Solutions to Midterm #1Statistics 401, Fall 2024The midterm says to round all answers to two decimal places, but I give more than thatin most of these solutions.1.(1 point)This is a convenience sample. A lot of the students who frequent Hill Centerare in the mathematical sciences. On the whole, these people are probably better atmath than the average undergraduate.Hence, I’d expect to overestimate the meanSAT score when using a sample of such individuals.2.(1 point)Without any additional information about the study, I can’t think of anyreason the study would be biased. The institution conducting the study is reputable,and there are no obvious conficts of interest.3.(1 point)General Mills is funding the study. This provides the authors of the studywith an incentive to make the results of the study favorable to General Mills.4.(a)(1 point)The subject of the study might be regarded as a sensitive issue bymembers of the population (particularly weight). This might help drive the non-response.(1 point)With respect to methodology, it’s possible that the survey gets buriedin people’s inboxes. This could also contribute to non-response.(b)(1 point)It appears as though a lot of the respondents are estimating their weights(people have a tendency to round like this when making estimates). Rather thanasking people for their weights, it would be better to weight them directly.5.(a)(1 point)By examining the histogram, it appears as thoughx >˜x.This isbecause of the presence of some high salaries in the sample (in the right tail ofthe distribution). The meanxis drawn towards these extreme observations whilethe median ˜xis not (the median is aresistantmeasure of center).(b)(1 point)I’d probably prefer the median ˜x. Since it’s unaffected by the unusuallyhigh salaries in the sample, ˜xgives a better impression of what a “typical” salarylooks like.6.(1 point)It’s a statistic since it’s computed from a sample.7.(a)(1 point)It’s a parameter since it’s a property of the population.Since mypopulation is very small, I can easily compute this parameter directly.1
(b)(1 point)It’s a statistic. In this case, the students in this class constitute a smallsample from the larger population ofallstudents.8. This is a binomial experiment withn= 20 trials and a probability of successp= 1/6(the probability of rolling a six). In the following, letXdenote the number of sixes Iget in all of the trials. ThenX∼Bin(n, p). Thus, for anyk= 0,1, . . . ,20,P(X=k) =nkpk(1−p)n−k=20k16k5620−k(1)(a)(1 point)Pluggingk= 0 into (1), we getP(X= 0) =2001605620≈0.0261(b)(1 point)Pluggingk= 1 into (1), we getP(X= 1) =2011615619≈0.104(c)(1 point)Using the result of parts (a) and (b), we getP(X= 0 orX= 1) =P(X= 0) +P(X= 1)≈0.0261 + 0.104≈0.1309.(a)(1 point)The probability that someone will purchase a small cup isP(Small) =P(Small and Decaf) +P(Small and Regular)= 0.14 + 0.20 =0.34(1 point)The probability that someone will purchase a decaf isP(Decaf) =P(Small and Decaf) +P(Medium and Decaf) +P(Large and Decaf)= 0.20 + 0.10 + 0.10 =0.40(b)(1 point)From part (a), we know thatP(Small) = 0.34. Hence,P(Decaf|Small) =P(Small and Decaf)P(Small)=0.200.34≈0.59(c)(1 point)From part (b), we know thatP(Decaf) = 0.40. Hence,P(Small|Decaf) =P(Small and Decaf)P(Decaf)=0.200.40=0.5010.(1 point)Since all males are equally likely to be selected,P(A|B) is the proportion ofprofessional basketball players who are over 6ft in height, andP(B|A) is the proportionof adult males over 6ft in height who are basketball players. Almost all professionalbasketball players are very tall, but only a small proportion of tall adult males arebasketball players. Thus I’d expect thatP(A|B)> P(B|A).2
11.(1 point)There are a total of(153)ways I can select three of the 15 bulbs,(42)waysI can select two of the four 23-watt bulbs, and(5+61)ways I can select the remainingbulb from among the 13- and 18-watt bulbs. Since all bulbs are equally likely to beselected, my probability isP(select three 23-watt bulbs) =(42)(5+61)(153)≈0.14512.(a)(1 point)We estimateµ= 51/5 = 10.2.(b)(1 point)For anyk= 0,1,2, . . .,P(X=k) =µke−µk!Substitutingk= 3, we getP(X= 3) =µ3e−µ3!=10.23e−10.23!≈0.006573