Question 5 of 25 0.2 Points [Exercise 'Ic] Discuss the benefits of using a box plot over a histogram for describing the distribution ofthe data.[Choose the FIVE best answers that apply] (Scored correct minus incorrect.) A. Box plots do not rely on arbitrarily chosen bins. Never see artificial gaps (no frequency area) or spikes (high frequency areas). B. Not affected by extreme values, unusual values, missing data, or not having values in a given range C. Very simple to interpret D. They hide overly detailed variations in data enabling gross characteristics to be quickly observed E. Enables visualizing the rate the data values increase over the range. F. Shows details needed to identify important characteristics such as most and least frequent areas and multiple-modes G. Shows summary measures of central tendency (median), variability (min, max, IQR), and shape (skew, kurtosis) without any effort H. Can directly visualize the particular shape of the distribution. Question 6 of 25 0.2 Points Question 10 of 25 0.2 Points [Exercise 4b] [Part 1] What is the highest and lowest expected frequencies(fe) of style and views? [part 2] What are the observed most likely and least likely (to) style and views (from the % of Totals table)? [Part 3] Are they the same? A. [Part 1] Highest fe is ranch with good view and lowest is ranch no view B. [Part 1] Highest fe is ranch with some view, lowest is twostory with no view C. [Part 2] Highest fo is ranch with some view and lowest is twostory with no view D. [Part 2] Highest f0 is ranch with good View and lowest is ranch no view E. [Part 3] They (f0 and fe) are the same! F. [Part 3] They are not the same! n. Innfinr' 1 1 nF'u: n 4 Uni-+1- Question 16 of 25 0.1 Points [Exercise 6a] [Part1] Discuss on principles the degree it is ok or not ok to assume home sales prices would be Normally distributed. [Part 2] How might we validate this assumption if we were to use this in practice? [Select exactly 5 of the best answers for Part1 and exactly 2 from Part 2] A. [Part 1] Not ok because home prices may be skewed to the right due to nonnegative prices but unlimited positive prices B. [Part 1] Not ok because you should never assume something is Normally distributed unless you can prove it C. [Part 1] 0k because in practice we can avoid negative prices if 0 is more than 3 standard deviations from the mean as the probability of a negative Normal value would be very low (less than 0.1%) D. [Part 1] ok because home price can be fractional, Normal values can be fractional (continuous numerical) E. [Part 1] Not ok because the a priori theory of home prices does not match the a priori theory of the Normal distribution. F. [Part1] Ok because home prices tend to fall around a middle, rarely being very high or very low. G. [Part 1] Not ok because home price is never negative, Normal values can be negative H. [Part 1] 0k because pretty much everything in business, especially demand for goods, is Normally distributed. Sales is a "normal" process hence Normally distributed. i Chrome File Edit View History Bookmarks Profiles Tab Window Help @3 & G) E >8 E]: 6? Q 6 WedOct5 9:38AM IOU O " O Q Laulima:BUS-310002 [MAN x Q -Discord|#general|BU8310 x 1 + v 71. h i laulima.hawaiiedu/portaIlsite/MAN.71683.202310/tool/Oece068bca7c4512ae36-841233c40919/jsf/delivery/beginTakin.\" Q (3 fr I] e Question 19 of 25 01 Points [Exercise 6c] Suppose our client's budget for purchasing a home is $320K. What percent of homes in this market would we expect to sell for less than $320K? [Part1: what is the random variable for sales price] [Part 2: What computation is needed] [Part 3: Interpretation of result] (One best answer per part.) A. [Part 1] P : sale price of home B. [Part1] P = frequency of sale price C. [Part 1] P = number of homes with sale price D. [Part 2] Prob[D > $320K] = 1 - Prob[D s $320K] = 1 - norm.dist(320.298,26,true) E, [Part 2] Prob[D m=i Question 20 of 25 01 Points F [Exercise 6d] How likely is it that the client will encounter a home that is too expensive for them? [Part 1: What computation is needed] [Part 2: Interpretation of result] (One best answer per part.) A. [Part 1] Prob[D > $320K] =1 - Prob[D s$320Kl =1 - norm.dist(320,298,26,true) B, [Part 1] Prob[D