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MATH 225N Week 4 quiz LATEST MATH 225N Week 4 quiz LATEST

Typology: Exams

2021/2022

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Download MATH 225N Week 4 quiz LATEST and more Exams Nursing in PDF only on Docsity! WEEK 4 QUIZ QUESTION 1 · 1/1 POINTS Alice sells boxes of candy at the baseball game and wants to know the mean number of boxes she sells. The numbers for the games so far are listed below. 16,14,14,21,1516,14,14,21,15 Find the mean boxes sold. That is correct! $$mean=16 boxes Answer Explanation Correct answers: $\text{mean=}16\text{ boxes}$mean=16 boxes Remember that the mean is the sum of the numbers divided by the number of numbers. There are 55 numbers in the list. So we find that the mean boxes sold is 16+14+14+21+155=805=1616+14+14+21+155=805=16 FEEDBACK Content attribution- Opens a dialog QUESTION 2 · 1/1 POINTS Given the following list of prices (in thousands of dollars) of randomly selected trucks at a car dealership, find the median. 20,46,19,14,42,26,3320,46,19,14,42,26,33 That is correct! $$median=26 thousands of dollars Answer Explanation Correct answers: $\text{median=}26\text{ thousands of dollars}$median=26 thousands of dollars It helps to put the numbers in order. 14,19,20,26,33,42,4614,19,20,26,33,42,46 Now, because the list has length 77, which is odd, we know the median number will be the middle number. In other words, we can count to item 44 in the list, which is 2626. So the median price (in thousands of dollars) of randomly selected trucks at a car dealership is 2626. FEEDBACK Content attribution- Opens a dialog QUESTION 3 · 1/1 POINTS Each person in a group shuffles a deck of cards and keeps selecting a card until a queen appears. Find the mode of the following number of cards drawn from a deck until a queen appears. 3,12,3,11,5,5,3,10,123,12,3,11,5,5,3,10,12 That is correct! A bar graph has a horizontal axis titled Values labeled from 2 to 18 in increments of 2 and a vertical axis titled Frequency labeled from 0 to 200 in increments of 50. 14 bars are plotted, above the numbers 2 to 16. From left to right, the heights of the bars are as follows: 1. 5. 10. 40, 75, 125, 190, 180, 130, 125, 60, 25,20, 10. All values are approximate. That is correct! The data are skewed to the left. The data are skewed to the right. The data are symmetric. Answer Explanation Correct answer: The data are symmetric. Note that the histogram appears to be roughly symmetric. So the data are symmetric. FEEDBACK Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog Content attribution- Opens a dialog QUESTION 21 · 1/1 POINTS A deck of cards contains red cards numbered 1,2,3,4,5,61,2,3,4,5,6, blue cards numbered 1,2,3,4,5,6,7,8,9,10,111,2,3,4,5,6,7,8,9,10,11 and green cards numbered 1,2,3,4,5,6,7,8,9,10,11,121,2,3,4,5,6,7,8,9,10,11,12. If a single card is picked at random, what is the probability that the card is red? Give your answer as a fraction. That is correct! $$629 Answer Explanation Correct answers: $\frac{6}{29}$629 Because there are 66 red cards, and 2929 cards total in the deck, the probability is 629629. FEEDBACK Content attribution- Opens a dialog QUESTION 22 · 1/1 POINTS The five-number summary for a set of data is given below. Min Q1 Median Q3 Max 454 5 494 9 5555 646 4 797 9 What is the interquartile range of the set of data? That is correct! $$Interquartile range: 15 Answer Explanation Correct answers: $\text{Interquartile range: }15$Interquartile range: 15 Remember that the interquartile range is the third quartile minus the first quartile. So we find that the interquartile range is 64−49=15.64−49=15. FEEDBACK $_2$_ $_3$_ $_14$_ $_15$_ $_22$_ Answer Explanation Correct answer: Min Q1 Median Q3 Max $_2$_ $_3$_ $_14$_ $_15$_ $_22$_ We can immediately see that the minimum value is $_2$_ and the maximum value is $_22$_. If we add up the frequencies in the table, we see that there are $_23$_ total values in the data set. Therefore, the median value is the one where there are $_11$_ values below it and $_11$_ values above it. By adding up frequencies, we see that this happens at the value $_14$_, so that is the median. Now, looking at the lower half of the data, there are $_11$_ values there, and so the median value of that half of the data is $_3$_. This is the first quartile. Similarly, the third quartile is the median of the upper half of the data, which is $_15$_. $_\color{blue}{2}$_, $_2$_, $_2$_, $_2$_, $_2$_, $_\color{blue}{3}$_, $_3$_, $_4$_, $_5$_, $_12$_, $_13$_, $_\color{blue}{14}$_, $_14$_, $_14$_, $_14$_, $_14$_, $_15$_, $_\color{blue}{15}$_, $_15$_, $_17$_, $_19$_, $_20$_, $_\color{blue}{22}$_ So, the five-number summary is Min Q1 Median Q3 Max $_2$_ $_3$_ $_14$_ $_15$_ $_22$_ FEEDBACK Content attribution- Opens a dialog QUESTION 25 · 0/1 POINTS Identify the parameter $_n$_ in the following binomial distribution scenario. A basketball player has a $_0.479$_ probability of making a free throw and a $_0.521$_ probability of missing. If the player shoots $_17$_ free throws, we want to know the probability that he makes more than $_9$_ of them. (Consider made free throws as successes in the binomial distribution.) Do not include '$_n=$_' in your answer. That's not right. $$8 Answer Explanation Correct answers: $17$17 The parameters $_p$_ and $_n$_ represent the probability of success on any given trial and the total number of trials, respectively. In this case, the total number of trials, or free throws, is $_n=17$_. FEEDBACK Content attribution- Opens a dialog QUESTION 26 · 0/1 POINTS The probability of buying a movie ticket with a popcorn coupon is $_0.717$_. If you buy $_21$_ movie tickets, what is the probability that exactly $_14$_ of the tickets have popcorn coupons? (Round your answer to $_3$_ decimal places if necessary.) That's not right. $$0.67 Answer Explanation Correct answers: $0.16$0.16 This probability can be found using the binomial distribution with success probability $_p=0.717$_ and $_21$_ trials. To find the probability that exactly $_14$_ of the tickets have popcorn coupons, use a calculator or computer: $_P{(X=14)}={\textrm{binompdf}} {(21,0.717,14)}=0.16$_.