T-Mobile LTE 3:10 PM 31%) Done Test-#1 (Ch. 1-6 with 40 Questions).docx 3. Frequ

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T-Mobile LTE 3:10 PM 31%) Done Test-#1 (Ch. 1-6 with 40 Questions).docx 3. Frequency polygons are particularly useful when a. data are qualitative. b. data are ungrouped. c. two or more frequency distributions are to be displayed on the same graph. d. the original frequency distribution is to be portrayed with complete accuracy 4. A stem and leaf display is ideal for summarizing distributions when you want to a. produce an idealized distribution. b. compare quantitative and qualitative data. c. emphasize the orderliness of data. d. preserve the identities of the individual scores. 5. An important characteristic of histograms, frequency polygons, and stem and leaf displays is a. S12e. b. total area. c. relative area. d. shape. 6 A frequency distribution of standardized IQ scores for equal numbers of male and female gradeschool children probably will be a. bimodal. b. normal. c. positively skewed. d. negatively skewed 7. When constructing a graph, you must first decide on the type of graph. This decision depends on a. the total number of observations. b. whether data are grouped or ungrouped. c. whether data are quan d. the titative or qualitative. impression that you wish to create. 8. Find the median for the following observations: a. 29, 54, 43, 38, 40, 71 b. 21 C. 40.5 d. 41.5 e 42.5

Explanation / Answer

Answers to all questions with proper explanations below:

3.They serve the same purpose as histograms, but are especially helpful for comparing sets of data.
So, option C is right

4. a. is wrong. It doen't show any idealistic version. b is also not right, we can't do this
in stem-leaf plot. c also is done by histogram, nothing unique with stemleaf plot
D, is unique to stem-leaf plot

5. d. is right. The stem-leaf plot displays the shape.
6. a. Bimodal is right. As there are 2 groups, each with one mode
7. b. is the right way to decide on which graph to decide
8. When there are even number of obs. then chosse the middle 2 obs and take average.
So, median = (40+43)/2 =41.5. D is right

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