9 h https://www.mathxl.com/Student/PlayerTestaspx?t estld-123899379¢erwin; y
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9 h https://www.mathxl.com/Student/PlayerTestaspx?t estld-123899379¢erwin; yes dean Shahan 2/11/16 5:09 PM Quiz: 08 Quiz Chapter 5 (part 2) Submil Quiz This Question: 3 pts lof4(°complete) This Quir. 17 pts possble Question Hep There were 53 competitors in a downhill sking event Ther tmes (in seconds) are shown below with a standard deviation of 4 52 seconds f the mean ime was 103 65 seconds, normal model is appropriate, what percent of tmes wll be less than 99.13 seconds? 9817 98.19 9868 9874 98 928 9951 9955 99 58 9958 100 01 100.11 100 23 100 42 10056 00 65 100 79 101.01 101 86 10194 101 96 102 06 102 27 102 47 102 62 102 74 102 79 102 87 102 91 103 07 03 18 103 24 103 43 103 65 103 74 103 91 104 03 104 84 104 95 104 97 05.19 105 27 105 32 105 48 105 66 09 76 110 11 112 13 112 48 11315 113.25 114 37 117 68 Round to the nearest integer as needed) b) What is the actual percent of tmes less than 99 13 seconds? o one decimal place as needed ) Do the two percentages agree? Why or why not O A Yes, because the normal model is appropnate B. O C. No, because the normal model is appropriae 0 Yes because he normal model is not approprat. No. because he normal modes net appropriate Click to select your answers) 8 0Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 99.13
u = mean = 103.65
s = standard deviation = 4.52
Thus,
z = (x - u) / s = -1
Thus, using a table/technology, the left tailed area of this is
P(z < -1 ) = 0.158655254 = 16% [ANSWER]
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b)
There are 5 items out of 53 less than 99.13.
Hence,
P = 5/53 = 0.094339623 = 9.4% [ANSWER]
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c)
As parts a and b are no equal, then
OPTION D: No, because the normal model is not appropriate. [ANSWER]
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