Question

Questions 24-32: based on the following sample data set that reports the number of boats
shipped per week by Ottertail Boats, Inc.

83, 115, 119, 120, 121, 122, 122, 126, 129,
130, 131, 132, 133, 134, 135, 135, 135, 190

24. The third quartile is

a. 135

b. 134

c. 134.25

d. 135.25

25. The 40th percentile is

a. 7.6

b. 122

c. 124.4

d. 126

26. The 56th percentile is

a. 10.64

b. 130

c. 131

d. 130.64

27. The first quartile is

a. 120

b. 120.75

c. 121

d. 4.75

28. The interquartile range is

a. 107

b. 13

c. 12

d. 13.5

29. The sample coefficient of variation is

a. 15.48%

b. 16.51%

c. 15.28%

d. 14.84%

30. The median is

a. 129

b. 130

c. 129.

5 d. 9.5

31. The sample Pearson’s coefficient of skewness is

a. 0.213

b. -0.167

c. 0.162

d. -0.162

32. The z-score of 135 is

a. 0.33

b. 1.33

c. 0.47

d. 0.89

Answer 1

24. Third quartile, Q3 = 0.75(n+1)th value = 14.25th value of sorted data = 134.25

Answer: c. 134.25

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25. The 40th percentile is = PERCENTILE.EXC(Range, 0.4) = 124.4

Answer: c. 124.4

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26. The 56th percentile is = PERCENTILE.EXC(A:A, 0.56) = 130.64

Answer : d. 130.64

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27. First quartile, Q1 = 0.25(n+1)th value = 4.75th value of sorted data = 120.75

Answer : b. 120.75

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28. IQR = Q3 - Q1 = 134.25 - 120.75 = 13.5

Answer : d. 13.5

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29.

∑x = 2312

∑x² = 303506

n = 18

Mean, x̅ = Ʃx/n = 2312/18 = 128.4444

Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(303506-(2312)²/18)/(18-1)] = 19.6176

Coefficient of variation, CV = s/x̅ = 0.1528 = 15.28%

Answer: c. 15.28%

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30. Median = 0.5(n+1)th value = 9.5th value of sorted data = 129.5

Answer : c. 129.5

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31. The sample Pearson’s coefficient of skewness = 3(median-Mean)/s = 0.162

Answer : c. 0.162

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32. The z-score of 135 is

(135-128.4444)/19.618 = 0.33

Answer : a. 0.33