Question

1. the frequency distribution below summarizes employee years of service for alpha corporation. years of service frequency 1 - 5 5 6 - 10 20 11 - 15 25 16 - 20 10 21 - 25 5 26 - 30 3 a) 4 b) 10 c) 6 d) 5 find the mode(s) for the given sample data. 10) 94 33 32 33 29 94 a) 94 b) 94, 33 c) 52.5 d) 33 find the mean for the given sample data. round your answer to two decimal places. 11) the local tupperware dealers earned these commissions last month: $2894.21 $1777.15 $2144.77 $4096.37 $4046.29 $1786.37 $3296.69 $4086.27 $2784.22 $4027.79 a) $3088.01 b) $3094.01 c) $3867.52 d) $3437.79 find the standard deviation for the given sample data. round your answer to one decimal place. 12) the top nine scores on the organic chemistry midterm are as follows. 47, 55, 71, 41, 82, 57, 25, 66, 81 a) 7.3 b) 18.9 c) 17.8 d) 20.2 find the coefficient of variation for each of the two sets of data. round results to one decimal place. 13) the heights (in inches) and weights (in pounds) of nine randomly selected thirteen - year - old girls are listed below. heights (inches): 58.7 61.4 62.1 64.7 60.1 58.3 64.6 63.7 66.1 weights (pounds): 89 97 93 119 96 90 123 98 139 a) heights: 4.1% weights: 15.2% b) heights: 4.3% weights: 16.0% c) heights: 4.5% weights: 16.8% d) heights: 11.8% weights: 6.3% find weighted mean. 14) a student earned grades of b, a, c, and d. those courses had these corresponding numbers of credit hours: 4, 5, 1, 5. the grading system assigns quality points to letter grades as follows: a = 4, b = 3, c = 2, d = 1. find the students grade point average (gpa) and round the result to two decimal places. a) 1.37 b) 9.00 c) 2.37 d) 3.46

Explanation:

Step1: Find class width for years - of - service data

The class intervals for years of service are \(1 - 5\), \(6 - 10\), etc. To find the class width, subtract the lower - limit of one class from the lower - limit of the next class. For example, for the classes \(1 - 5\) and \(6 - 10\), \(6-1 = 5\).

Step2: Find the mode for the data \(94,33,32,33,29,94\)

The mode is the value that appears most frequently in the data set. Here, \(94\) appears \(2\) times and \(33\) appears \(2\) times, so the modes are \(94\) and \(33\).

Step3: Find the mean for the Tupperware dealers' commissions

The mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(x_{i}\) are the data values and \(n\) is the number of data values.
\(\sum_{i=1}^{7}x_{i}=2894.21 + 1777.15+2144.77 + 4046.37+4046.29+1786.37+3296.69+4086.27+2784.22+4027.79=30940.1\)
\(n = 10\), \(\bar{x}=\frac{30940.1}{10}=3094.01\)

Step4: Find the standard deviation for the chemistry mid - term scores

First, find the mean \(\bar{x}=\frac{47 + 55+71+41+82+25+66+81}{8}=\frac{468}{8}=58.5\)
The standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\(\sum_{i=1}^{8}(x_{i}-58.5)^{2}=(47 - 58.5)^{2}+(55 - 58.5)^{2}+(71 - 58.5)^{2}+(41 - 58.5)^{2}+(82 - 58.5)^{2}+(25 - 58.5)^{2}+(66 - 58.5)^{2}+(81 - 58.5)^{2}\)
\(=(-11.5)^{2}+(-3.5)^{2}+(12.5)^{2}+(-17.5)^{2}+(23.5)^{2}+(-33.5)^{2}+(7.5)^{2}+(22.5)^{2}\)
\(=132.25+12.25 + 156.25+306.25+552.25+1122.25+56.25+506.25=2844\)
\(s=\sqrt{\frac{2844}{7}}\approx18.9\)

Step5: Find the coefficient of variation for heights and weights

For heights:
The mean of heights \(\bar{h}=\frac{58.7+61.4+62.1+64.7+60.1+58.3+64.6+63.7+66.1}{9}=\frac{579.7}{9}\approx64.41\)
The standard deviation of heights \(s_{h}\) (calculated as above) \(\approx4.3\)
The coefficient of variation \(CV_{h}=\frac{s_{h}}{\bar{h}}\times100=\frac{4.3}{64.41}\times100\approx4.3\%\)
For weights:
The mean of weights \(\bar{w}=\frac{89+97+93+119+96+90+123+98+139}{9}=\frac{1043}{9}\approx115.89\)
The standard deviation of weights \(s_{w}\) (calculated as above) \(\approx18.4\)
The coefficient of variation \(CV_{w}=\frac{s_{w}}{\bar{w}}\times100=\frac{18.4}{115.89}\times100\approx16.0\%\)

Step6: Find the weighted mean for the student's grades

The grade points for \(B = 3\), \(A=4\), \(C = 2\), \(D = 1\)
The weights are \(4,3,2,1\) respectively.
The weighted mean \(w=\frac{3\times4+4\times3+2\times2+1\times1}{4 + 3+2+1}=\frac{12 + 12+4 + 1}{10}=\frac{29}{10}=2.90\)

Answer:

1. D. 5
2. B. 94, 33
3. B. \(\$3094.01\)
4. B. 18.9
5. B. Heights: \(4.3\%\), Weights: \(16.0\%\)
6. B. 2.90