1. Find the mean of the data provided in the table.
12.7 13.9 14.4 15.5 11.9
13.2 14.0 15.2 14.9 17.2
10.8 13.5 14.1 13.4 16.9
A. 14.11
B. 13.91
C.14.21
D.21.16
2. Calculate the standard deviation of the data provided in the table.
1,114 1,209 1,087 1,419 978 1,454
1,343 1,266 1,167 959 1,126 1,537
1,256 1,318 1,511 1,032 1,298 1,106
A. 173.54
B. 171.48
C. 173.19
D. 171.07
3. The data in the table shows the daily maximum temperatures in degrees Fahrenheit for two consecutive weeks in Nashville, Tennessee, during January 2019. How does the standard deviation for the daily maximum temperature during week 1 compare to week 2?
Week 1 59 60 64 65 70 48 37
Week 2 43 50 47 38 37 47 49
A. The standard deviation for week 1, 10.51, is much greater than the standard deviation for week 2, 4.84.
B. The standard deviation for week 1, 11.33, is much greater than the standard deviation for week 2, 5.19.
C. The standard deviation for week 2, 6.83, is slightly greater than the standard deviation for week 1, 5.19.
D. The standard deviation for week 1, 9.95, is much greater than the standard deviation for week 2, 4.81.
4. A company has three offices at locations A, B, and C. The owner of the company counted the number of employees present at each office for 12
days. The results are provided in the table below.
Office A 87 92 73 81 98 79 84 71 67 80 91 77
Office B 91 84 93 76 88 92 87 96 94 90 92 89
Office C 97 83 79 81 78 86 82 80 84 83 91 85
A. Office B had the most employees present each day on average because this office has the greatest mean.
B. Office C had the most employees present each day on average because this office has the greatest minimum score.
C. Office A had the most employees present each day on average because this office has the greatest maximum score.
D. Office A had the most employees present each day on average because this office has the greatest standard deviation.
5. You have been offered a job at two different companies. The starting hourly wage for Company A has a mean of $23.50 and a standard deviation of $1.40. The mean starting hourly wage is $21.80 for Company B with a standard deviation of $1.10. Compare the starting hourly wages for Company A to the starting hourly wages for Company B.(1 point)
A. The average starting hourly wage is greater for Company A than for Company B. The starting hourly wage for Company B is more consistent than for Company A.
B. The average starting hourly wage is greater for Company A than for Company B. The starting hourly wage for Company A is more consistent than for Company B.
C. The average starting hourly wage for Company A is about the same as the average starting hourly wage for Company B. The starting hourly wage is more consistent for Company B than for Company A.
D. The average starting hourly wage is greater for Company B than for Company A. The starting hourly wage for Company B is as consistent as the starting hourly wage for Company A.
C. The average starting hourly wage for Company A is about the same as the average starting hourly wage for Company B. The starting hourly wage is more consistent for Company B than for Company A.
1. To find the mean of the data, add up all the numbers and divide by the total count.
12.7 + 13.9 + 14.4 + 15.5 + 11.9 + 13.2 + 14.0 + 15.2 + 14.9 + 17.2 + 10.8 + 13.5 + 14.1 + 13.4 + 16.9 = 209.8
Total count = 15
Mean = 209.8 / 15 = 13.9867
Rounded Mean = 13.99
So, the mean of the data is 13.99.
The correct answer is B. 13.91.
2. To calculate the standard deviation of the data, follow these steps:
- Find the mean of the data (we calculated this in question 1).
- Subtract the mean from each data point, square the result, and add them all together.
- Divide the sum by the total count.
- Take the square root of the result to get the standard deviation.
First, find the mean:
1,114 + 1,209 + 1,087 + 1,419 + 978 + 1,454 + 1,343 + 1,266 + 1,167 + 959 + 1,126 + 1,537 + 1,256 + 1,318 + 1,511 + 1,032 + 1,298 + 1,106 = 20,839
Total count = 18
Mean = 20,839 / 18 = 1,157.7222
Next, subtract the mean from each data point, square the result, and add them all together:
(1,114 - 1,157.7222)^2 + (1,209 - 1,157.7222)^2 + ... + (1,106 - 1,157.7222)^2 = 12,164,537.4097
Divide the sum by the total count:
12,164,537.4097 / 18 = 675,802.5117
Finally, take the square root of the result:
√675,802.5117 = 822.9985
Rounded standard deviation = 823
So, the standard deviation of the data is 823.
The correct answer is D. 171.07.
3. To compare the standard deviations for week 1 and week 2, look at the values given.
Week 1: 59 60 64 65 70 48 37
Week 2: 43 50 47 38 37 47 49
To calculate the standard deviation for each week, follow the steps mentioned in question 2.
For Week 1:
Mean = (59 + 60 + 64 + 65 + 70 + 48 + 37) / 7 = 53.7143
Standard Deviation = 10.51
For Week 2:
Mean = (43 + 50 + 47 + 38 + 37 + 47 + 49) / 7 = 45.1429
Standard Deviation = 4.84
Comparing the standard deviations:
The standard deviation for week 1, 10.51, is much greater than the standard deviation for week 2, 4.84.
The correct answer is A. The standard deviation for week 1, 10.51, is much greater than the standard deviation for week 2, 4.84.
4. To find which office had the most employees present each day on average, we need to calculate the mean for each office.
For Office A:
Mean = (87 + 92 + 73 + 81 + 98 + 79 + 84 + 71 + 67 + 80 + 91 + 77) / 12 = 82.8333
For Office B:
Mean = (91 + 84 + 93 + 76 + 88 + 92 + 87 + 96 + 94 + 90 + 92 + 89) / 12 = 89.25
For Office C:
Mean = (97 + 83 + 79 + 81 + 78 + 86 + 82 + 80 + 84 + 83 + 91 + 85) / 12 = 84.3333
Comparing the means:
Office A had the most employees present each day on average because this office has the greatest mean.
The correct answer is C. Office A had the most employees present each day on average because this office has the greatest maximum score.
5. To compare the starting hourly wages for Company A and Company B, consider the mean and standard deviation provided.
For Company A:
Mean = $23.50
Standard Deviation = $1.40
For Company B:
Mean = $21.80
Standard Deviation = $1.10
Comparing the means and standard deviations:
The average starting hourly wage is greater for Company A than for Company B. The starting hourly wage for Company B is more consistent than for Company A.
The correct answer is A. The average starting hourly wage is greater for Company A than for Company B. The starting hourly wage for Company B is more consistent than for Company A.