1. Travis is testing how far he can throw a baseball to prepare himself for the season. He makes 16
throws and records the length of each throw in feet. The results are provided in the accompanying table.
236 240 232 242 238 235 228 245
247 239 234 238 241 227 243 238
Travis says that the histogram provided below could be used to represent the data.
Show whether the histogram Travis created is correct and, if not, explain how the histogram could be corrected.
3 answers
It is not possible to determine if Travis' histogram is correct without seeing the actual histogram. However, if the histogram accurately represents the data, then it should show the frequency of each range of throw lengths. One way to correct the histogram would be to adjust the range of each bar so that they accurately represent the frequency of the data. Additionally, Travis could label the x-axis with the range of throw lengths and the y-axis with the frequency or number of throws.
1. For a project, Fay kept track of the total annual amount of precipitation in inches for two cities over the last 12
years. The results are shown in the table below.
City A 41.50 69.43 48.15 39.23 37.03 66.02 47.62 68.38 38.69 52.48 70.03 43.74
City M 48.14 61.59 44.05 33.14 32.41 72.92 48.49 50.21 33.70 48.69 49.06 44.47
Find the mean and standard deviation for each city, and then compare the means and standard deviations. Round the means and standard deviations to two decimal places.
years. The results are shown in the table below.
City A 41.50 69.43 48.15 39.23 37.03 66.02 47.62 68.38 38.69 52.48 70.03 43.74
City M 48.14 61.59 44.05 33.14 32.41 72.92 48.49 50.21 33.70 48.69 49.06 44.47
Find the mean and standard deviation for each city, and then compare the means and standard deviations. Round the means and standard deviations to two decimal places.
To find the mean for each city, we will add up all the precipitation amounts for that city and divide by the number of years (12).
For City A:
Mean = (41.50 + 69.43 + 48.15 + 39.23 + 37.03 + 66.02 + 47.62 + 68.38 + 38.69 + 52.48 + 70.03 + 43.74) / 12
Mean = 52.30 inches
For City M:
Mean = (48.14 + 61.59 + 44.05 + 33.14 + 32.41 + 72.92 + 48.49 + 50.21 + 33.70 + 48.69 + 49.06 + 44.47) / 12
Mean = 46.26 inches
To find the standard deviation for each city, we can use the following formula:
s = √[(Σx - μ)² / (n - 1)]
where s is the standard deviation, Σx is the sum of all the data points, μ is the mean, and n is the number of data points.
For City A:
s = √[((41.50 - 52.30)² + (69.43 - 52.30)² + ... + (43.74 - 52.30)²) / (12 - 1)]
s = 13.97 inches
For City M:
s = √[((48.14 - 46.26)² + (61.59 - 46.26)² + ... + (44.47 - 46.26)²) / (12 - 1)]
s = 9.96 inches
Comparing the means, we can see that City A has a higher average annual precipitation than City M.
Comparing the standard deviations, we can see that City A has a larger variability in annual precipitation than City M.
For City A:
Mean = (41.50 + 69.43 + 48.15 + 39.23 + 37.03 + 66.02 + 47.62 + 68.38 + 38.69 + 52.48 + 70.03 + 43.74) / 12
Mean = 52.30 inches
For City M:
Mean = (48.14 + 61.59 + 44.05 + 33.14 + 32.41 + 72.92 + 48.49 + 50.21 + 33.70 + 48.69 + 49.06 + 44.47) / 12
Mean = 46.26 inches
To find the standard deviation for each city, we can use the following formula:
s = √[(Σx - μ)² / (n - 1)]
where s is the standard deviation, Σx is the sum of all the data points, μ is the mean, and n is the number of data points.
For City A:
s = √[((41.50 - 52.30)² + (69.43 - 52.30)² + ... + (43.74 - 52.30)²) / (12 - 1)]
s = 13.97 inches
For City M:
s = √[((48.14 - 46.26)² + (61.59 - 46.26)² + ... + (44.47 - 46.26)²) / (12 - 1)]
s = 9.96 inches
Comparing the means, we can see that City A has a higher average annual precipitation than City M.
Comparing the standard deviations, we can see that City A has a larger variability in annual precipitation than City M.
1 answer