vareriya wants to determine the likely voting outcome for the new town council election. whick sample is likely yo yield mist ba
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<span>he box plots below show attendance at a local movie theater and high school basketball games:
two box plots shown. The top one is labeled Movies. Minimum at 60, Q1 at 65, median at 95, Q3 at 125, maximum at 150. The bottom box plot is labeled Basketball games. Minimum at 90, Q1 at 95, median at 125, Q3 at 145, maximum at 150.
Which of the following best describes how to measure the spread of the data?
The IQR is a better measure of spread for movies than it is for basketball games.
The standard deviation is a better measure of spread for movies than it is for basketball games.
The IQR is the best measurement of spread for games and movies.
The standard deviation is the best measurement of spread for games and movies.</span>
The standard deviation is 9.27. The typical heart rate for the data set varies from the mean by an average of 9.27 beats per minute.
<h3>How to determine the standard deviation of the data set?</h3>The dataset is given as:
Heart Rate Frequency
60 1
65 3
70 4
75 12
80 8
85 15
90 9
95 5
100 3
Calculate the mean using
Mean = Sum/Count
So, we have
Mean = (60 * 1 + 65 * 3 + 70 * 4 + 75 * 12 + 80 * 8 + 85 * 15 + 90 * 9 + 95 * 5 + 100 * 3)/(1 + 3 + 4 + 12 + 8 + 15 + 9 + 5 + 3)
Evaluate
Mean = 82.25
The standard deviation is
So, we have:
SD = √[1 * (60 - 82.25)^2 + 3 * (65 - 82.25)^2 + 4 * (70 - 82.25)^2 + 12 * (75 - 82.25)^2 + 8 * (80 - 82.25)^2 + 15 * (85 - 82.25)^2 + 9 * (90 - 82.25)^2 + 5 * (95 - 82.25)^2 + 3 * (100 - 82.25)^2)]/[(1 + 3 + 4 + 12 + 8 + 15 + 9 + 5 + 3 - 1)]
This gives
SD = √85.9533898305
Evaluate
SD = 9.27
Hence. the standard deviation is 9.27. The typical heart rate for the data set varies from the mean by an average of 9.27 beats per minute.
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