Below are data collected by a researcher who measured the number of hours college students spend interacting with social media: X 3 3 4 4 5 5 8 Calculate the mean for the sample AND explain what the number tells you (how do you interpret this statistic?) Calculate the standard deviation AND explain what the number tells you (how do you interpret this statistic?) Using z-scores and the unit normal table, calculate the probability of spending 6 hours or MORE interacting with social media [ p (x > 6)|

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Below are data collected by a researcher who measured the number of hours college students spend interacting
with social media:
X
3
3
4
4
5
5
8
Calculate the mean for the sample AND explain what the number tells you (how do you interpret this statistic?)
Calculate the standard deviation AND explain what the number tells you (how do you interpret this statistic?)
Using z-scores and the unit normal table, calculate the probability of spending 6 hours or MORE interacting
with social media [ p (x > 6)|
Transcribed Image Text:Below are data collected by a researcher who measured the number of hours college students spend interacting with social media: X 3 3 4 4 5 5 8 Calculate the mean for the sample AND explain what the number tells you (how do you interpret this statistic?) Calculate the standard deviation AND explain what the number tells you (how do you interpret this statistic?) Using z-scores and the unit normal table, calculate the probability of spending 6 hours or MORE interacting with social media [ p (x > 6)|
Expert Solution
Step 1

From the provided data the mean and standard deviation are computed as,

Mean=x=ixin=3610=3.6

Thus, the mean is 3.6 which implies that on an average college student spends 3.6 hours interacting with social media.

Standard deviation=s=i(xi-x¯)2n-1=(0-3.6)2+(2-3.6)2+(2-3.6)2+(3-3.6)2+(3-3.6)2+(4-3.6)2(4-3.6)2+(5-3.6)2(5-3.6)2+(8-3.6)210-1=4.711111 =2.170509

Thus, standard deviation is, 2.170509, which implies the average distance between the data points and the mean. There is an average variation of 2.1705 hours from the mean.

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