The distribution of rhesus monkey tail lengths is bell-shaped, unimodal, and approximately symmetric. The average tail length is 6.8 cm and the standard deviation is 0.44 cm. Roscoe has a tail that is 10.2 cm long. What conclusion can we make based on the information given?
a. We can apply the empirical rule to conclude that Roscoe is a potential outlier because he falls more than three standard deviations away from the mean.
b. We can apply the empirical rule to conclude that Roscoe is not a potential outlier because he falls within three standard deviations away from the mean.
c. We cannot apply the empirical rule because the distribution does not fit the criteria for the empirical rule.
d. There is not enough information given to make any conclusions about potential outliers
Here' the answer. Option A is correct. But, please read through the answer for detailed explanation. Let me know in case you want any help. Please dont hesitate to give a "thumbs up" for the answer, in case you're satisfied with it. Have a good day!
Its clearly given that the distribution is UNIMODAL, nearly SYMMETRICAL and BELL)SHAPED.
All the 3 characters of the distribution in BOLD, lead us to say that the distribution is normal. ie.. the empirical rule can be applied here.
SO, Mean = 6.8
Stdev = .44
Roscoe' tail is 10.2 cm long. Lets normalize it using the distribution' params of Mean, and Stdev
Hence, Z of Roscoe = (X-Mu)/Sigma = (10.2-6.8)/.44 = 7.73 , which in itself is more than 3
We can apply the empirical rule, which states that a Z score of more than 3 means a potential outlier
Here a Z score of 7 is more than 3, and hence Roscoe' tails is an OUTLIER
Option a. is CORRECT
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