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Sagot :
The approximately 99.7% of the students have commute distances between 13.9 miles and 24.1 miles, the area within two standard deviation is 95%
What is the standard deviation?
It is defined as the measure of data disbursement, It gives an idea about how much is the data spread out.
a) We have:
mean u = 19.0 miles,
[tex]\rm \sigma = 5.1 \ miles[/tex]
According to the empirical rule, the area between [tex]\mu - \sigma[/tex] and [tex]\mu + \sigma[/tex] is 99.7%
[tex]\mu - \sigma = 19- 5.1 = 13.9[/tex]
[tex]\mu + \sigma = 19+5.1 = 24.1[/tex]
b) = P(8.8 < X < 29.2)
[tex]\rm P(\dfrac{8.8-\mu}{\sigma } < \dfrac{X-\mu}\sigma < \dfrac{29.2-\mu}{\sigma})[/tex]
[tex]\rm P(\dfrac{8.8-19}{5.1} < Z < \dfrac{29.2-19}{5.1})[/tex]
P(-2 < Z < 2)
Thus, the approximately 99.7% of the students have commute distances between 13.9 miles and 24.1 miles, the area within two standard deviation is 95%
Learn more about the standard deviation here:
brainly.com/question/12402189
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