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At one university, the mean distance commuted to campus by students is 19.0 miles, with a standard deviation of 5.1 miles. Suppose that the commute
distances are normally distributed. Complete the following statements.
(ə) Approximately 99.7% of the students have commute distances between__miles
and__miles
(b) Approximately ??
of the students have commute distances between 8.8 miles and 29.2 miles.

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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