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At one university, the mean distance commuted to campus by students is 19.0 miles, with a standard deviation of 4.2 miles. Suppose that the commutedistances are normally distributed. Complete the following statements.(a) Approximately 95% of the students have commute distances between ? milesand ? miles(b) Approximately ? of the students have commute distances between 6.4 miles and 31.6 miles.

At one university, the mean distance commuted to campus by students is 19.0 miles-example-1
User Mike Cornell
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1 Answer

10 votes
10 votes
Part a.

From the given information, we know that the mean and standard deviation are, respectively,


\begin{gathered} \mu=19 \\ \sigma=4.2 \end{gathered}

From the 68-95-99 rule, we know that approximately 95% falls between 2 standard deviation of the mean, that is,


-2=(x-19)/(4.2)...(A)

and


2=(x-19)/(4.2)...(B)

From equation (A), we have


x-19=-8.4

then


x=10.6

Now, from equation (B), we get


\begin{gathered} x-19=8.4 \\ then \\ x=27.4 \end{gathered}

Therefore, the answer for part a is: Approximately 95% of the students have commute between 10.6 and 27.4 miles

Part b.

In this case, we need to find the z score value for 6.4 miles and 31.6 miles and then obtain the corresponding probabilty from the z-table.

For 6.4 miles, the z score is


z=(6.4-19)/(4.2)=-3

and for 31.6 miles, the z score is


z=(31.6-19)/(4.2)=3

Now, we need to find the corresponding probability between z=-3 and z=3, which is 0.9973

Therefore, the answer for part b is: Approximately. 0.9973 of the students have commute distances between 6.4 miles and 31.6 miles

At one university, the mean distance commuted to campus by students is 19.0 miles-example-1
User Garo Yeriazarian
by
3.0k points
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