1. ## Simple quick question

Samuel has 2 choices of travel routes. The travel time for both routes are normally distributed. The travel time for Route A has a mean of 28 min, and std deviation of 5 min, whereas Route B has a mean of 30 min and std deviation of 2 min.

If Samuel and his brother start at 7.00 am, where Samuel takes Route A while his brother takes Route B. They have to reach before7.30 am. Find who has a higher probability of reaching earlier.
I did:

P(X<30), and my answer was Samuel has a higher prob. then his brother. However the answer given is Samuel's brother. Any help would be appreciated. TIA.

2. Route A (Samuel)

A: $\displaystyle \mu = 28, \sigma = 5$

$\displaystyle P(X<30) = P\left(Z<\frac{28-30}{5}\right) = P(Z<-0.4)$

Route B (Samuel's Brother)

B: $\displaystyle \mu = 30, \sigma = 2$

$\displaystyle P(X<30) = P\left(Z<\frac{30-30}{2}\right) = P(Z<0)$

Now

$\displaystyle P(Z<0) > P(Z<-0.4)$

therefore Samuel's brother has a better chance of arriving first.

3. Originally Posted by smmxwell
I did:

P(X<30), and my answer was Samuel has a higher prob. then his brother. However the answer given is Samuel's brother. Any help would be appreciated. TIA.

And no need for calculations since 30 is greater than the mean for route A while 30 is equal to the mean for route B.

4. Originally Posted by pickslides
Route A (Samuel)

A: $\displaystyle \mu = 28, \sigma = 5$

$\displaystyle P(X<30) = P\left(Z<\frac{28-30}{5}\right) = P(Z<-0.4)$ Mr F says: Sorry, but this Z-value is wrong. It should be 0.4.

Route B (Samuel's Brother)

B: $\displaystyle \mu = 30, \sigma = 2$

$\displaystyle P(X<30) = P\left(Z<\frac{30-30}{2}\right) = P(Z<0)$

Now

$\displaystyle P(Z<0) > P(Z<-0.4)$

therefore Samuel's brother has a better chance of arriving first.
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