1. ## standard deviation

data was collected outside a popular new restaruant to determine the ean waiting time to be seated at a table

mean,45 min
standard deviation,12 min

determine the probability that a person has to wait less than twenty minutes.

also,determine the probability a person has to wait more than an hour.

do I have to calculate the z scores then look on the table,I've always hated standard deviation stuff I never get it

2. $Z= \frac{X-\mu}{\sigma}$

$P(X<20) = P\left(Z<\frac{20-25}{12}\right) = P(Z<-0.42)$

$P(X>60) = P\left(Z>\frac{60-25}{12}\right) = P(Z>2.92)$

3. mean,45 min....not 25
And I guess the underlying distribution is normal, otherwise I have no idea how to proceed.

4. Originally Posted by matheagle
mean,45 min....not 25
$Z= \frac{X-\mu}{\sigma}$

$P(X<20) = P\left(Z<\frac{20-45}{12}\right) = P(Z<-2.08)$

$P(X>60) = P\left(Z>\frac{60-45}{12}\right) = P(Z>1.25)$

Originally Posted by matheagle
And I guess the underlying distribution is normal, otherwise I have no idea how to proceed.
most importantly! The mention of z-scores was the only hint.

5. Originally Posted by pickslides
$Z= \frac{X-\mu}{\sigma}$

$P(X<20) = P\left(Z<\frac{20-45}{12}\right) = P(Z<-2.08)$

$P(X>60) = P\left(Z>\frac{60-45}{12}\right) = P(Z>1.25)$

most importantly! The mention of z-scores was the only hint.
z-scores were mentioned by the student (almost as a hopeful stab in the dark), not in the question itself. So it's not quite in the league of a hint.

6. Originally Posted by mr fantastic
z-scores were mentioned by the student (almost as a hopeful stab in the dark), not in the question itself. So it's not quite in the league of a hint.
fair call.