# standard deviation

• May 29th 2009, 07:21 PM
dan123
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
• May 30th 2009, 02:55 AM
pickslides
$Z= \frac{X-\mu}{\sigma}$

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

$P(X>60) = P\left(Z>\frac{60-25}{12}\right) = P(Z>2.92)$
• May 30th 2009, 02:26 PM
matheagle
mean,45 min....not 25
And I guess the underlying distribution is normal, otherwise I have no idea how to proceed.
• May 31st 2009, 02:08 PM
pickslides
Quote:

Originally Posted by matheagle
mean,45 min....not 25

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

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

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

Quote:

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.
• May 31st 2009, 08:04 PM
mr fantastic
Quote:

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.
• May 31st 2009, 08:49 PM
pickslides
Quote:

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.