Question

Suppose the average wait time for a sandwich at a local deli is 55 min with a standard deviation of 33 min. Is it unusual to wait over 1212 minutes to get a sandwich?

Answer 1

Answer:

The wait  time is unusual because the probability value obtained is less than 2 standard deviation above the mean

Step-by-step explanation:

From the question we are told that

     The average wait time is  $\mu = 55 \ min$

     The standard deviation is  $\sigma = 33 \ min$

Generally the probability of  waiting over 1212 minutes is

     $P(X > 1212) = P(\frac{X - \mu}{ \sigma } > \frac{ 1212 - 55}{ 33} )$

     $P(X > 1212) = P(\frac{X - \mu}{ \sigma } > \frac{ 1212 - 55}{ 33} )$

$\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )$

So

       $P(X > 1212) = P(Z > 35 )$

From z table  the probability of  (Z  >  35 ) is  

        $P(Z > 35 ) = 0$

Hence

    $P(X > 1212) = 0$

The  time is unusual because the probability value obtained is less than 2 standard deviation above the mean