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

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

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((X - \mu)/( \sigma ) > ( 1212 - 55)/( 33) )`

`P(X > 1212) = P((X - \mu)/( \sigma ) > ( 1212 - 55)/( 33) )`

`(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