Question

A study found that the mean amount of time cars spent in drive-throughs of a certain fast-food restaurant was 148.4 seconds. Assuming drive-through times are normally distributed with a standard deviation of 26 seconds, complete parts (a) through (d) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Your mur pur us. (d) Would it be unusual for a car to spend more than 3 minutes in the restaurant's drive-through? Why? The probability that a car spends more than 3 minutes in the restaurant's drive-through is so it would not be unusual, since the probability is greater than 0.05. (Round to four decimal places as needed.)

Answer 1

Solution :

Given that ,

mean = $\mu$ = 148.4 seconds

standard deviation = $\sigma$ = 26 seconds

3 minutes = 180 seconds

P(x > 180) = 1 - p( x< 180)

=1- p P[(x - $\mu$ ) / $\sigma$ < (180 - 148.4) / 26]

=1- P(z < 1.22)

Using z table,

= 1 - 0.8888

= 0.1112