a) What is the probability that a randomly selected car spends more than 2 minutes (i.e., 120 seconds) in the drive-through?
To solve this problem, we need to standardize the drive-through time using the formula z = (x - μ) / σ, where x is the drive-through time, μ is the mean, and σ is the standard deviation. Then we can look up the area under the standard normal curve corresponding to a z-score of z using a standard normal distribution table or a calculator.
In this case, we want to find the probability that a randomly selected car spends more than 2 minutes in the drive-through, which corresponds to a drive-through time of x > 120 seconds. To standardize this value, we use:
z = (x - μ) / σ = (120 - 137.5) / 27 = -0.648
Looking up the area under the standard normal curve corresponding to a z-score of -0.648, we find:
P(z > -0.648) = 0.7406
Therefore, the probability that a randomly selected car spends more than 2 minutes in the drive-through is approximately 0.7406 or 74.06%.
b) What is the probability that a randomly selected car spends between 90 and 150 seconds in the drive-through?
To find the probability that a randomly selected car spends between 90 and 150 seconds in the drive-through, we need to standardize these values using the formula z = (x - μ) / σ, where x is the drive-through time, μ is the mean, and σ is the standard deviation. Then we can find the area under the standard normal curve between the corresponding z-scores using a standard normal distribution table or a calculator.
To standardize 90 seconds and 150 seconds, we use:
z1 = (90 - 137.5) / 27 = -1.7593
z2 = (150 - 137.5) / 27 = 0.4629
Using a standard normal distribution table or a calculator to find the area under the standard normal curve between z1 = -1.7593 and z2 = 0.4629, we get:
P(-1.7593
(I hope that helps you)