When Ava commutes to work, the amount of time it takes her to arrive is normally distributed with a mean of 35 minutes and a standard deviation of 3.5 minutes. What is the probability that on a given day, her commute will be between 33 and 42 minutes, to the nearest thousandth?
To find the probability that Ava's commute will be between 33 and 42 minutes, we need to calculate the z-scores for both endpoints and then use the standard normal distribution table.
First, we find the z-score for 33 minutes:
z = (33 - 35) / 3.5 = -2 / 3.5 = -0.5714
Next, we find the z-score for 42 minutes:
z = (42 - 35) / 3.5 = 2 / 3.5 = 0.5714
Using the standard normal distribution table, we find the cumulative probability for the z-scores:
P(z < -0.5714) = 0.2852
P(z < 0.5714) = 0.7148
Therefore, the probability that Ava's commute will be between 33 and 42 minutes is given by:
P(-0.5714 < z < 0.5714) = P(z < 0.5714) - P(z < -0.5714) = 0.7148 - 0.2852 = 0.4296
So, the probability that Ava's commute will be between 33 and 42 minutes is approximately 0.430 (to the nearest thousandth).