The time a group of high school students arrive home from school each day was found to be normally distributed. The mean time was 3:15pm and the times had a standard deviation of 15 minutes. What is the probability that a student chosen at random arrives home from school before 2:30pm?

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Caculate Z score for this time.

Z = (x - μ)/SD

x =2:30, μ = 3:15, SD = 15

Look up the Z score in the back of your statistics text on a table of areas under the normal distribution to get your probability.

I hope this helps. Thanks for asking.

To find the probability that a student chosen at random arrives home from school before 2:30pm, we need to find the area under the normal distribution curve to the left of the given time.

First, we need to standardize the time using the z-score formula:

z = (x - μ) / σ

where:
x = 2:30pm
μ = mean = 3:15pm
σ = standard deviation = 15 minutes

Converting 2:30pm to minutes, we get 2:30pm = 2 * 60 + 30 = 150 minutes.

Now we can calculate the z-score:

z = (150 - 195) / 15 = -3

Next, we will use a z-table (also known as a standard normal distribution table) to find the corresponding cumulative probability. The cumulative probability gives us the probability of an event occurring up to a certain z-score.

Looking up the z-score of -3 in the z-table, we find that the cumulative probability is approximately 0.0013.

Therefore, the probability that a student chosen at random arrives home from school before 2:30pm is approximately 0.0013, or 0.13%.