John drives to work each morning and the trip takes an average of m = 38 minutes. The distribution of driving times is approximately normal with a standard deviation of s = 5 minutes. For a randomly selected morning, what is the probability that John's drive to work will take less than 35 minutes?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
.2743
To find the probability that John's drive to work will take less than 35 minutes, we need to calculate a z-score and then use a standard normal distribution table.
The formula to calculate the z-score is:
z = (x - μ) / σ
Where:
x = the value we want to find the probability for (in this case, 35 minutes)
μ = the mean of the distribution (given as 38 minutes)
σ = the standard deviation of the distribution (given as 5 minutes)
Substituting the given values into the formula:
z = (35 - 38) / 5
z = -3 / 5
z = -0.6
Now we need to find the probability associated with this z-score using a standard normal distribution table or calculator.
Looking up the z-score of -0.6 in the table, we find that the probability is approximately 0.2743.
Therefore, the probability that John's drive to work will take less than 35 minutes is approximately 0.2743.
To find the probability that John's drive to work will take less than 35 minutes, we need to use the concept of Z-scores and the standard normal distribution.
Step 1: Convert the given value of 35 minutes to a Z-score.
The formula for calculating Z-score is: Z = (X - μ)/σ
where X is the value we are interested in (35 minutes), μ is the mean (38 minutes), and σ is the standard deviation (5 minutes).
Z = (35 - 38) / 5
Z = -0.6
Step 2: Use the Z-score to find the corresponding probability.
We can use a standard normal distribution table or a calculator to find the probability. For a Z-score of -0.6, the table or calculator will give us the area to the left of this Z-score.
Using a standard normal distribution table or calculator, we find that the area to the left of -0.6 is approximately 0.2743.
Therefore, the probability that John's drive to work will take less than 35 minutes is approximately 0.2743 or 27.43%.
Note: The standard normal distribution assumes that the driving times are normally distributed, and this approximation may not hold exactly in the real world.