John drives to work each morning and the trip takes an average of μ = 38 minutes.
The distribution of driving times is approximately normal with a standard deviation of σ = 5
minutes. For a randomly selected morning, what is the probability that John’s drive to work
will take less than 35 minutes?
p=0.2743
-0.6---> z-score = 0.2743
Use z-scores.
z = (x - mean)/sd
With your data:
z = (35 - 38)/5
Finish the calculation, then check a z-table for the probability of less than 35 minutes using the z-score.
I hope this will help get you started.
500
To find the probability that John's drive to work will take less than 35 minutes, we can use the information given about the average (mean) driving time, the standard deviation, and the assumption that the distribution is approximately normal.
First, we need to standardize the value of 35 minutes using the z-score formula:
z = (x - μ) / σ
where x is the specific value (35 minutes), μ is the mean (38 minutes), and σ is the standard deviation (5 minutes).
Substituting the values into the formula:
z = (35 - 38) / 5
z = -0.6
We can now use a standard normal distribution table or calculator to find the probability associated with a z-value of -0.6. This will give us the probability of John's drive to work taking less than 35 minutes.
Using a standard normal distribution table, we can look up the value corresponding to -0.6. The table gives the area under the curve to the left of a given z-value. In this case, the table indicates 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%.