Suppose that the travel time from your home to your office is normally distributed
with a mean of 40 minutes and standard deviation of 7 minutes. If you want to be
95% certain that you will not be late for an office appointment at 1:00pm, what is the
latest time that you should leave home?
From a normal distribution table (example
http://www.math.unb.ca/~knight/utility/NormTble.htm
)
look up the z-value for 95% probability, which is 1.645.
μ=40 minutes, σ=7 minutes.
Calculate n from the transformation of x to z as follows:
z=(x-μ)/σ
x=σz+μ
Everything on the right hand side is know, so calculate x required.
0.5
To determine the latest time you should leave home to be 95% certain that you will not be late, we need to calculate the z-score corresponding to the desired level of confidence.
The z-score is calculated using the formula:
z = (x - μ) / σ
Where:
- x is the desired travel time in minutes,
- μ is the mean travel time (40 minutes in this case),
- σ is the standard deviation (7 minutes in this case).
To find the z-score corresponding to a 95% confidence level, we'll use the standard normal distribution (also called the Z-distribution).
The standard normal distribution has a mean of 0 and a standard deviation of 1.
Since we want a 95% confidence level, we need to find the z-score that corresponds to a cumulative probability of 0.95.
Using a standard normal distribution table or a statistical software, we find that the z-score corresponding to a cumulative probability of 0.95 is approximately 1.645.
Now we can calculate the latest time to leave home by rearranging the z-score formula:
x = z * σ + μ
x = 1.645 * 7 + 40
x ≈ 11.515 + 40
x ≈ 51.515
Therefore, you should leave home no later than 51.515 minutes before your office appointment at 1:00 pm. In practical terms, you should plan to leave home around 12:08 pm to be 95% certain that you will not be late.
To determine the latest time you should leave home to be 95% certain that you will not be late for your appointment, you need to calculate the upper limit of the travel time distribution.
First, let's find the z-score corresponding to a 95% confidence level. The z-score represents how many standard deviations away from the mean (40 minutes) we need to go to capture a certain percentage of the data. Since we want to be 95% certain, we need to find the z-score that captures 95% of the data.
The z-score is calculated using the standard normal distribution table or a statistical calculator. The z-score corresponding to a 95% confidence level is approximately 1.96.
Next, we'll use the formula to calculate the upper limit (UL) of the travel time distribution:
UL = mean + (z * standard deviation)
UL = 40 + (1.96 * 7)
UL ≈ 40 + 13.72
UL ≈ 53.72
Therefore, the latest time you should leave home is 53.72 minutes before your office appointment. If your appointment is at 1:00pm, you should leave home no later than 1:00pm - 53.72 minutes, which is approximately 12:06pm.