Dave drives to work each morning at about the same time. His commute time is normally distributed with a mean of 50 minutes and a standard deviation of 6 minutes.  What percentage of the time will his commute time be greater than 58 minutes?

To answer this question, we need to use the properties of the normal distribution.

Step 1: Calculate the Z-score. The Z-score is a measure of how many standard deviations an element is from the mean. The formula for the Z-score is:

Z = (X - μ) / σ

where: X is the value we are interested in (in this case, 58 minutes), μ is the mean (in this case, 50 minutes), and σ is the standard deviation (in this case, 6 minutes).

So, Z = (58 - 50) / 6 = 1.33

Step 2: Look up the Z-score in the Z-table. The Z-table tells us what percentage of the population is below a certain Z-score. For a Z-score of 1.33, the Z-table tells us that about 90.77% of the population is below this score.

Step 3: Since we want to know the percentage of the time his commute time will be greater than 58 minutes, we need to find the percentage of the population above this Z-score. To do this, we subtract the percentage below the Z-score from 100%.

So, 100% - 90.77% = 9.23%

Therefore, Dave's commute time will be greater than 58 minutes about 9.23% of the time.

To answer this question, we need to use the properties of the normal distribution.

Step 1: Calculate the Z-score The Z-score is a measure of how many standard deviations an element is from the mean. We can calculate the Z-score for 58 minutes using the formula:

Z = (X - μ) / σ

where: X = 58 minutes μ = mean = 50 minutes σ = standard deviation = 6 minutes

So, Z = (58 - 50) / 6 = 1.33

Step 2: Look up the Z-score in the Z-table The Z-table tells us what percentage of the population is below a certain Z-score. For a Z-score of 1.33, the value is approximately 0.9082. This means that 90.82% of the time, Dave's commute will be less than 58 minutes.

Step 3: Find the percentage of the time the commute will be greater than 58 minutes Since the total percentage must be 100%, the percentage of the time that the commute will be greater than 58 minutes is 100% - 90.82% = 9.18%.

So, Dave's commute time will be greater than 58 minutes about 9.18% of the time.

To answer this question, we need to use the properties of the normal distribution.

Step 1: Calculate the Z-score The Z-score is a measure of how many standard deviations an element is from the mean. We can calculate the Z-score for 58 minutes using the formula:

Z = (X - μ) / σ

where: X = 58 minutes μ = mean = 50 minutes σ = standard deviation = 6 minutes

So, Z = (58 - 50) / 6 = 1.33

Step 2: Look up the Z-score in the Z-table The Z-table tells us what percentage of the population is below a certain Z-score. Looking up a Z-score of 1.33 in the Z-table gives us a value of 0.9082. This means that 90.82% of the time, Dave's commute will be less than 58 minutes.

Step 3: Find the percentage of the time the commute will be greater than 58 minutes Since we know that 90.82% of the time the commute will be less than 58 minutes, we can subtract this from 100% to find out what percentage of the time the commute will be greater than 58 minutes.

100% - 90.82% = 9.18%

So, Dave's commute will be greater than 58 minutes about 9.18% of the time.