The average travel time to work for a person living and working in​ Kokomo, Indiana is 17 minutes. Suppose the standard deviation of travel time to work is 4.5 minutes and the distribution of travel time is approximately normally distributed.

What is the travel time to work that separates the bottom​ 2.5% of people with the median or​ 50%? Round to the nearest tenth of a minute.

50% in a normal distribution = median = mean

Z = (score-mean)/SD

Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability (.025) and its Z score. Insert into equation above and solve for score.

Subtract one score from the other.

Im still not understanding

To find the travel time that separates the bottom 2.5% of people with the median (50%), we need to calculate the z-score for the given percentile and then convert it back to the original travel time using the formula:

z = (x - μ) / σ

where:
z = z-score
x = travel time to work
μ = mean travel time
σ = standard deviation

In this case, we're given the mean travel time (μ) as 17 minutes and the standard deviation (σ) as 4.5 minutes. The z-score corresponding to the bottom 2.5% is approximately -1.96 (as it is on the left side of the distribution). Let's calculate the travel time (x):

-1.96 = (x - 17) / 4.5

To solve for x, we can rearrange the equation:

x - 17 = -1.96 * 4.5
x - 17 = -8.82
x = 17 - 8.82
x ≈ 8.18

Therefore, the travel time to work that separates the bottom 2.5% of people with the median or 50% is approximately 8.18 minutes.

To find the travel time that separates the bottom 2.5% of people, we need to use the z-score formula and the standard normal distribution table.

The z-score formula is given by: z = (X - μ) / σ

Where:
- X is the travel time we want to find
- μ is the mean travel time (17 minutes)
- σ is the standard deviation (4.5 minutes)
- z is the z-score

Since we want to find the travel time that separates the bottom 2.5% of people, we need to find the z-score corresponding to that percentile. The bottom 2.5% corresponds to a cumulative probability of 0.025.

Looking up the z-score corresponding to a cumulative probability of 0.025 in the standard normal distribution table, we find it to be approximately -1.96.

Now we can rearrange the z-score formula to solve for X:
X = μ + (z * σ)

Plugging in the values:
X = 17 + (-1.96 * 4.5)

Calculating this, we get:
X ≈ 8.32

Therefore, the travel time to work that separates the bottom 2.5% of people with the median or 50% is approximately 8.32 minutes.