The average middle-distance runner at a local high school runs the mile in 4.5 minutes, with a standard deviation of 0.3 minute. The percentage of a runners that will run the mile in less than 4 minutes is %.

.5/.3 = 1.67 standard deviations below mean

Look at table for normal distribution
My quick table is very rough
for z = -1.6 F(z) = .055
for z = -1.7 F(z) = .045
so about F(1.67) = about .05 or 5 %

By the way, I do not believe it. Suspect sigma much lower than 0.3 min or possibly not a normal distribution at all.

Well, let me put it this way. Running a mile in less than 4 minutes is pretty impressive. It's like trying to eat a whole pizza by yourself while playing the piano!

Now, to calculate the percentage of runners who will achieve this speedy feat, we can use some statistics. Fortunately, I just happen to have my trusty calculator-bot right here!

First, we need to find the z-score of the desired time (4 minutes) using the formula:

z = (x - μ) / σ

where x is the desired time, μ is the mean (4.5 minutes), and σ is the standard deviation (0.3 minute).

So, plugging in the values:

z = (4 - 4.5) / 0.3

Now, let's ask my calculator-bot to crunch the numbers... beeping and booping... and we get a z-score of -1.67.

To find the percentage of runners who will achieve this time, we can consult a standard normal distribution table. Looking up the z-score of -1.67, we find that roughly 4.43% of the runners will run the mile in less than 4 minutes.

So, we can conclude that approximately 4.43% of the middle-distance runners at the local high school will be sprinting across the finish line in less than 4 minutes. Hats off to those speedsters!

To find the percentage of runners that will run the mile in less than 4 minutes, we need to calculate the z-score and look up the corresponding value in the standard normal distribution table.

First, we need to calculate the z-score using the formula:

z = (x - μ) / σ

where "x" is the value we want to find the percentage for, "μ" is the mean, and "σ" is the standard deviation.

In this case, x = 4 minutes, μ = 4.5 minutes, and σ = 0.3 minutes.

z = (4 - 4.5) / 0.3
z = -0.5 / 0.3
z ≈ -1.67

Next, we use the z-score (-1.67) to find the probability associated with it in the standard normal distribution table. This probability represents the percentage of runners who will run the mile in less than 4 minutes.

Looking up -1.67 in the standard normal distribution table gives us a probability of approximately 0.0475 or 4.75%.

So, the percentage of runners that will run the mile in less than 4 minutes is approximately 4.75%.

To find the percentage of runners who will run the mile in less than 4 minutes, we need to convert the given information into a standardized form and then use a normal distribution table or calculator.

Step 1: Convert the given information into Z-scores.
A Z-score measures how many standard deviations a data point is from the mean. We can calculate the Z-score using the formula:
Z = (X - μ) / σ

where:
X is the value we want to calculate the probability for (in this case, 4 minutes),
μ is the population mean (average time, which is 4.5 minutes), and
σ is the population standard deviation (0.3 minutes).

So, in this case:
Z = (4 - 4.5) / 0.3

Step 2: Calculate the Z-score.
Z = (4 - 4.5) / 0.3
Z = -0.5 / 0.3
Z = -1.67 (rounded to two decimal places)

Step 3: Find the percentage using a normal distribution table or calculator.
Using a normal distribution table or calculator, we can find the percentage of runners who will run the mile in less than 4 minutes when the Z-score is -1.67.

By referring to a normal distribution table or using a calculator, we find that the percentage of runners who will run the mile in less than 4 minutes is approximately 5.47%.

Please note that the values provided above are approximate and may vary slightly based on the specific table or calculator used.