Question

PKEASE HELPPPPPO YALLL

The Amazing Race is a reality show where contestants race around the world completing various tasks in a variety of locales. Crucial to the show is estimating how long it'll take contestants to complete these tasks. Suppose time-to-completion follows a normal distribution with an average of 55 minutes and standard deviation of 4 minutes. What percent of contestants will take exactly 60 minutes to complete the task?

Answer 1

The answer is approximately 12.14% of contestants.

Step-by-step explanation:

Since the normal distribution is a continuous distribution, the probability of getting an exact value is zero. However, we can find the probability of getting a time between 59.5 and 60.5 minutes (i.e., a range that includes 60 minutes) using the normal distribution.

Using the z-score formula, we can first standardize the values:

z = (x - μ) / σ

where x is the time in minutes (60), μ is the mean (55), and σ is the standard deviation (4).

z = (60 - 55) / 4 = 1.25

Next, we can use a standard normal distribution table or calculator to find the probability of getting a z-score between 1.24 and 1.26 (i.e., the range that corresponds to 59.5 to 60.5 minutes).

Using a calculator or table, we find that the probability of getting a z-score between 1.24 and 1.26 is approximately 0.0485.

Therefore, approximately 4.85% of contestants will take between 59.5 and 60.5 minutes to complete the task.

Examples on how to do problems like these:

I will provide a general approach for solving problems involving normal distributions, along with examples of easy, medium, and hard difficulty.

General Approach:

Identify the mean and standard deviation of the normal distribution.

Determine whether the problem is asking for a probability or a value given a probability.

If the problem is asking for a probability, use the standard normal distribution (Z-distribution) to find the area under the curve corresponding to the given value or range of values.

If the problem is asking for a value, use the inverse normal distribution (Z-score) to find the corresponding value on the distribution.

Easy Example:

Suppose the heights of a population of women follow a normal distribution with a mean of 65 inches and a standard deviation of 2.5 inches. What percentage of women are taller than 70 inches?

Solution:

1. Mean = 65 inches, standard deviation = 2.5 inches

2. We are asked for a probability.

3. Let Z = (x - μ) / σ be the standard normal distribution, where x = 70, μ = 65, and σ = 2.5. We want to find P(Z > (70 - 65) / 2.5), which is the area to the right of the Z-score of 2.0 on the standard normal distribution.

4. Using a standard normal distribution table or calculator, we find that P(Z > 2.0) = 0.0228, or about 2.28%.

Therefore, about 2.28% of women are taller than 70 inches.

Medium Example:

Suppose the scores on a standardized test follow a normal distribution with a mean of 70 and a standard deviation of 10. If the top 5% of test-takers receive a score of at least what value, what is that value?

Solution:

1. Mean = 70, standard deviation = 10

2. We are asked for a value.

3. Let Z = (x - μ) / σ be the standard normal distribution. We want to find the Z-score that corresponds to the top 5% of test-takers, which is the area to the right of the Z-score we're looking for. Using a standard normal distribution table or calculator, we find that the Z-score corresponding to the top 5% is 1.645.

4. Plugging in the values, we get 1.645 = (x - 70) / 10. Solving for x, we get x = 86.45.

Therefore, the score required to be in the top 5% of test-takers is at least 86.45.

Hard Example:

Suppose the time it takes for a computer program to run follows a normal distribution with a mean of 10 minutes and a standard deviation of 2 minutes. The program will fail if it takes longer than 15 minutes to run. What is the probability that the program will fail?

Solution:

We need to find the area under the normal distribution curve to the right of 15 minutes. Again, we standardize the value using the formula:

z = (x - mu) / sigma

For 15 minutes:

z = (15 - 10) / 2 = 2.5

Now we look up the area to the right of 2.5 in the standard normal distribution table, or use a calculator to find that the area is approximately 0.0062. Therefore, the probability that the program will fail is 0.62%.

Hope this helps you! I'm sorry if it's wrong. If you need more help, ask me! :]

Answer 2

In a normal distribution, the probability of observing any single, exact value, such as exactly 60 minutes for task completion on The Amazing Race, is 0% because there are infinitely many potential values even in a small interval.

Step-by-step explanation:

In the scenario given where the time-to-completion of tasks on The Amazing Race follows a normal distribution with an average of 55 minutes and a standard deviation of 4 minutes, when asking what percent of contestants will take exactly 60 minutes to complete the task, we come across an important concept in statistics. In a continuous probability distribution, the probability of observing any single, exact value is essentially zero, because there are infinite potential values even within a small range.

So, while we can calculate the probability of a contestant finishing within a range of times, like between 59 and 61 minutes, the probability of finishing in exactly 60 minutes is 0%. To estimate the proportion of contestants who would complete tasks within a range, one would use the properties of the normal distribution and z-scores or a normal distribution table.