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Running a mile A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normal with mean 7.11 minutes and standard deviation 0.74 minute. \({ }^{10}\) Choose a student at random from this group and call his time for the mile \(Y\). Find \(P(Y<6)\) and interpret the result.

Short Answer

Expert verified
The probability is approximately 0.0668, meaning about 6.68% of students run a mile in under 6 minutes.

Step by step solution

01

Understand the Problem

We need to find the probability that a randomly chosen male student runs a mile in less than 6 minutes. The time is normally distributed with a mean of 7.11 minutes and a standard deviation of 0.74 minutes. So, we are looking for \( P(Y < 6) \) where \( Y \) follows the normal distribution \( N(7.11, 0.74) \).
02

Standardize the Variable

We need to convert the mile run time into a standard normal variable \( Z \) using the formula: \[ Z = \frac{Y - \mu}{\sigma} \] where \( \mu = 7.11 \) and \( \sigma = 0.74 \). This will help us use the standard normal table. Let's compute \( Z \): \[ Z = \frac{6 - 7.11}{0.74} \].
03

Compute the Z-Score

Calculate the Z-score: \[ Z = \frac{6 - 7.11}{0.74} = \frac{-1.11}{0.74} \approx -1.50 \]. So, the probability we want to find is \( P(Z < -1.50) \).
04

Use the Standard Normal Distribution Table

Look up the Z-score of \( -1.50 \) in the standard normal distribution table. The table gives us \( P(Z < -1.50) \approx 0.0668 \).
05

Interpret the Result

The probability that a randomly selected student runs a mile in less than 6 minutes is approximately 0.0668, or 6.68%. This means that about 6.68% of students run the mile in under 6 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Normal Distribution
The standard normal distribution is a crucial concept in statistics, especially when dealing with probabilities related to normal distributions. It is a type of normal distribution that has a mean of zero and a standard deviation of one. This specific distribution is often denoted by the symbol \( N(0,1) \). The reason it is so useful is because it provides a standardized way to relate and convert any normal distribution to this common framework.
When dealing with problems involving normally distributed data, such as in our exercise, the use of the standard normal distribution allows us to make comparisons and compute probabilities using tables or statistical software. By standardizing data, we simplify the process of determining probabilities, no matter the original mean and standard deviation of the dataset.
By using the standard normal distribution, we can easily determine how likely or unlikely a particular observation is. This is achieved through the calculation of Z-scores, which we will explore further in the next section.
Z-Score
The Z-score is a measure that describes a data point's position relative to the mean, measured in terms of standard deviations. It is an essential concept when working with the standard normal distribution, as it allows us to convert any normal distribution to the standard form for easier analysis.
To calculate a Z-score, use the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Here, \( X \) represents the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. This score tells us how many standard deviations away from the mean the data point is.
In the context of our exercise, the Z-score was computed to determine the likelihood that a randomly chosen student could run a mile in under 6 minutes. By calculating
\[ Z = \frac{6 - 7.11}{0.74} \approx -1.50 \]
we transformed the original variable into the standardized form, allowing probabilistic analysis using the standard normal distribution table. The interpretation of a Z-score is straightforward: a score of -1.50, for instance, indicates a value that is 1.5 standard deviations below the mean.
Probability Calculation
Probability calculation involves determining the likelihood of a particular event occurring. With the standard and normal distributions, this often involves finding the area under the curve to the left of a specific Z-score on the standard normal distribution. This area represents the cumulative probability.
In the provided example, once the Z-score is calculated, the next step is to look up this score in a standard normal distribution table or use statistical software to find the probability \( P(Z < -1.50) \). The table or software provides the probability that a standard normal variable is less than the given Z-score.
For our Z-score of \(-1.50\), we find that \( P(Z < -1.50) \approx 0.0668 \), which means that there is approximately a 6.68% chance that a randomly selected student from the group can complete the mile in less than 6 minutes.
Probability calculations are instrumental in statistics as they provide quantitative measures that can be used for decision-making or interpreting results. Understanding how to compute and interpret these probabilities is key to analyzing data spread and likelihood within given distributions.

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