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Triathlon times, Part II. In Exercise 4.4 we saw two distributions for triathlon times: \(N(\mu=4313, \sigma=\) 583 ) for Men, Ages \(30-34\) and \(N(\mu=5261, \sigma=807)\) for the Women, Ages \(25-29\) group. Times are listed in seconds. Use this information to compute each of the following: (a) The cutoff time for the fastest \(5 \%\) of athletes in the men's group, i.e. those who took the shortest \(5 \%\) of time to finish. (b) The cutoff time for the slowest \(10 \%\) of athletes in the women's group.

Short Answer

Expert verified
Men: 3354 seconds, Women: 6296 seconds.

Step by step solution

01

Understanding the Men's Fastest 5%

We need to find the cutoff time for the fastest 5% of male athletes. This equates to finding the 5th percentile in a normal distribution with mean \(\mu = 4313\) and standard deviation \(\sigma = 583\).
02

Calculate the Z-score for Fastest 5% (Men)

The Z-score for the 5th percentile can be found using a standard normal distribution table or calculator. For the 5th percentile, the Z-score is approximately \(z = -1.645\).
03

Apply the Z-score Formula (Men)

Use the Z-score formula to find the cutoff time: \[x = \mu + (z \cdot \sigma)\]Plug in the values:\[x = 4313 + (-1.645 \cdot 583)\approx 3354\]
04

Understanding the Women's Slowest 10%

Next, we find the cutoff time for the slowest 10% of female athletes, which is the 90th percentile in a normal distribution with mean \(\mu = 5261\) and standard deviation \(\sigma = 807\).
05

Calculate the Z-score for Slowest 10% (Women)

Using a standard normal distribution table or calculator, the Z-score for the 90th percentile is approximately \(z = 1.282\).
06

Apply the Z-score Formula (Women)

Use the Z-score formula to find the cutoff time: \[x = \mu + (z \cdot \sigma)\]Plug in the values:\[x = 5261 + (1.282 \cdot 807) \approx 6296\]

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Key Concepts

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

Percentiles
Percentiles are a way to understand the position of a value within a data set. They divide the data into 100 equal parts. Each percentile represents a ranking in which a certain percentage of the data falls below a specific value. For example, the 5th percentile marks the point below which 5% of the data can be found. This is useful for identifying cutoff points, such as the time for the fastest 5% of athletes.
Percentiles can help compare an individual's performance to that of a group:
  • Useful in exams - a score in the 90th percentile indicates you did better than 90% of test-takers.
  • In health, children in the 50th percentile for height are taller than half of their peers.
Understanding percentiles gives insight into how data is distributed and helps in decision-making. It's an important tool in statistics, allowing us to communicate information about relative standing within a group.
Z-score
A Z-score, also known as a standard score, indicates how many standard deviations an element is from the mean of the data set. It's a measure that helps compare different data points in a standard way. Z-scores can be positive or negative, indicating whether the value is above or below the mean, respectively.
Calculating a Z-score follows a simple formula:
  • First, subtract the mean (\( \mu \) ) from the data point (x).
  • Then, divide the result by the standard deviation (\( \sigma \)).
Mathematically, it is expressed as:\[ z = \frac{x - \mu}{\sigma} \]
By finding the Z-score, you can determine how unusual or typical a value is within a specific context. A Z-score of 0 means the data point is exactly average, while a Z-score of 2 implies it's 2 standard deviations above the mean. Z-scores are instrumental in finding percentiles and probabilities in normal distributions.
Standard Deviation
Standard deviation is a statistical measure that shows how much variation or dispersion there is in a set of values. It tells us how spread out the numbers are in a data set, with a larger standard deviation indicating greater variability among data points.
Here's how standard deviation is helpful:
  • Low standard deviation: Data points tend to be close to the mean, implying consistency across the data set.
  • High standard deviation: Data points are more spread out, indicating a wider range of values.
The formula to calculate standard deviation involves several steps:
  • Calculate the mean of the data set.
  • Subtract the mean from each data point and square each result.
  • Find the average of these squared differences.
  • Take the square root of this average to get the standard deviation.
Understanding standard deviation helps grasp how a data set is structured and how every data point relates to the mean of the data set. It's fundamental in predicting outcomes in statistics and assessing distribution patterns in a dataset.

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Most popular questions from this chapter

CFLs. A manufacturer of compact fluorescent light bulbs advertises that the distribution of the lifespans of these light bulbs is nearly normal with a mean of 9,000 hours and a standard deviation of 1.000 hours. (a) What is the probability that a randomly chosen light bulb lasts more than 10.500 hours? (b) Describe the distribution of the mean lifespan of 15 light bulbs. (c) What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10.500 hours? (d) Sketch the two distributions (population and sampling) on the same scale. (e) Could you estimate the probabilities from parts (a) and (c) if the lifespans of light bulbs had a skewed distribution?

Eye color, Part I. A husband and wife both have brown eyes but carry genes that make it possible for their children to have brown eyes (probability 0.75\()\), blue eyes \((0.125),\) or green eyes (0.125) . (a) What is the probability the first blue-eyed child they have is their third child? Assume that the eye colors of the children are independent of each other. (b) On average, how many children would such a pair of parents have before having a blue-eyed child? What is the standard deviation of the number of children they would expect to have until the first blue-eyed child?

Spray paint, Part II. As described in Exercise 4.14 , the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. (a) What is the probability that the area covered by a can of spray paint is more than 27 square feet? (b) Suppose you want to spray paint an area of 540 square feet using 20 cans of spray paint. On average, how many square foet must each can be able to cover to spray paint all 540 square feet? (c) What is the probability that you can cover a 540 square feet area using 20 cans of spray paint? (d) If the area covered by a can of spray paint had a slightly skewed distribution, could you still calculate the probabilities in parts (a) and (c) using the normal distribution?

Is it Bernoulit? Determine if each trial can be considered an independent Bernoulli trial for the following sitmations. (a) Cards dealt in a hand of poker. (b) Outcome of each roll of a die.

Lemonade at The Cafe. Drink pitchers at The Cafe are intended to hold about 64 ounces of lemonade and glasses hold about 12 ounces. However, when the pitchers are filled by a server, they do not always fill it with exactly 64 ounces. There is some variability. Similarly, when they pour out some of the lemonade, they do not pour exactly 12 ounces. The amount of lemonade in a pitcher is normally distributed with mean 64 ounces and standard deviation 1.732 ounces. The amount of lemonade in a glass is normally distributed with mean 12 ounces and standard deviation 1 ounce. (a) How much lemonade would you expect to be left in a pitcher after pouring one glass of lemonade? (b) What is the standard deviation of the amount left in a pitcher after pouring one glass of lemonade? (c) What is the probability that more than 50 ounces of lemonade is left in a pitcher after pouring one glass of lemonade?

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