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Maria and Ellen both specialize in throwing the javelin. Maria throws the javelin a mean distance of 200 feet with a standard deviation of 10 feet, whereas Ellen throws the javelin a mean distance of 210 feet with a standard deviation of 12 feet. Assume that the distances cach of these athletes throws the javelin are normally distributed with these population means and standard deviations. If Maria and Ellen each throw the javelin once, what is the probability that Maria's throw is longer than Ellen's?

Short Answer

Expert verified
The probability that Maria's javelin throw is longer than Ellen’s is approximately 0.261 or 26.1%.

Step by step solution

01

Calculate the Difference in Mean Distances

Subtract Maria's mean distance (200 feet) from Ellen's mean distance (210 feet) to get the difference in the mean distances. The result (-10 feet) is the mean of the differences that will be used during the standardization process.
02

Calculate the Standard Deviation of the Difference

The standard deviation of the difference can be calculated using the following formula: \(\sqrt{(\text{Maria's Standard Deviation})^2 + (\text{Ellen's Standard Deviation})^2}\). This results in \(\sqrt{(10)^2 + (12)^2} = 15.6 \) feet. This is the standard deviation of the differences.
03

Calculate the Z-Score

The Z score can now be calculated using the formula: \(\frac{(\text{Difference in Mean}) - 0}{\text{Standard Deviation of Difference}}\). Substituting from our previous steps, this would be \(\frac{-10 - 0}{15.6} = -0.641\). This value, Z, represents how many standard deviations away our data point is from the mean.
04

Find the Probability

To find the probability, we look up this Z score in the standard normal distribution table, or use a standard normal distribution function in a calculator or programming language. The probability associated with Z = -0.641 corresponds to 0.261. This is the probability that Maria’s javelin throw is longer than Ellen’s.

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Key Concepts

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

Understanding Normal Distribution
Normal distribution is a crucial concept in probability theory and statistics. It represents a continuous probability distribution characterized by a symmetric bell-shaped curve. In this curve, most values cluster around a central mean, with probabilities for values tapering off similarly on both sides.

The basic features of a normal distribution include:
  • A single peak at the mean, which is also the median and mode.
  • Symmetry about the mean, meaning the left and right sides of the curve are mirror images.
  • A predictable distribution of data, often referred to as the 68-95-99.7 (empirical) rule, which states that approximately 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.
In the context of our exercise, both Maria's and Ellen's javelin throw distances follow a normal distribution, allowing us to use statistical methods to compare their performances.
The Role of Standard Deviation
Standard deviation measures the amount of variation or dispersion in a set of values. A smaller standard deviation means the values tend to be closer to the mean, while a larger standard deviation signifies values are spread out over a wider range.

In our exercise:
  • Maria's throws have a standard deviation of 10 feet, indicating less variability in her performance.
  • Ellen's throws, with a standard deviation of 12 feet, are slightly more variable than Maria's.
This metric becomes essential when comparing their performances to determine the likelihood that one throw might exceed another. By calculating the standard deviation of the differences between their throws, we can more accurately ascertain the probability of such events.
Deciphering the Z-Score
A Z-score, or standard score, is a measure that describes a value's relationship to the mean of a group of values. It is expressed in terms of standard deviations away from the mean. In other words, the Z-score tells us how far and in what direction a single data point deviates from the mean.

To calculate the Z-score in this context, you take the difference between the means and divide it by the standard deviation of the differences. This is expressed by the formula: \[ Z = \frac{(\text{Difference in Mean}) - 0}{\text{Standard Deviation of Difference}} \]For Maria and Ellen's throws, the Z-score of -0.641 indicates that Maria's throw is 0.641 standard deviations below the mean difference.

This score helps us to use standard normal distribution tables or statistical software to find the associated probability that determines how likely one outcome is compared to another.
The Importance of Mean Difference
Mean difference is simply the difference between the mean values of two groups or datasets. In probability theory and statistics, it plays a vital role in hypothesis testing and comparisons.

For Maria and Ellen's scenario:
  • The mean difference is calculated by subtracting Maria's mean throw (200 feet) from Ellen's (210 feet), resulting in -10 feet.
  • This negative value indicates that on average, Ellen's throws are further than Maria's.
Understanding the mean difference is pivotal as it provides the basis for standardizing the data, allowing us to apply the Z-score and subsequently find probabilities. This is particularly valuable in assessing relative performance in sports or other comparative situations.

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Most popular questions from this chapter

Employees of a large corporation are concerned about the declining quality of medical services provided by their group health insurance. A random sample of 100 office visits by employees of this corporation to primary care physicians during 2004 found that the doctors spent an average of 19 minutes with each patient. This year a random sample of 108 such visits showed that doctors spent an average of \(15.5\) minutes with each patient. Assume that the standard deviations for the two populations are \(2.7\) and \(2.1\) minutes, respectively. a. Construct a \(95 \%\) confidence interval for the difference between the two population means for these two years. b. Using a \(2.5 \%\) level of significance, can you conclude that the mean time spent by doctors with each patient is lower for this year than for \(2004 ?\) c. What would your decision be in part \(\mathrm{b}\) if the probability of making a Type I error were zero? Explain.

According to a report in The New York Times, in the United States, accountants and auditors earn an average of \(\$ 70,130\) a year and loan officers carn \(\$ 67,960\) a year (Jessica Silver-Greenberg, The New York Times, April 22,2012 ). Suppose that these estimates are based on random samples of 1650 accountants and auditors and 1820 loan officers. Further assume that the sample standard deviations of the salaries of the two groups are \(\$ 14,400\) and \(\$ 13,600\), respectively, and the population standard deviations are equal for the two groups. a. Construct a \(98 \%\) confidence interval for the difference in the mean salaries of the two groupsaccountants and auditors, and loan officers. b. Using a \(1 \%\) significance level, can you conclude that the average salary of accountants and auditors is higher than that of loan officers?

According to a Randstad Global Work Monitor survey, \(52 \%\) of men and \(43 \%\) of women said that working part-time hinders their career opportunities (USA TODAY, October 6,2011 ). Suppose that these results are based on random samples of 1350 men and 1480 women. a. Let \(p_{1}\) and \(p_{2}\) be the proportions of all men and all women, respectively, who will say that working part-time hinders their career opportunities. Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) b. Using a \(2 \%\) significance level, can you conclude that \(p_{1}\) and \(p_{2}\) are different? Use both the critical-value and the \(p\) -value approaches.

What is the shape of the sampling distribution of \(\hat{p}_{1}-\hat{p}_{2}\) for two large samples? What are the mean and standard deviation of this sampling distribution?

The management at New Century Bank claims that the mean waiting time for all customers at its branches is less than that at the Public Bank, which is its main competitor. A business consulting firm took a sample of 200 customers from the New Century Bank and found that they waited an average of \(4.5\) minutes before being served. Another sample of 300 customers taken from the Public Bank showed that these customers waited an average of \(4.75\) minutes before being served. Assume that the standard deviations for the two populations are \(1.2\) and \(1.5\) minutes, respectively. a. Make a \(97 \%\) confidence interval for the difference between the population means. b. Test at a 2.5\% significance level whether the claim of the management of the New Century Bank is truc. c. Calculate the \(p\) -value for the test of part b. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=01 ?\) What if \(\alpha=05\) ?

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