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Chapter 3: Problem 16

Use \(z\) scores to compare the given values. In the 87th Academy Awards, Eddie Redmayne won for best actor at the age of 33 and Julianne Moore won for best actress at the age of 54 . For all best actors, the mean age is 44.1 years and the standard deviation is \(8.9\) years. For all best actresses, the mean age is \(36.2\) years and the standard deviation is \(11.5\) years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the Oscar: Eddie Redmayne or Julianne Moore? Explain.

Short Answer

Expert verified
Julianne Moore had the more extreme age when winning the Oscar.

Step by step solution

01

- Understand the z-score formula

The z-score formula is given by: \[ z = \frac{x - \mu}{\sigma} \] where \( x \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. The z-score indicates how many standard deviations a value is from the mean.
02

- Calculate Eddie Redmayne's z-score

For Eddie Redmayne, who won at the age of 33, the mean age for best actors is \( 44.1 \) years and the standard deviation is \( 8.9 \) years. Substituting these values into the z-score formula: \[ z_{Eddie} = \frac{33 - 44.1}{8.9} \approx -1.25 \]
03

- Calculate Julianne Moore's z-score

For Julianne Moore, who won at the age of 54, the mean age for best actresses is \( 36.2 \) years and the standard deviation is \( 11.5 \) years. Substituting these values into the z-score formula: \[ z_{Julianne} = \frac{54 - 36.2}{11.5} \approx 1.55 \]
04

- Compare the z-scores

The z-score for Eddie Redmayne is \( -1.25 \), and the z-score for Julianne Moore is \( 1.55 \). The absolute value of Julianne Moore's z-score (1.55) is greater than the absolute value of Eddie Redmayne's z-score (1.25). A higher absolute value of z-score indicates a more extreme age relative to their genders.
05

- Conclude the comparison

Since Julianne Moore has a higher absolute z-score, this indicates that Julianne Moore had the more extreme age relative to the mean age of best actresses compared to Eddie Redmayne relative to the mean age of best actors.

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Key Concepts

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

mean age
The concept of *mean* or *average* is essential in understanding statistics.
It is calculated by adding up all the values and dividing by the number of values.
In our example, the mean age for best actors is 44.1 years and for best actresses, it is 36.2 years.
This means that, on average, best actors win their awards at age 44.1 and best actresses at age 36.2.
Understanding the mean helps us see what is typical for a group.
standard deviation
Standard deviation measures the spread or variability of a set of data.
It tells us how much the ages of the winners deviate from the mean age.
A smaller standard deviation means the ages are close to the mean.
A larger standard deviation means they are more spread out.
In our case, the standard deviation for best actors is 8.9 years, whereas for best actresses it is 11.5 years.
This implies more variability in the ages of best actress winners compared to best actor winners.
probability distributions
In statistics, a *probability distribution* shows the possible values a random variable can take and how often they occur.
For our example, the ages of winners follow a certain distribution.
To compare one winner's age to this distribution, we use the mean age and standard deviation.
This helps us see how typical or unusual a particular age is.
In our exercise, we use the z-score to compare Eddie Redmayne's age of 33 and Julianne Moore's age of 54 to their respective gender groups.
comparative analysis
Comparative analysis is the process of comparing two or more items to determine their differences and similarities.
In this exercise, we compare the ages of Eddie Redmayne and Julianne Moore relative to the mean ages of previous winners of their genders.
By calculating and comparing their z-scores, we can see who had a more atypical age when they won their Oscars.
Comparative analysis helps us make sense of how individuals stand out in a larger group.
z-score formula
The *z-score* formula, given by \(z = \frac{x - \mu}{\sigma} \), is a tool to measure how many standard deviations a value (x) is from the mean (碌).
Let's break down the formula:
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Most popular questions from this chapter

Refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where \(x\) represents the class midpoint, \(f\) represents the class frequency, and \(n\) represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 11.5 years; (Exercise 38) 8.9 years; (Exercise 39) 59.5; (Exercise 40) 65.4. $$s=\sqrt{\frac{n\left[\Sigma\left(f \cdot x^{2}\right)\right]-[\Sigma(f \cdot x)]^{2}}{n(n-1)}}$$ $$ \begin{array}{c|c} \hline \begin{array}{c} \text { Blood Platelet } \\ \text { Count of } \\ \text { Females } \end{array} & \text { Frequency } \\ \hline 100-199 & 25 \\ \hline 200-299 & 92 \\ \hline 300-399 & 28 \\ \hline 400-499 & 0 \\ \hline 500-599 & 2 \\ \hline \end{array} $$

Find the coefficient of variation for each of the two samples; then compare the variation. (The same data were used in Section 3-1.) Theft Listed below are amounts (in millions of dollars) collected from parking meters by Brinks and others in New York City during similar time periods. A larger data set was used to convict five Brinks employees of grand larceny. The data were provided by the attorney for New York City, and they are listed on the DASL Website. Do the two samples appear to have different amounts of variation? \(\begin{array}{lllllllllll} \text { Collection Contractor Was Brinks } & 1.3 & 1.5 & 1.3 & 1.5 & 1.4 & 1.7 & 1.8 & 1.7 & 1.7 & 1.6 \\ \text { Collection Contractor Was Not Brinks } & 2.2 & 1.9 & 1.5 & 1.6 & 1.5 & 1.7 & 1.9 & 1.6 & 1.6 & 1.8 \end{array}\)

Use \(z\) scores to compare the given values. Based on Data Set 4 "Births" in Appendix B, newborn males have weights with a mean of \(3272.8 \mathrm{~g}\) and a standard deviation of \(660.2 \mathrm{~g} .\) Newborn females have weights with a mean of \(3037.1 \mathrm{~g}\) and a standard deviation of \(706.3 \mathrm{~g}\). Who has the weight that is more extreme relative to the group from which they came: a male who weighs \(1500 \mathrm{~g}\) or a female who weighs \(1500 \mathrm{~g}\) ?

Find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as "minutes") in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions. Listed below are the amounts (dollars) it costs for marriage proposal packages at the different Major League Baseball stadiums. Five of the teams don't allow proposals. Are there any outliers, and are they likely to have much of an effect on the measures of variation? \(5 \quad 200\) \(\begin{array}{rrrrrrrrrrrr}39 & 50 & 50 & 50 & 55 & 55 & 75 & 85 & 100 & 115 & 175 & 175 \\ 209 & 250 & 250 & 350 & 400 & 450 & 500 & 500 & 500 & 500 & 1500 & 2500\end{array}\)

Watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, \((b)\) median, (c) mode, (d) midrange, and then answer the given question. Listed below are the jersey numbers of 11 players randomly selected from the roster of the Seattle Seahawks when they won Super Bowl XLVIII. What do the results tell us? $$ \begin{array}{lllllllllll} 89 & 91 & 55 & 7 & 20 & 99 & 25 & 81 & 19 & 82 & 60 \end{array} $$
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