/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 The table below shows data (from... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 56

The table below shows data (from a 2004 Bureau of the Census report) on the number of times 20 - to 24 -year-old men have been married. $$\begin{array}{crr} \text { No. Times } & \text { Count } & \text { Percentage } \\ \hline 0 & 8418 & 84.0 \\ 1 & 1594 & 15.9 \\ 2 & 10 & 0.1 \\ \text { Total } & \mathbf{1 0 0 2 2} & \mathbf{1 0 0 . 0} \end{array}$$ a. Verify that the mean number of times men have been married is 0.16 and that the standard deviation is 0.37 . b. Find the actual percentages of observations within \(1,2,\) and 3 standard deviations of the mean. How do these compare to the percentages predicted by the empirical rule? c. How do you explain the results in part b?

Short Answer

Expert verified
The mean is 0.16; standard deviation is 0.37. The actual percentages are higher than predicted by the empirical rule due to the skewness of the data towards zero marriages.

Step by step solution

01

Calculate Mean

To calculate the mean of the number of marriages, use the formula for the weighted mean: \( \mu = \frac{\sum (x_i \cdot f_i)}{n} \), where \( x_i \) is the number of times and \( f_i \) is the frequency.Calculating:\[ \mu = \frac{(0 \times 8418 + 1 \times 1594 + 2 \times 10)}{10022} = \frac{1614}{10022} \approx 0.16 \] The mean is correct: 0.16.
02

Calculate Standard Deviation

The standard deviation \( \sigma \) is the square root of the variance. First, find the variance:\[ \sigma^2 = \frac{\sum ((x_i - \mu)^2 \cdot f_i)}{n} = \frac{(0.16^2 \times 8418 + 0.84^2 \times 1594 + 1.84^2 \times 10)}{10022} \]Calculating each term:- \( 0.16^2 \times 8418 = 215.60 \)- \( 0.84^2 \times 1594 = 1125.81 \)- \( 1.84^2 \times 10 = 33.86 \)\[ \sigma^2 = \frac{215.60 + 1125.81 + 33.86}{10022} \approx 0.1359 \]\[ \sigma = \sqrt{0.1359} \approx 0.37 \]The standard deviation is correct: 0.37.
03

Calculate Observations within 1 Standard Deviation

Mean \( \mu = 0.16 \) and standard deviation \( \sigma = 0.37 \).For 1 standard deviation, calculate the interval: \( 0.16 \pm 0.37 = [-0.21, 0.53] \).Observations within this interval are those who have been married 0 times: 8418.Percentage: \( \frac{8418}{10022} \times 100 \approx 84.0\% \).
04

Calculate Observations within 2 Standard Deviations

For 2 standard deviations, calculate the interval: \( 0.16 \pm 2(0.37) = [-0.58, 0.90] \).Observations within this interval include those married 0 and 1 times: 8418 + 1594 = 10012.Percentage: \( \frac{10012}{10022} \times 100 \approx 99.9\% \).
05

Calculate Observations within 3 Standard Deviations

For 3 standard deviations, calculate the interval: \( 0.16 \pm 3(0.37) = [-1.01, 1.19] \).All observations (0, 1, and 2 times married) fall within this interval: 10022.Percentage: \( \frac{10022}{10022} \times 100 = 100\% \).
06

Compare to Empirical Rule Predictions

The empirical rule suggests approximately 68% within 1 standard deviation, 95% within 2, and 99.7% within 3. Our calculations show 84.0%, 99.9%, and 100%, respectively. The calculated percentages are higher due to the skewed distribution of marriages and effectively cut-off data at higher counts decreasing variability.
07

Explain Differences in Part b

The differences occur because the empirical rule applies to normal (bell-shaped) distributions, while this data distribution is heavily skewed towards 0 times married. Therefore, more values fall close to the mean, resulting in higher percentages even within a lesser standard deviation coverage.

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Key Concepts

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

Mean Calculation
The mean, in statistics, is a measure of the central tendency of a set of values. It provides an average of the data set, giving us an idea of where the data points tend to cluster. Calculating the mean for the number of times 20 to 24-year-old men have been married involves using the formula for the weighted mean. We calculate it based on the frequency of each data point in the distribution.

In this example, to find the mean number of marriages, you add up the products of each value with its frequency, then divide by the total count of data points. Here鈥檚 the step-by-step process:
  • Multiply each data point by its frequency: 0 x 8418 = 0, 1 x 1594 = 1594, and 2 x 10 = 20.
  • Add these products together: 0 + 1594 + 20 = 1614.
  • Divide this sum by the total number of observations: \(\frac{1614}{10022} \approx 0.16.\)
The calculated mean is approximately 0.16, indicating that on average, the men in this age group have been married just a fraction of a time, reflecting a predominance of zero or very low marriage counts in the population.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the numbers in a dataset differ from the mean. A smaller standard deviation implies that the data points are clustered around the mean, while a larger one suggests more spread out values.

In our dataset, we calculated the standard deviation to understand the variability in marriage occurrences among young men. Here's how:
  • First, calculate the variance, which is the average of the squared differences from the Mean. Using the formula \(\sigma^2 = \frac{\sum((x_i - \mu)^2 \cdot f_i)}{n},\) we compute the squared deviations for each frequency, sum them, and then divide by the total count (10022).
  • Next, apply the square root to the variance to get the standard deviation, \(\sigma \approx 0.37.\)
This low standard deviation tells us that most young men in this dataset have similar experiences regarding marriage frequency, heavily concentrated around few (or no) marriages.
Empirical Rule
The empirical rule, often referred to as the 68-95-99.7 rule, applies to datasets that are normally distributed (bell-shaped). According to this rule, approximately 68% of data falls within one standard deviation from the mean, 95% within two, and 99.7% within three.

However, when applied to our dataset on marriage occurrences:
  • Around 84.0% of observations lie within one standard deviation of the mean.
  • 99.9% are within two standard deviations.
  • 100% fall within three standard deviations.
These percentages deviate from the empirical rule's predictions due to the dataset's skewed nature, heavily leaning towards zero. While in a normal distribution, data tails off symmetrically from the mean, our data鈥檚 concentration at low values (no marriages) causes more observations to lie near the mean, hence the higher percentages even within one or two standard deviations.

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