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According to the South Dakota Department of Health, the mean number of hours of TV viewing per week is higher among adult women than men. A recent study showed women spent an average of 34 hours per week watching \(\mathrm{TV}\) and men 29 hours per week (www.state.sd.us/DOH/Nutrition/TV.pdt). Assume that the distribution of hours watched follows the normal distribution for both groups, and that the standard deviation among the women is 4.5 hours and it is 5.1 hours for the men. a. What percent of the women watch TV less than 40 hours per week? b. What percent of the men watch TV more than 25 hours per week? c. How many hours of TV do the one percent of women who watch the most TV per week watch? Find the comparable value for the men.

Short Answer

Expert verified
a. 90.82% of women watch less than 40 hours. b. 78.22% of men watch more than 25 hours. c. Women 44.5 hours, Men 40.88 hours.

Step by step solution

01

Understand the Problem

We have information about the TV viewing habits of men and women, given their average viewing hours per week and standard deviations. Using the normal distribution, we need to find probabilities and specific percentiles related to TV viewing hours for each gender.
02

Calculate z-score for Women (Part a)

For women watching less than 40 hours, calculate the z-score using the formula: \( z = \frac{X - \mu}{\sigma} \), where \( X = 40 \), \( \mu = 34 \), and \( \sigma = 4.5 \). Substitute these values to get:\[ z = \frac{40 - 34}{4.5} = \frac{6}{4.5} \approx 1.33 \].
03

Determine Percentage for Women (Part a)

Using the z-score calculated in Step 2, find the probability from a standard normal distribution table or calculator. A z-score of 1.33 corresponds to approximately 90.82%. Thus, 90.82% of women watch less than 40 hours of TV per week.
04

Calculate z-score for Men (Part b)

For men watching more than 25 hours, use the z-score formula: \( z = \frac{X - \mu}{\sigma}, \) where \( X = 25 \), \( \mu = 29 \), and \( \sigma = 5.1 \). This gives:\[ z = \frac{25 - 29}{5.1} = \frac{-4}{5.1} \approx -0.78 \].
05

Determine Percentage for Men (Part b)

A z-score of -0.78 means that approximately 21.78% of men watch less than 25 hours of TV. Therefore, using the complement, about 78.22% of men watch more than 25 hours.
06

Calculate 99th Percentiles (Part c)

For women, find the value of \( z \) corresponding to the 99th percentile, which is approximately 2.33. Use the formula: \( X = \mu + z\sigma \). For women:\[ X_{women} = 34 + 2.33 \times 4.5 \approx 44.5 \].For men, using the same percentile \( z = 2.33 \):\[ X_{men} = 29 + 2.33 \times 5.1 \approx 40.88 \].

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Key Concepts

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

Z-score Calculation
The z-score is a powerful tool in statistics that allows us to measure how far away a particular value is from the mean in terms of standard deviations. It provides insight into the relative position of data within a normal distribution. The formula for calculating a z-score is:
\[ z = \frac{X - \mu}{\sigma} \]
where:
  • \( X \) is the data point for which you are finding the z-score.
  • \( \mu \) is the mean of the dataset.
  • \( \sigma \) is the standard deviation of the dataset.
By using z-scores, you can transform a normally distributed random variable into the standard normal distribution, allowing you to compare scores across different distributions. For example, if a woman watches 40 hours of TV per week, her z-score would be calculated as follows, with given mean \( \mu = 34 \) and standard deviation \( \sigma = 4.5 \):
\[ z = \frac{40 - 34}{4.5} \approx 1.33 \]
This z-score tells us how far 40 is away from the average 34 in units of the standard deviation, meaning it's about 1.33 standard deviations above the mean. This helps to understand her viewing habits in comparison to the general trend.
Percentiles
Percentiles are important indicators in statistics that tell us where a particular value stands relative to the rest of the data. They are often used to interpret raw scores. For example, if a score is at the 90th percentile, it means that it is higher than 90% of all other scores in the data set.
In the context of TV viewing, if we want to determine the top 1% of women in terms of weekly TV hours, we look for the 99th percentile. To calculate the hours for the top 1% of female TV watchers, given a mean \( \mu = 34 \) and standard deviation \( \sigma = 4.5 \), and knowing the z-score for the 99th percentile is about 2.33, we proceed as follows:
\[ X = \mu + z\sigma = 34 + 2.33 \times 4.5 \approx 44.5 \]
This tells us that the one percent of women who watch the most TV per week watch about 44.5 hours. Using a similar method, you can find percentile values for men or any other dataset, showing how percentiles help to compare where individual scores fall within the broader distribution.
Standard Deviation
Standard deviation is a critical concept in statistics that measures the amount of variation or dispersion in a set of values. A low standard deviation indicates that the data points are close to the mean, whereas a high standard deviation suggests a wider spread out from the mean.
In TV viewing habits, the standard deviation helps to describe how consistent the viewing time is among men and women. For instance, women have a standard deviation of 4.5 hours, while men have a slightly larger standard deviation of 5.1 hours. These values signify that men's viewing hours are more spread out around their mean compared to women.
Here's why standard deviation is useful:
  • It provides a quantitative measure of uncertainty.
  • It allows for comparison between the variability of different data sets.
  • It helps in identifying outliers or unusual observations.
Understanding standard deviation is crucial when interpreting data, as it affects how you perceive the average behavior and predict future trends. In our example, this influences how we interpret deviations in TV viewing habits between genders efficiently.

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Most popular questions from this chapter

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