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Chapter 6: Problem 43

Assume the cholesterol levels of adult American women can be described by a Normal model with a mean of \(188 \mathrm{mg} / \mathrm{dL}\) and a standard deviation of \(24 .\) a) Draw and label the Normal model. b) What percent of adult women do you expect to have cholesterol levels over \(200 \mathrm{mg} / \mathrm{dL}\) ? c) What percent of adult women do you expect to have cholesterol levels between 150 and \(170 \mathrm{mg} / \mathrm{dL}\) ? d) Estimate the IQR of the cholesterol levels. e) Above what value are the highest \(15 \%\) of women's cholesterol levels?

Short Answer

Expert verified
a) Normal curve with mean 188 and SD 24. b) 30.85%. c) 16.95%. d) 32.4 mg/dL. e) 212.9 mg/dL.

Step by step solution

01

Draw and Label the Normal Model

To draw the Normal model, sketch a symmetric bell-shaped curve. Mark the mean \( \mu = 188 \) mg/dL in the center. Label the horizontal axis with standard deviations (\( \sigma = 24 \) mg/dL) from the mean: \( \mu - \sigma = 164 \), \( \mu + \sigma = 212 \), and continue labeling \( \mu \pm 2\sigma \) and \( \mu \pm 3\sigma \). This depicts the spread and standard deviations of the data.
02

Percent of Women with Cholesterol Levels Over 200 mg/dL

To find the percentage, calculate the Z-score for \( 200 \) mg/dL: \( Z = \frac{200 - 188}{24} \approx 0.5 \). Using the Z-table, find the probability associated with \( Z = 0.5 \), which is approximately \( 0.6915 \). The probability for levels over \( 200 \) mg/dL is \( 1 - 0.6915 = 0.3085 \), so about \( 30.85\% \) of women have levels over \( 200 \) mg/dL.
03

Percent of Women with Cholesterol Levels Between 150 and 170 mg/dL

Calculate Z-scores for \( 150 \) and \( 170 \) mg/dL: \( Z_{150} = \frac{150 - 188}{24} = -1.58 \) and \( Z_{170} = \frac{170 - 188}{24} = -0.75 \). Using the Z-table, find probabilities \( P(Z_{150}) \approx 0.0571 \) and \( P(Z_{170}) \approx 0.2266 \). The percent between \( 150 \) and \( 170 \) mg/dL is \( 0.2266 - 0.0571 = 0.1695 \) or about \( 16.95\% \).
04

Estimate the Interquartile Range (IQR)

The IQR can be estimated by finding the Z-scores corresponding to the 25th percentile (\( Q1 \)) and 75th percentile (\( Q3 \)). The Z-score for \( Q1 \) is \(-0.6745\) and for \( Q3 \) is \(0.6745\). Calculate \( Q1 = \mu + Z_{Q1}\sigma = 188 + (-0.6745)(24) \approx 171.8 \) mg/dL and \( Q3 = 188 + 0.6745(24) \approx 204.2 \) mg/dL. Thus, \( \text{IQR} = Q3 - Q1 = 204.2 - 171.8 = 32.4 \) mg/dL.
05

Find the Cholesterol Level for Highest 15%

To find the cholesterol level above which the top 15% fall, determine the Z-score that corresponds to the 85th percentile (\(100\% - 15\% = 85\%\)). The Z-score is approximately \(1.036\). Calculate the value: \( 188 + 1.036(24) \approx 212.9 \) mg/dL. Therefore, the highest 15% have cholesterol levels above \( 212.9 \) mg/dL.

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Key Concepts

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

Z-score
The Z-score is a measure that helps to understand how far, and in what direction, a data point deviates from the mean of a dataset, within a normal distribution. In simpler terms, it's a way to say how many standard deviations a particular value is from the average (mean).
  • To calculate a Z-score, you subtract the mean from your data point and then divide that by the standard deviation.
  • If a Z-score is positive, it indicates that the data point is above the mean. Conversely, a negative Z-score indicates that the data point is below the mean.
  • Z-scores can help identify where a value lies in relation to other values in the dataset. For example, if the Z-score of a cholesterol level is 2, it means the cholesterol level is 2 standard deviations above the mean.
This tool is highly useful in statistics because it allows any value in a dataset to be compared with others using the common metric of standard deviations, making it easier to draw conclusions about the data.
Percentile
A percentile ranks data points within a dataset by showing the relative standing of a value. It provides insight into the distribution of values and indicates how one sample compares with all other samples.
  • A percentile rank is the percentage of scores in its frequency distribution that are the same or lower, making it easy to see how a particular value compares within a broader group.
  • For instance, if a cholesterol level falls in the 70th percentile, it means that 70% of the other cholesterol levels are the same or lower.
  • Percentiles are commonly used to interpret data that follow a normal distribution, as they help quantify the skewness and spread of the data.
By using percentiles, statisticians and researchers can easily communicate the rank or positioning of specific data points, which is especially useful when comparing populations or subgroup data.
Interquartile Range
The interquartile range (IQR) is a measure of statistical dispersion and is the difference between the 75th percentile (Q3) and the 25th percentile (Q1). It describes the middle 50% of the data.
  • The IQR is a great measure for indicating variability when data is skewed or contains outliers since it focuses only on the middle and disregards extreme values.
  • In the context of cholesterol levels, if the IQR is 32.4 mg/dL, this means that the middle 50% of women have cholesterol levels within this range starting from Q1 to Q3.
  • It's calculated by finding the difference: IQR = Q3 - Q1.
Through the IQR, it's possible to have a better understanding of the variation present in the data which complements other statistical measures like mean and standard deviation.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us, on average, how far each value in the data set is from the mean.
  • A smaller standard deviation means that the data points are closer to the mean, while a larger standard deviation indicates a more spread-out dataset.
  • In our example of cholesterol levels, the standard deviation is 24 mg/dL, showing the typical amount by which individual levels differ from the mean of 188 mg/dL.
  • Standard deviation is fundamental in understanding normal distributions because it allows us to determine the probability of a data point occurring within a particular range.
This measure is crucial in fields like finance, science, and engineering because it helps in assessing risks, variations, and inconsistencies within any data set.

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Most popular questions from this chapter

NFL data from the 2006 football season reported the number of yards gained by each of the league's 167 wide receivers: a) According to the Normal model, what percent of receivers would you expect to gain fewer yards than 2 standard deviations below the mean number of yards? b) For these data, what does that mean? c) Explain the problem in using a Normal model here.

A high school senior uses the Internet to get information on February temperatures in the town where hell be going to college. He finds a Web site with some statistics, but they are given in degrees Celsius. The conversion formula is \({ }^{\circ} \mathrm{F}=9 / 5^{\circ} \mathrm{C}+32\). Determine the Fahrenheit equivalents for the summary information below. Maximum temperature \(=11^{\circ} \mathrm{C} \quad\) Range \(=33^{\circ}\) Mean \(=1^{\circ} \quad\) Standard deviation \(=7^{\circ}\) Median \(=2^{\circ} \quad\) IQR \(=16^{\circ}\)

Each year thousands of high school students take either the SAT or the ACT, standardized tests used in the college admissions process. Combined SAT Math and Verbal scores go as high as 1600, while the maximum ACT composite score is 36 . Since the two exams use very different scales, comparisons of performance are difficult. A convenient rule of thumb is \(S A T=40 \times A C T+150\); that is, multiply an ACT score by 40 and add 150 points to estimate the equivalent SAT score. An admissions officer reported the following statistics about the ACT scores of 2355 students who applied to her college one year. Find the summaries of equivalent SAT scores. Lowest score \(=19 \quad\) Mean \(=27\) Standard deviation \(=3\) \(\mathrm{Q} 3=30\) Median \(=28 \quad \mathrm{IQR}=6\)

The Virginia Cooperative Extension reports that the mean weight of yearling Angus steers is 1152 pounds. Suppose that weights of all such animals can be described by a Normal model with a standard deviation of 84 pounds. a) How many standard deviations from the mean would a steer weighing 1000 pounds be? b) Which would be more unusual, a steer weighing 1000 pounds or one weighing 1250 pounds?

Here are the summary statistics for the weekly payroll of a small company: lowest salary \(=\$ 300\), mean salary \(=\$ 700\), median \(=\$ 500\), range \(=\$ 1200, \mathrm{IQR}=\) \(\$ 600\), first quartile \(=\$ 350\), standard deviation \(=\$ 400\). a) Do you think the distribution of salaries is symmetric, skewed to the left, or skewed to the right? Explain why. b) Between what two values are the middle \(50 \%\) of the salaries found? c) Suppose business has been good and the company gives every employee a \(\$ 50\) raise. Tell the new value of each of the summary statistics. d) Instead, suppose the company gives each employee a \(10 \%\) raise. Tell the new value of each of the summary statistics.
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