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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

The first Stats exam had a mean of 65 and a standard deviation of 10 points; the second had a mean of 80 and a standard deviation of 5 points. Derrick scored an 80 on both tests. Julie scored a 70 on the first test and a 90 on the second. They both totaled 160 points on the two exams, but Julie claims that her total is better. Explain.

. Exercise 26 proposes modeling IQ scores with \(N(100,16) .\) What IQ would you consider to be unusually high? Explain.

In the Normal model \(N(100,16)\), what cutoff value bounds a) the highest \(5 \%\) of all IQs? b) the lowest \(30 \%\) of the IQs? c) the middle \(80 \%\) of the IQs?

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?

A forester measured 27 of the trees in a large woods that is up for sale. He found a mean diameter of \(10.4\) inches and a standard deviation of \(4.7\) inches. Suppose that these trees provide an accurate description of the whole forest and that a Normal model applies. a) Draw the Normal model for tree diameters. b) What size would you expect the central \(95 \%\) of all trees to be? c) About what percent of the trees should be less than an inch in diameter? d) About what percent of the trees should be between \(5.7\) and \(10.4\) inches in diameter? e) About what percent of the trees should be over 15 inches in diameter?

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