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Chapter 4: Problem 10

Find the SD. Cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 185 milligrams per deciliter \((\mathrm{mg} / \mathrm{dl})\). Women with cholesterol levels above \(220 \mathrm{mg} / \mathrm{dl}\) are considered to have high cholesterol and about \(18.5 \%\) of women fall into this category. Find the standard deviation of this distribution.

Short Answer

Expert verified
The standard deviation is approximately 39.33 mg/dl.

Step by step solution

01

Understanding the Problem

We are given that cholesterol levels follow a normal distribution with a mean of 185 mg/dl. We know that 18.5% of women have cholesterol levels above 220 mg/dl. We need to find the standard deviation of this distribution.
02

Identify the Z-Score

The problem specifies 18.5% of women have cholesterol levels higher than 220 mg/dl. Using the standard normal distribution table or a Z-table, we find the Z-score which leaves 18.5% in the upper tail. The Z-score corresponding to 81.5% from the left (100% - 18.5% = 81.5%) is approximately 0.89.
03

Set up the Z-Score Formula

The Z-score formula is \(Z = \frac{X - \mu}{\sigma}\), where \(X = 220\), \(\mu = 185\), and the unknown \(\sigma\) is the standard deviation. As determined, \(Z = 0.89\).
04

Solve for the Standard Deviation

Substitute the known values into the equation: \(0.89 = \frac{220 - 185}{\sigma}\). Simplify to find \(\sigma\):\[\sigma = \frac{35}{0.89} \approx 39.33\].
05

Verify the Calculation

We double-check the calculation by substituting \(\sigma\) back into the Z-score formula: \(Z = \frac{220 - 185}{39.33} \approx 0.89\), which matches our original determination from the Z-table.

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics, often represented as a bell-shaped curve. It describes the spread and central tendency of a set of data, where most observations cluster around the mean, or average, value.
Key characteristics include:
  • The mean, median, and mode are all equal and located at the center of the distribution.
  • The curve is symmetric about the mean, indicating equal spread on either side.
  • The total area under the curve equals 1, representing the entirety of the probability distribution.
  • Standard deviation determines the width of the curve. A smaller standard deviation results in a steeper curve; a larger one makes it flatter.
Understanding how data follows a normal distribution helps in making informed predictions and assumptions about real-world phenomena.
Z-Score
The Z-score is a statistical measure that describes a value's relation to the mean of a group of values. Specifically, it indicates how many standard deviations a particular datum lies from the average of the data set.
A Z-score can help understand the positioning of data points within a normal distribution.
  • If the Z-score is zero, it means the data point's score is identical to the mean score.
  • A positive Z-score indicates the data point is above the mean, while a negative Z-score signifies it is below the mean.
  • The magnitude of the Z-score indicates how far the data point deviates from the mean.
Using the Z-score, one can calculate the probability of a particular score occurring within a distribution, or compare scores from different normal distributions.
Cholesterol Levels
Cholesterol, a vital substance needed for building cells, exists at varying concentrations in people's blood. Cholesterol levels are typically measured in milligrams per deciliter (mg/dl). While cholesterol is necessary for normal bodily functions, abnormal levels can pose health risks.
In analyzing cholesterol data:
  • A normal range indicates a low risk of cardiovascular diseases.
  • A cholesterol level above 220 mg/dl, as in this exercise, often signals a potential health concern that requires monitoring or intervention.
Statistical methods, like those utilizing a normal distribution, help professionals understand and anticipate health trends among different populations.
Statistical Calculation
When performing statistical calculations, like finding the standard deviation, it's important to break down the process into understandable steps. In this scenario, the goal is to find the spread of cholesterol levels that follow a normal distribution pattern.
The given problem uses a Z-score to translate real-world cholesterol data into the language of the normal distribution.
The steps to solve the exercise were straightforward:
  • Identify the percentage given (18.5% with cholesterol above 220 mg/dl).
  • Convert this percentage to a Z-score using a Z-table, accounting for the right tail of the distribution.
  • Use the Z-score formula: \( Z = \frac{X - \mu}{\sigma} \) to find the standard deviation \( \sigma \).
  • Substitute back to ensure completeness and accuracy of the calculations.
These steps illustrate how statistics can turn complex health data into actionable insights.

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Most popular questions from this chapter

GRE scores, Part 1. Sophia who took the Graduate Record Examination (GRE) scored 160 on the Verbal Reasoning section and 157 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section for all test takers was 151 with a standard deviation of \(7,\) and the mean score for the Quantitative Reasoning was 153 with a standard deviation of 7.67 . Suppose that both distributions are nearly normal. (a) What is Sophia's Z-score on the Verbal Reasoning section? On the Quantitative Reasoning section? Draw a standard normal distribution curve and mark these two Z-scores. (b) What do these Z-scores tell you? (c) Relative to others, which section did she do better on? (d) Find her percentile scores for the two exams. (e) What percent of the test takers did better than her on the Verbal Reasoning section? On the Quantitative Reasoning section? (f) Explain why simply comparing raw scores from the two sections could lead to an incorrect conclusion as to which section a student did better on. (g) If the distributions of the scores on these exams are not nearly normal, would your answers to parts (b) \- (e) change? Explain your reasoning.

Distribution of \(\mu\). Suppose the true population proportion were \(p=0.6\) and a researcher takes a simple random sample of size \(n=50\). (a) Find and interpret the standard deviation of the sample proportion \(\hat{p} .\) (b) Calculate the probability that the sample propartion will be larger than 0.65 for a random sample of size 50 .

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Stats final scores. Each year about 1500 students take the introductory statistics course at a large university. This year scores on the final exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the final) but a few students scored below 20 points. (a) Is the distribution of scores on this final exam symmetric, right skewed, or left skewed? (b) Would you expect most students to have scored above or below 70 points? (c) Can we calculate the probability that a randomly chosen student scored abave 75 using the normal distribution? (d) What is the probability that the average score for a random sample of 40 students is above \(75 ?\) (e) How would eutting the sample size in half affect the standard deviation of the mean?

With and without replacement. In the following situations assume that half of the specified population is male and the other half is female. (a) Suppose you're sampling from a room with 10 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (b) Now suppose you're sampling from a stadium with 10,000 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement? (c) We often treat individuals who are sampled from a large population as independent. Using your findings from parts (a) and (b), explain whether or not this assumption is reasonable.
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