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The plasma cholesterol level (mg/dL) for patients with no prior evidence of heart disease who experience chest pain is normally distributed with mean 200 and standard deviation 35. Consider randomly selecting an individual of this type. What is the probability that the plasma cholesterol level a. Is at most 250 ? b. Is between 300 and 400 ? c. Differs from the mean by at least \(1.5\) standard deviations?

Short Answer

Expert verified
a. 0.9236; b. 0.0021; c. 0.1336

Step by step solution

01

Understand the Problem

The plasma cholesterol levels are normally distributed with a mean \( \mu = 200 \) mg/dL and a standard deviation \( \sigma = 35 \) mg/dL. We need to calculate probabilities related to this distribution.
02

Find Probability for Cholesterol \( \leq 250 \)

To find \( P(X \leq 250) \), convert 250 to a standard normal random variable using \( Z = \frac{X - \mu}{\sigma} \). The Z-score for 250 is \( \frac{250 - 200}{35} \approx 1.43 \). Using the standard normal distribution table, \( P(Z \leq 1.43) \approx 0.9236 \).
03

Find Probability for Cholesterol Between 300 and 400

Convert the values 300 and 400 to the corresponding Z-scores: \( Z_{300} = \frac{300 - 200}{35} \approx 2.86 \) and \( Z_{400} = \frac{400 - 200}{35} \approx 5.71 \). Using the standard normal distribution table, \( P(Z \leq 2.86) \approx 0.9979 \) and \( P(Z \leq 5.71) \approx 1 \). Thus, \( P(300 \leq X \leq 400) = P(Z \leq 5.71) - P(Z \leq 2.86) = 1 - 0.9979 = 0.0021 \).
04

Find Probability Cholesterol Differs From Mean by at Least 1.5 SDs

Values differ by at least 1.5 standard deviations from the mean when \( X \leq \mu - 1.5\sigma \) or \( X \geq \mu + 1.5\sigma \). Calculating these limits: \( \mu \pm 1.5\sigma = 200 \pm 1.5 \times 35 = 200 \pm 52.5 \). \( P(X \leq 147.5) \) or \( P(X \geq 252.5) \). Convert these to Z scores: \( Z_{147.5} = \frac{147.5 - 200}{35} \approx -1.5 \) and \( Z_{252.5} = \frac{252.5 - 200}{35} \approx 1.5 \). \( P(Z \leq -1.5) \approx 0.0668 \); \( P(Z \geq 1.5) = 1 - P(Z \leq 1.5) = 1 - 0.9332 = 0.0668 \). Therefore, \( P(X \leq 147.5 \text{ or } X \geq 252.5) = 0.0668 + 0.0668 = 0.1336 \).

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Key Concepts

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

Probability Calculations
When dealing with normal distributions, one of the primary tasks is to calculate probabilities or the chance of certain events happening. In the given exercise, we are tasked with finding the probabilities linked to specific cholesterol levels, given that the levels follow a normal distribution. This probability calculation often requires transforming the actual data points using a Z-score formula and then using the standard normal distribution to find probabilities.
  • Finding the probability: We determine how likely a specific outcome is, say, whether a cholesterol value is less than or greater than some threshold.

  • Probabilities with Z-scores: Once we calculate the Z-score, we can use standard tables or software to find the probabilities associated with those scores.

This process simplifies finding the probability of any given distribution, as long as the data is normally distributed. For part (a) of the exercise, we found that there is a 92.36% chance the cholesterol level is at most 250 mg/dL.
Standard Normal Distribution
The standard normal distribution is a fundamental concept in statistics, representing a normal distribution with a mean of 0 and a standard deviation of 1. It is the basis for the Z-score calculation and serves as a universal standard for comparing different data sets. By converting any normal distribution to a standard form, we enable easy probability calculations regardless of the original mean or standard deviation.
  • Standard transformation: This transformation involves converting any real-life normal distribution to standard normal form using the formula: \( Z = \frac{X - \mu}{\sigma} \).

  • Use of standard tables: Once you have the Z-score, you can find probabilities using tables that represent areas under the standard curve, which reflect probabilities.

This approach allows us to work universally across various problems, such as in part (c) of the exercise, where we calculate the probabilities for values differing from the mean by a specific number of standard deviations.
Z-score
The Z-score is a numerical measurement that describes how many standard deviations a data point is from the mean. It's a crucial tool for evaluating normal distributions, allowing us to standardize different data points and make comparisons across different datasets.
  • Purpose of the Z-score: It converts various datasets into a common standard, making it easier to calculate probabilities and compare different distributions.

  • Calculation: Given by \( Z = \frac{X - \mu}{\sigma} \), where \(X\) is the data value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

This conversion helps find the probability of a particular score in a normal distribution. For instance, in the exercise, we used this method to determine how extreme or typical certain cholesterol levels were relative to the average.

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