/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Cerebral blood flow (CBF) in the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Cerebral blood flow (CBF) in the brains of healthy people is normally distributed with a mean of 74 and a standard deviation of 16 a. What proportion of healthy people will have CBF readings between 60 and \(80 ?\) b. What proportion of healthy people will have CBF readings above \(100 ?\) c. If a person has a CBF reading below \(40,\) he is classified as at risk for a stroke. What proportion of healthy people will mistakenly be diagnosed as "at risk"?

Short Answer

Expert verified
Answer: Approximately 45.54% of healthy people will have CBF readings between 60 and 80, 5.16% will have readings above 100, and 1.66% will have readings below 40.

Step by step solution

01

Understanding the given information and calculating z-scores for the intervals

We have the mean μ = 74 and the standard deviation σ = 16. We need to find the z-scores for the given CBF values in each part of the problem. The formula for calculating the z-score is: $$z = \frac{x - \mu}{\sigma}$$ where x is the CBF value, μ is the mean, and σ is the standard deviation.
02

Solving Part a - Proportion of people with CBF readings between 60 and 80

First, we need to find the z-scores for both 60 and 80 using the formula mentioned in step 1. For x=60: $$z_1 = \frac{60 - 74}{16} = -0.875$$ For x=80: $$z_2 = \frac{80 - 74}{16} = 0.375$$ Now we have to find the probability between these two z-scores. P(60 < CBF < 80) = P(-0.875 < Z < 0.375) We can look up these z-scores in the standard normal distribution table and find the probabilities. P(Z < -0.875) ≈ 0.1909 P(Z < 0.375) ≈ 0.6463 We subtract the two probabilities to find the proportion of people with CBF between 60 and 80. P(-0.875 < Z < 0.375) ≈ 0.6463 - 0.1909 = 0.4554 Approximately 45.54% of healthy people will have CBF readings between 60 and 80.
03

Solving Part b - Proportion of people with CBF readings above 100

First, find the z-score for x=100: $$z = \frac{100 - 74}{16} = 1.625$$ Now, we find the probability of having a CBF reading above 100. P(CBF > 100) = P(Z > 1.625) We look up the z-score in the standard normal distribution table: P(Z < 1.625) ≈ 0.9484 Since we want the probability above 1.625, we can calculate: P(Z > 1.625) = 1 - P(Z < 1.625) = 1 - 0.9484 = 0.0516 Approximately 5.16% of healthy people will have CBF readings above 100.
04

Solving Part c - Proportion of people with CBF readings below 40 and risk of stroke diagnosis

First, find the z-score for x=40: $$z = \frac{40 - 74}{16} = -2.125$$ Now, we find the probability of having a CBF reading below 40. P(CBF < 40) = P(Z < -2.125) We look up the z-score in the standard normal distribution table: P(Z < -2.125) ≈ 0.0166 Approximately 1.66% of healthy people will have CBF readings below 40. This means that 1.66% of healthy people will mistakenly be diagnosed as "at risk" for a stroke.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Z-Score
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If you are given a z-score of 0, it means that the data point is exactly at the mean. A positive z-score indicates that the data point is above the mean, while a negative z-score signifies that it is below the mean.

For example, in the exercise about cerebral blood flow (CBF), z-scores are calculated to determine how a person's CBF reading compares to the average healthy individual's reading. A z-score is calculated using the formula:
\[z = \frac{x - \mu}{\sigma}\]
where \(x\) is the value of interest, \(\mu\) is the mean, and \(\sigma\) is the standard deviation of the dataset. In this way, z-scores allow comparison between different data points in a distribution, or between a data point and the mean of a distribution, providing insight on how 'unusual' a reading is.
The Role of Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the data points tend to be close to the mean, whereas a high standard deviation indicates that the data is spread out over a wider range of values.

Using our CBF example, the standard deviation is 16. This tells us that the CBF readings of healthy people typically vary by 16 units above or below the mean, which is 74. To put this into perspective, consider that a CBF value that is one standard deviation away from the mean (either above or below) would fall within the range most people are expected to be. Readings that are two or three standard deviations away from the mean are much rarer and could be considered outliers. The standard deviation is crucial when calculating z-scores, as it helps standardize different data points for comparison.
Probability in Normal Distribution
Probability, in the context of a normal distribution, refers to the likelihood of a certain range of outcomes occurring. The total area under a normal distribution curve represents a probability of 1 (or 100%). Probabilities for specific intervals are represented as areas under the curve between two points, often calculated using z-scores.

For instance, when we want to find the proportion of people with CBF readings between 60 and 80, we need to calculate the probability that a person’s CBF falls within this interval. This is done by finding the z-scores for 60 and 80 and then determining the area under the standard normal distribution curve between these z-scores. This area represents the probability we are looking for and provides insight into how common or rare that range of readings is among the healthy population.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The diameters of Douglas firs grown at a Christmas tree farm are normally distributed with a mean of 4 inches and a standard devia- tion of 1.5 inches. a. What proportion of the trees will have diameters between 3 and 5 inches? b. What proportion of the trees will have diameters less than 3 inches? c. Your Christmas tree stand will expand to a diameter of 6 inches. What proportion of the trees will not fit in your Christmas tree stand?

The meat department at a local supermarket specifically prepares its " 1 -pound" packages of ground beef so that there will be a variety of weights, some slightly more and some slightly less than 1 pound. Suppose that the weights of these "1pound" packages are normally distributed with a mean of 1.00 pound and a standard deviation of .15 pound. a. What proportion of the packages will weigh more than 1 pound? b. What proportion of the packages will weigh between .95 and 1.05 pounds? c. What is the probability that a randomly selected package of ground beef will weigh less than .80 pound? d. Would it be unusual to find a package of ground beef that weighs 1.45 pounds? How would you explain such a large package?

Let \(x\) be a binomial random variable with \(n=100\) and \(p=.2 .\) Find approximations to these probabilities: a. \(P(x>22)\) b. \(P(x \geq 22)\) c. \(P(20

Calculate the area under the standard normal curve between these values: a. \(z=-2.0\) and \(z=2.0\) b. \(z=-2.3\) and -1.5

The life span of oil-drilling bits depends on the types of rock and soil that the drill encounters, but it is estimated that the mean length of life is 75 hours. Suppose an oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 75 hours and a standard deviation equal to 12 hours. a. What proportion of the company's drill bits will fail before 60 hours of use? b. What proportion will last at least 60 hours? c. What proportion will have to be replaced after more than 90 hours of use?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.