/*! 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 26 Porphyrin is a pigment in blood ... [FREE SOLUTION] | 91Ó°ÊÓ

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Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Let \(x\) be a random variable that represents the number of milligrams of porphyrin per deciliter of blood. In healthy adults, \(x\) is approximately normally distributed with mean \(\mu=38\) and standard deviation \(\sigma=12\) (see reference in Problem 25 ). What is the probability that : (a) \(x\) is less than \(60 ?\) (b) \(x\) is greater than \(16 ?\) (c) \(x\) is between 16 and \(60 ?\) (d) \(x\) is more than \(60 ?\) (This may indicate an infection, anemia, or another type lness.)

Short Answer

Expert verified
(a) 0.9664, (b) 0.9664, (c) 0.9328, (d) 0.0336.

Step by step solution

01

Understand the Distribution

The random variable \(x\) is normally distributed with mean \(\mu = 38\) and standard deviation \(\sigma = 12\). We will use this information to find probabilities for different ranges.
02

Find the Z-score for Part (a)

To find \(P(x < 60)\), we must first convert \(x = 60\) to a z-score using the formula \(z = \frac{x - \mu}{\sigma}\). Thus, \(z = \frac{60 - 38}{12} = \frac{22}{12} \approx 1.83\).
03

Calculate Probability for Part (a)

Using the standard normal distribution table or calculator, find \(P(z < 1.83)\). From the z-table, \(P(z < 1.83) \approx 0.9664\). Therefore, \(P(x < 60) = 0.9664\).
04

Find the Z-score for Part (b)

For \(P(x > 16)\), convert \(x = 16\) to its z-score: \(z = \frac{16 - 38}{12} = \frac{-22}{12} \approx -1.83\).
05

Calculate Probability for Part (b)

Find \(P(z > -1.83)\). From the z-table, \(P(z < -1.83) \approx 0.0336\). Therefore, \(P(z > -1.83) = 1 - 0.0336 = 0.9664\). Thus, \(P(x > 16) = 0.9664\).
06

Combine Probabilities for Part (c)

For \(P(16 < x < 60)\), we use the fact that it equals \(P(x < 60) - P(x < 16)\). From previous calculations, \(P(x < 16) = 0.0336\) and \(P(x < 60) = 0.9664\). Thus, \(P(16 < x < 60) = 0.9664 - 0.0336 = 0.9328\).
07

Use Complement Rule for Part (d)

For \(P(x > 60)\), use \(P(x > 60) = 1 - P(x < 60)\). Since \(P(x < 60) = 0.9664\), \(P(x > 60) = 1 - 0.9664 = 0.0336\).

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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 key tool for understanding how a particular value compares to the average of a dataset, especially when dealing with normal distributions. In simple terms, a Z-score tells you how many standard deviations away a given point is from the mean of the dataset. - **Formula**: The Z-score is calculated using the formula: \[ z = \frac{x - \mu}{\sigma} \] where: - \(x\) is the value being standardized, - \(\mu\) is the mean of the dataset, - \(\sigma\) is the standard deviation.By converting an observation to a Z-score, you can determine the likelihood of observing a value below or above a certain threshold, using standard normal distribution tables or software. For example, a Z-score of 1.83 means the value is 1.83 standard deviations above the mean. Understanding Z-scores can thus provide significant insights when evaluating probabilities in a normal distribution.
Mean and Standard Deviation
Mean and Standard Deviation are foundational concepts in statistics that help describe a dataset's characteristics. The **mean** represents the average of all data points, acting as a central point of the data distribution. It is calculated by summing all values and dividing by the number of observations. When the mean is used with normally distributed data, it serves as a balancing point, with half the data on either side. The **standard deviation** measures the spread or variability within the dataset, showing how much individual data points deviate from the mean. A smaller standard deviation indicates that the data points tend to be close to the mean, while a larger standard deviation suggests the data is more spread out. In the context of a normal distribution, approximately 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three, making these parameters critical for understanding data behavior. Together, knowing the mean and standard deviation allows for detailed probability calculations and insights into the dataset's distribution.
Probability Calculations
Probability calculations in statistics help determine the likelihood that a particular event will occur, and they can be particularly powerful when applied to a normally distributed dataset. For normally distributed variables: - **Using Z-scores**: Once you've converted your variable to a Z-score, you can use standard normal distribution tables to find probabilities associated with certain values. This tells you the probability that a variable will fall below a particular observed value. - **Range probabilities**: To find the probability that a variable falls within a certain range, calculate the probabilities of each endpoint using Z-scores and subtract. For instance, to find the probability of a variable falling between two values, subtract the probability of being below the lower value from the probability of being above the higher one. Understanding these calculations allows you to make informed predictions about your data based on statistical probabilities, aiding in decision-making and hypothesis testing.
Complement Rule
The complement rule in probability is a straightforward concept, yet it is extremely useful, especially in the context of calculating probabilities for normal distributions. The rule states that the probability of an event not occurring is equal to one minus the probability of the event occurring. Mathematically, this is expressed as:- \( P(A^c) = 1 - P(A) \) where \(A^c\) is the complement event of \(A\).This rule is particularly helpful because it often simplifies probability calculations, avoiding cumbersome direct calculations. For instance, if you know the probability of a random variable being less than a certain value in a normal distribution, you can easily find the probability of it being greater than that value by subtracting from one. Using the complement rule can thus make it more straightforward to solve complex statistical problems, enhancing the efficiency and intuitiveness of probability calculations.

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Most popular questions from this chapter

Let \(x\) be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12 -hour fast. Assume that for people under 50 years old, \(x\) has a distribution that is approximately normal, with mean \(\mu=85\) and estimated standard deviation \(\sigma=25\) (based on information from Diagnostic Tests with Nursing Applications, edited by S. Loeb, Springhouse). A test result \(x<40\) is an indication of severe excess insulin, and medication is usually prescribed. (a) What is the probability that, on a single test, \(x<40 ?\) (b) Suppose a doctor uses the average \(\bar{x}\) for two tests taken about a week apart. What can we say about the probability distribution of \(\bar{x} ?\) Hint: See Theorem \(6.1 .\) What is the probability that \(\bar{x}<40 ?\) (c) Repeat part (b) for \(n=3\) tests taken a week apart. (d) Repeat part (b) for \(n=5\) tests taken a week apart. (c) Interpretation Compare your answers to parts (a), (b), (c), and (d). Did the probabilities decrease as \(n\) increased? Explain what this might imply if you were a doctor or a nurse. If a patient had a test result of \(\bar{x}<40\) based on five tests, explain why either you are looking at an extremely rare event or (more likely) the person has a case of excess insulin.

Attendance at large exhibition shows in Denver averages about 8000 people per day, with standard deviation of about \(500 .\) Assume that the daily attendance figures follow a normal distribution. (a) What is the probability that the daily attendance will be fewer than 7200 people? (b) What is the probability that the daily attendance will be more than 8900 people? (c) What is the probability that the daily attendance will be between 7200 and 8900 people?

Is \(\hat{p}\) an unbiased estimator for \(p\) when \(n p>5\) and \(n q>5 ?\) Recall that a statistic is an unbiased estimator of the corresponding parameter if the mean of the sampling distribution equals the parameter in question.

Consider two \(\bar{x}\) distributions corresponding to the same \(x\) distribution. The first \(\bar{x}\) distribution is based on samples of size \(n=100\) and the second is based on samples of size \(n=225 .\) Which \(\bar{x}\) distribution has the smaller standard error? Explain.

The visibility standard index (VSI) is a measure of Denver air pollution that is reported each day in the Denver Post. The index ranges from 0 (excellent air quality) to 200 (very bad air quality). During winter months, when air pollution is higher, the index has a mean of about 90 (rated as fair) with a standard deviation of approximately 30\. Suppose that for 15 days, the following VSI measures were reported each day: $$\begin{array}{l|ccccccccc} \hline \text { Day } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \text { VSI } & 80 & 115 & 100 & 90 & 15 & 10 & 53 & 75 & 80 \\ \hline & & & & & & & & & \\ \hline \text { Day } & 10 & 11 & 12 & 13 & 14 & 15 & & & \\ \hline \text { VSI } & 110 & 165 & 160 & 120 & 140 & 195 & & & \\ \hline \end{array}$$ Make a control chart for the VSI, and plot the preceding data on the control chart. Identify all out-of-control signals (high or low) that you find in the control chart by type (I, II, or III).

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