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Chapter 4: Problem 69

Assume that \(X\) is normally distributed with a mean of 5 and a standard deviation of \(4 .\) Determine the following: (a) \(P(X<11)\) (b) \(P(X>0)\) (c) \(P(3

Short Answer

Expert verified
(a) 0.9332, (b) 0.8944, (c) 0.3830, (d) 0.8012, (e) 0.5468

Step by step solution

01

Standardize the Random Variable

To find probabilities related to a normal distribution, we need to standardize (convert to a Z-score) using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where \( \mu = 5 \) and \( \sigma = 4 \). This will help to use standard normal distribution tables or a calculator later.
02

Calculate P(X

For \( P(X<11) \), compute the Z-score using:\[ Z = \frac{11 - 5}{4} = 1.5 \]Using the standard normal distribution table or calculator, find \( P(Z<1.5) \). This value is approximately 0.9332.
03

Calculate P(X>0)

For \( P(X>0) \), compute the Z-score using:\[ Z = \frac{0 - 5}{4} = -1.25 \]Using the standard normal distribution table or calculator, find \( P(Z>-1.25) = 1 - P(Z< -1.25) \). This value is approximately 0.8944.
04

Calculate P(3

For \( P(3<X<7) \), calculate two Z-scores:\[ Z_1 = \frac{3 - 5}{4} = -0.5 \] and \[ Z_2 = \frac{7 - 5}{4} = 0.5 \]Thus, \( P(-0.5<Z<0.5) \) can be found by \( P(Z<0.5) - P(Z<-0.5) \), which is approximately 0.6915 - 0.3085 = 0.3830.
05

Calculate P(-2

For \( P(-2<X<9) \), calculate two Z-scores:\[ Z_1 = \frac{-2 - 5}{4} = -1.75 \] and \[ Z_2 = \frac{9 - 5}{4} = 1.0 \]Thus, \( P(-1.75<Z<1.0) \) can be found by \( P(Z<1.0) - P(Z<-1.75) \), which is approximately 0.8413 - 0.0401 = 0.8012.
06

Calculate P(2

For \( P(2<X<8) \), calculate two Z-scores:\[ Z_1 = \frac{2 - 5}{4} = -0.75 \] and \[ Z_2 = \frac{8 - 5}{4} = 0.75 \]Thus, \( P(-0.75<Z<0.75) \) can be found by \( P(Z<0.75) - P(Z<-0.75) \), which is approximately 0.7734 - 0.2266 = 0.5468.

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Key Concepts

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

Z-score
Understanding Z-scores is essential in statistics, especially when dealing with a normal distribution. A Z-score represents how many standard deviations a particular value (X) is from the mean (\mu). This helps to standardize different data points for comparison. The formula for calculating the Z-score is:
  • \[ Z = \frac{X - \mu}{\sigma} \]
Where:
  • \(X\) is the value from the dataset.
  • \(\mu\) is the mean of the dataset.
  • \(\sigma\) is the standard deviation.
By converting raw scores into standardized scores (Z-scores), we can easily determine probability using the standard normal distribution table. A Z-score of 0 indicates the score is exactly at the mean, while positive or negative values tell us how far and in what direction the score deviates from the mean.
Standard Normal Distribution
The Standard Normal Distribution is a special case of the normal distribution. Here, the mean is 0, and the standard deviation is 1. This helps statisticians and researchers use Z-scores in a unified framework to calculate probabilities and interpret data. The bell-shaped curve of a standard normal distribution allows for evaluating data using probabilities derived from Z-scores.
  • A Z-score helps you locate the position of sample data within this distribution.
  • The table associated with the standard normal distribution contains cumulative probability values for corresponding Z-scores.
By referring to this standard, it becomes simple to compute probabilities, which are essential when making predictions or hypotheses about a dataset.
Probability Calculation
Probability calculations help us understand the likelihood of an event happening in a normally distributed dataset. After converting a score to a Z-score, we use it to check the probability value.For example, when calculating the probability \(P(X<11)\):
  • First, find the Z-score by standardizing it.
  • Then, use the standard normal distribution table to locate the cumulative probability associated with that Z-score.
  • In this case, compute \(P(Z<1.5)\), which gives a probability of approximately 0.9332.
Understanding this process is vital in various applications like quality control, risk assessment, and prediction models. By knowing the probability, one can make informed decisions based on the data distribution.
Standard Deviation
Standard deviation is a measure of how much variation exists within a dataset. It tells us how spread out the data points are from the mean. A small standard deviation indicates that the data points are clustered close to the mean, and a large standard deviation suggests a wide spread of values.In formulas like the Z-score calculation, standard deviation serves as a scaling factor, ensuring that data from different scales can be compared universally. With a standard deviation \(\sigma\), you gain insights into the reliability and consistency of your data:
  • It informs about data variability, crucial for understanding the data's nature.
  • Helps in assessing risks and identifying data anomalies.
It is a cornerstone concept in statistics, essential for interpreting data and making data-driven decisions.

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Most popular questions from this chapter

Consider the regional right ventricle transverse wall motion in patients with pulmonary hypertension (PH). The right ventricle ejection fraction (EF) is approximately normally distributed with standard deviation of 12 for \(\mathrm{PH}\) subjects, and with mean and standard deviation of 56 and \(8,\) respectively, for control subjects. (a) What is the EF for control subjects exceeded with \(99 \%\) probability? (b) What is the mean for PH subjects such that the probability is \(1 \%\) that the EF of a PH subject is greater than the value in part (a)? (c) Comment on how well the control and PH subjects [with the mean determined in part (b)] can be distinguished by EF measurements.

Without an automated irrigation system, the height of plants two weeks after germination is normally distributed with a mean of 2.5 centimeters and a standard deviation of 0.5 centimeter. (a) What is the probability that a plant's height is greater than 2.25 centimeters? (b) What is the probability that a plant's height is between 2.0 and 3.0 centimeters?

Suppose that \(X\) has a Weibull distribution with \(\beta=0.2\) and \(\delta=100\) hours. Determine the following: (a) \(P(X<10,000)\) (b) \(P(X>5000)\)

The life (in hours) of a magnetic resonance imaging machine (MRI) is modeled by a Weibull distribution with parameters \(\beta=2\) and \(\delta=500\) hours. Determine the following: (a) Mean life of the MRI (b) Variance of the life of the MRI (c) Probability that the MRI fails before 250 hours.

The length of time (in seconds) that a user views a page on a Web site before moving to another page is a lognormal random variable with parameters \(\theta=0.5\) and \(\omega^{2}=1\). (a) What is the probability that a page is viewed for more than 10 seconds? (b) By what length of time have \(50 \%\) of the users moved to another page? (c) What are the mean and standard deviation of the time until a user moves from the page?
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