/*! 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 32 If \(X\) is a normal rv with mea... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If \(X\) is a normal rv with mean 80 and standard deviation 10 , compute the following probabilities by standardizing: a. \(P(X \leq 100)\) b. \(P(X \leq 80)\) c. \(P(65 \leq X \leq 100)\) d. \(P(70 \leq X)\) e. \(P(85 \leq X \leq 95)\) f. \(P(|X-80| \leq 10)\)

Short Answer

Expert verified
a) 0.9772, b) 0.5, c) 0.9104, d) 0.8413, e) 0.2417, f) 0.6826.

Step by step solution

01

Introduction to Standardization

To find probabilities involving a normal random variable, we first convert the variable to a standard normal distribution. This involves calculating the Z-score using the formula: \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
02

Calculate Z for Part a

For \( P(X \leq 100) \):\[Z = \frac{100 - 80}{10} = 2\]Consult a standard normal distribution table or use a calculator to find \( P(Z \leq 2) = 0.9772 \).
03

Calculate Z for Part b

For \( P(X \leq 80) \):\[Z = \frac{80 - 80}{10} = 0\]Since \( P(Z \leq 0) = 0.5 \) by the symmetry of the normal distribution, this is the desired probability.
04

Calculate Z for Part c

For \( P(65 \leq X \leq 100) \):First, find the Z-score for \( X=65 \):\[Z = \frac{65 - 80}{10} = -1.5\]Next, use the Z-score for \( X=100 \) from Step 1 (which is 2). Thus, the probability is:\( P(-1.5 \leq Z \leq 2) = P(Z \leq 2) - P(Z \leq -1.5) = 0.9772 - 0.0668 = 0.9104 \).
05

Calculate Z for Part d

For \( P(70 \leq X) \):Find the Z-score for \( X=70 \):\[Z = \frac{70 - 80}{10} = -1\]Therefore, \( P(Z \geq -1) = 1 - P(Z \leq -1) = 1 - 0.1587 = 0.8413 \).
06

Calculate Z for Part e

For \( P(85 \leq X \leq 95) \):Find the Z-score for \( X=85 \):\[Z = \frac{85 - 80}{10} = 0.5\]Find the Z-score for \( X=95 \):\[Z = \frac{95 - 80}{10} = 1.5\]Thus, \( P(0.5 \leq Z \leq 1.5) = P(Z \leq 1.5) - P(Z \leq 0.5) = 0.9332 - 0.6915 = 0.2417 \).
07

Calculate Range for Part f

For \( P(|X-80| \leq 10) \), this means \( P(70 \leq X \leq 90) \).Find Z-scores for \( X=70 \) and \( X=90 \):\[Z_{70} = \frac{70 - 80}{10} = -1\]\[Z_{90} = \frac{90 - 80}{10} = 1\]Probability is \( P(-1 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq -1) = 0.8413 - 0.1587 = 0.6826 \).

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.

Z-score
The Z-score is a key concept when dealing with normal distributions. It helps us understand how a particular value compares to the mean of a distribution. The formula for calculating the Z-score is:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here, \(X\) is the observed value, \(\mu\) represents the mean of the distribution, and \(\sigma\) stands for the standard deviation.

Z-scores tell us how many standard deviations an element is from the mean. A Z-score of 2.0 indicates that the data point is 2 standard deviations above the mean.

This calculation standardizes our data, making it possible to find probabilities using the standard normal distribution tables or calculators. This conversion to standard normal form simplifies complex probability questions into more manageable forms.
Standard Normal Distribution
The Standard Normal Distribution, often portrayed as a bell curve, is a special case of the normal distribution. Its key characteristics include:
  • A mean of 0
  • A standard deviation of 1
Using this distribution simplifies the process of calculating probabilities since standard normal distribution tables are readily available for reference.

By transforming a normal random variable into a standard normal variable through the Z-score, we shift and scale the original distribution to match this standard form. The shape remains a perfect bell curve, providing an easy framework for finding probabilities through standard tables. This approach is essential in statistical methods, instead of needing separate calculations for each distinct normal distribution.
Probability Calculations
Once we have the Z-scores, probability calculations become straightforward. Here’s how it works:
  • Look up the Z-score in the standard normal distribution table. This gives the cumulative probability for values less than or equal to the desired point.
  • For probabilities greater than a certain point, subtract the table value from 1.
  • To find probabilities between two points, find the difference between their cumulative probabilities.

This method is efficient since it uses pre-calculated values in standard tables or statistical software. Calculating probabilities with Z-scores saves time and avoids potential errors in manual computation.
Standardization
Standardization is a vital concept in statistics when comparing different data sets. The process involves converting individual data points into standard scores (Z-scores). This transformation has several benefits:
  • Allows comparison across different data sets with differing units or scales.
  • Prepares the data for advanced statistical operations, like hypothesis testing.
  • Transforms complex distributions into a common standard form, facilitating easier interpretation.

By standardizing, we abstract individual data points into a uniform scale, making analyses more coherent and allowing for more straightforward comparisons and conclusions across varied contexts. It’s pivotal in making the daunting task of probability finding simpler and systematic.

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 lifetime \(X\) (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=3\). Compute the following: a. \(E(X)\) and \(V(X)\) b. \(P(X \leq 6)\) c. \(P(1.5 \leq X \leq 6)\) (This Weibull distribution is suggested as a model for time in service in "On the Assessment of Equipment Reliability: Trading Data Collection Costs for Precision," J. Engr. Manuf., 1991: 105-109.)

Let \(X\) have a standard beta density with parameters \(\alpha\) and \(\beta\). a. Verify the formula for \(E(X)\) given in the section. b. Compute \(E\left[(1-X)^{m}\right]\). If \(X\) represents the proportion of a substance consisting of a particular ingredient, what is the expected proportion that does not consist of this ingredient?

The time \(X(\min\) ) for a lab assistant to prepare the equipment for a certain experiment is believed to have a uniform distribution with \(A=25\) and \(B=35\). a. Determine the pdf of \(X\) and sketch the corresponding density curve. b. What is the probability that preparation time exceeds 33 min? c. What is the probability that preparation time is within 2 min of the mean time? [Hint: Identify \(\mu\) from the graph of \(f(x) .]\) d. For any \(a\) such that \(25

Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 104 and standard deviation 5 (information in the article "Mathematical Model of Chloride Concentration in Human Blood," J. of Med. Engr. and Tech., 2006: 25-30, including a normal probability plot as described in Section 4.6, supports this assumption). a. What is the probability that chloride concentration equals 105 ? Is less than 105 ? Is at most 105 ? b. What is the probability that chloride concentration differs from the mean by more than 1 standard deviation? Does this probability depend on the values of \(\mu\) and \(\sigma\) ? c. How would you characterize the most extreme . \(1 \%\)

There are two machines available for cutting corks intended for use in wine bottles. The first produces corks with diameters that are normally distributed with mean \(3 \mathrm{~cm}\) and standard deviation \(.1 \mathrm{~cm}\). The second machine produces corks with diameters that have a normal distribution with mean \(3.04 \mathrm{~cm}\) and standard deviation \(.02 \mathrm{~cm}\). Acceptable corks have diameters between \(2.9 \mathrm{~cm}\) and \(3.1 \mathrm{~cm}\). Which machine is more likely to produce an acceptable cork?

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.