/*! 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 9 Assume that \(x\) has a normal d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 9

Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. $$ P(8 \leq x \leq 12) ; \mu=15 ; \sigma=3.2 $$

Short Answer

Expert verified
The probability is approximately 0.1605.

Step by step solution

01

Understand the Normal Distribution

We have a normal distribution variable \( x \) with mean \( \mu = 15 \) and standard deviation \( \sigma = 3.2 \). We need to find the probability \( P(8 \leq x \leq 12) \).
02

Convert to Standard Normal Distribution

First, we convert the values 8 and 12 into their respective z-scores using the formula \( z = \frac{x - \mu}{\sigma} \). Calculate \( z_1 \) for \( 8 \): \[ z_1 = \frac{8 - 15}{3.2} = -2.1875 \]. Next, calculate \( z_2 \) for \( 12 \): \[ z_2 = \frac{12 - 15}{3.2} = -0.9375 \].
03

Find Probabilities from Z-Scores

Using a standard normal distribution table or calculator, find \( P(Z \leq -2.1875) \) and \( P(Z \leq -0.9375) \). These are the probabilities for the z-scores.
04

Calculate Probability Between the Z-Scores

Use the standard normal table to find \( P(Z \leq -2.1875) \), which is approximately 0.0143, and \( P(Z \leq -0.9375) \), which is approximately 0.1748. The probability \( P(-2.1875 \leq Z \leq -0.9375) \) is the difference: \( 0.1748 - 0.0143 = 0.1605 \).
05

Conclusion

Thus, the probability \( P(8 \leq x \leq 12) \) is approximately 0.1605 for the given normal distribution with \( \mu = 15 \) and \( \sigma = 3.2 \).

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 91影视!

Key Concepts

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

Understanding Z-Scores
When working with normal distributions, z-scores are incredibly helpful. A z-score measures how many standard deviations a specific value is from the mean of the distribution. To calculate a z-score, you use the formula: \( z = \frac{x - \mu}{\sigma} \). Here, \( x \) represents the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Z-scores help in converting any normal distribution to a standard normal distribution, which has a mean of 0 and a standard deviation of 1. This makes it easier to find probabilities using standard normal tables or calculators. In our example, the z-score for \( x = 8 \) was calculated as \( -2.1875 \), and for \( x = 12 \), it was \( -0.9375 \).

  • A negative z-score indicates a value below the mean.
  • A positive z-score indicates a value above the mean.
  • Z-scores can help compare values from different distributions.
Role of Standard Deviation
Standard deviation, denoted by \( \sigma \), is a vital statistical measure that tells us how much data points in a distribution deviate from the mean. In simpler terms, it reflects the spread or dispersion of a set of data points.

A smaller standard deviation means that data points are close to the mean, while a larger standard deviation indicates more spread out data. In the exercise, the standard deviation is 3.2, which defines how spread out the values of \( x \) are around their mean of 15.

  • Helps to determine the variability of the dataset.
  • Critical for transforming the normal distribution into a standard normal distribution.
  • Essential for calculating z-scores.
Understanding Probability in Normal Distribution
Probability in the context of normal distribution deals with finding the likelihood that a variable falls within a specific range. Given a normal distribution defined by its mean and standard deviation, the probability \( P(a \leq x \leq b) \) is the area under the distribution curve between two points \( a \) and \( b \).

To find this probability, converting the normal distribution to a standard normal distribution using z-scores is essential. Then, using the standard normal distribution table or a calculator, you can identify the probabilities associated with these z-scores. In our scenario, the probabilities for \( z = -2.1875 \) and \( z = -0.9375 \) were found to be approximately 0.0143 and 0.1748, respectively.

  • Sum of probabilities for all possibilities is always 1.
  • The total area under the normal distribution curve equals 1.
  • Probabilities help us understand the behavior of a random variable within the distribution.

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

What is a population parameter? Give three examples.

Templeton World is a mutual fund that invests in both U.S. and foreign markets. Let \(x\) be a random variable that represents the monthly percentage return for the Templeton World fund. Based on information from the Morningstar Guide to Mutual Funds (available in most libraries), \(x\) has mean \(\mu=1.6 \%\) and standard deviation \(\sigma=0.9 \%\). (a) Templeton World fund has over 250 stocks that combine together to give the overall monthly percentage return \(x\). We can consider the monthly return of the stocks in the fund to be a sample from the population of monthly returns of all world stocks. Then we see that the overall monthly return \(x\) for Templeton World fund is itself an average return computed using all 250 stocks in the fund. Why would this indicate that \(x\) has an approximately normal distribution? Explain. Hint: See the discussion after Theorem \(7.2\). (b) After 6 months, what is the probability that the average monthly percentage return \(\bar{x}\) will be between \(1 \%\) and \(2 \%\) ? Hint: See Theorem \(7.1\), and assume that \(x\) has a normal distribution as based on part (a). (c) After 2 years, what is the probability that \(\bar{x}\) will be between \(1 \%\) and \(2 \%\) ? (d) Compare your answers to parts (b) and (c). Did the probability increase as \(n\) (number of months) increased? Why would this happen? (e) If after 2 years the average monthly percentage return \(\bar{x}\) was less than \(1 \%\), would that tend to shake your confidence in the statement that \(\mu=1.6 \%\) ? Might you suspect that \(\mu\) has slipped below \(1.6 \%\) ? Explain.

Sketch the areas under the standard normal curve over the indicated intervals, and find the specified areas. Between \(z=-2.42\) and \(z=-1.77\)

When we use a normal distribution to approximate a binomial distribution, why do we make a continuity correction?

Distribution Suppose we have a binomial experiment in which success is defined to be a particular quality or attribute that interests us. (a) Suppose \(n=33\) and \(p=0.21\). Can we approximate the \(\hat{p}\) distribution by a normal distribution? Why? What are the values of \(\mu_{\hat{p}}\) and \(\sigma_{\hat{p}} ?\) (b) Suppose \(n=25\) and \(p=0.15 .\) Can we safely approximate the \(\hat{p}\) distribution by a normal distribution? Why or why not? (c) Suppose \(n=48\) and \(p=0.15 .\) Can we approximate the \(\hat{p}\) distribution by a normal distribution? Why? What are the values of \(\mu_{\hat{p}}\) and \(\sigma_{\hat{p}} ?\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation
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.