/*! 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 50 Use the following information to... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the following information to answer the next four exercises: X ~ N(54, 8) Find the probability that \(x>56\)

Short Answer

Expert verified
The probability that \(x>56\) is approximately 0.4013.

Step by step solution

01

Identify the Normal Distribution Parameters

The given problem states that the random variable \(X\) follows a normal distribution with a mean (\(\mu\)) of 54 and a standard deviation (\(\sigma\)) of 8. We represent this distribution as \(X \sim N(54, 8)\).
02

Standardize the Variable

We need to find the probability that \(x > 56\). To do this, we first convert the variable \(x\) to a standard normal variable (\(z\)) using the formula: \(z = \frac{x - \mu}{\sigma}\). Substituting the given values, we have: \(z = \frac{56 - 54}{8} = \frac{2}{8} = 0.25\).
03

Find the Probability Using the Standard Normal Distribution

Now that we have standardized the variable, we need to find \(P(Z > 0.25)\), where \(Z\) is the standard normal variable. This is equivalent to \(1 - P(Z \leq 0.25)\). Using a standard normal distribution table or a calculator, we find that \(P(Z \leq 0.25) \approx 0.5987\).
04

Calculate the Required Probability

Compute the probability of \(Z > 0.25\) by subtracting \(P(Z \leq 0.25)\) from 1: \(P(Z > 0.25) = 1 - 0.5987 = 0.4013\). Thus, the probability that \(x > 56\) is approximately 0.4013.

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.

Standard Normal Distribution
The standard normal distribution is a specific type of normal distribution. It is a continuous probability distribution that is symmetrical around its mean. For a standard normal distribution, the mean is 0 and the standard deviation is 1.

This standardization helps make calculations easier and more universal because it allows statisticians to use a single standard normal distribution table. This table provides probabilities for different z-values, which represent the number of standard deviations a data point is from the mean.

In simpler terms, any normal distribution can be transformed into a standard normal distribution by converting the data points into z-scores. This standardized form is what makes the standard normal distribution such a fundamental tool in statistics, enabling easier probability calculations.
Probability Calculation
In statistics, probability calculation involves determining the likelihood of a certain event occurring within a given distribution. When dealing with a normal distribution, such calculations often require the use of z-scores.

To find the probability that a random variable takes on a particular value or falls within a certain range, we reference the standard normal distribution table. Once a variable is standardized into a z-score, the table can provide the probability of that z-score or less occurring.

For instance, in the original exercise, after converting the value of interest ( 76) into a z-score of 0.25, we used the standard normal distribution table to find the probability that a z-score is less than or equal to 0.25. This probability was then subtracted from 1 to determine the probability of the value being greater than 0.25, which is the goal of many practical probability calculations.
Standardization
Standardization is a crucial process in statistics that transforms different data series into a common format. This is achieved by re-scaling the data so that they can be compared on a uniform basis.

In the context of normal distributions, standardization is the process of converting a normal random variable into a standard normal variable (z-score).
  • The formula used for standardization is:
    \[ z = \frac{x - \mu}{\sigma} \]
  • Here, \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation of the original distribution.
By standardizing, we effectively set the mean to 0 and the standard deviation to 1, allowing for the use of the standard normal distribution table to find probabilities. In practical terms, this process helps in comparing different datasets and simplifies the calculation of probabilities for a wide range of applications.

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

Suppose a normal distribution has a mean of six and a standard deviation of \(1.5 .\) What is the \(z\) -score of \(x=5.5 ?\)

Use the following information to answer the next two exercises: The life of Sunshine CD players is normally distributed with mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for three years. We are interested in the length of time a CD player lasts. Define the random variable X in words. X = _______________.

Suppose \(X \sim N(-1,2) .\) What is the \(z\) -score of \(x=2 ?\)

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. In 2005, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean ? = 520 and standard deviation ? = 115. a. Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence. b. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score? c. For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took?

Suppose \(X \sim N(2,3) .\) What value of \(x\) has a \(z\) -score of \(-0.67 ?\)

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.