/*! 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 51 A random variable \(X\) is assum... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 51

A random variable \(X\) is assumed to have a standard normal distribution. Find the observed significance level if the random variable is equal to the given value in an experiment. Observed value of \(0.93\)

Short Answer

Expert verified
The observed significance level is 0.1762.

Step by step solution

01

- Understand the standard normal distribution

The standard normal distribution is a special case of the normal distribution with a mean \(\mu = 0\) and standard deviation \(\sigma = 1\).
02

- Define the observed value

The observed value given in the problem is \(X = 0.93\).
03

- Calculate the cumulative probability

To find the observed significance level, convert the observed value \(X = 0.93\) to a cumulative probability using the standard normal distribution table or a calculator. This will give the probability that a standard normal variable is less than or equal to 0.93.
04

- Look up in standard normal table

Using a standard normal distribution table or a calculator, find \(P(Z \leq 0.93)\). According to the table, \(P(Z \leq 0.93) = 0.8238\).
05

- Calculate the observed significance level

The observed significance level is the probability that a standard normal variable is greater than the observed value. Hence, subtract the cumulative probability from 1. \(| = 1 - P(Z \leq 0.93) = 1 - 0.8238 = 0.1762\).

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.

cumulative probability
When dealing with a standard normal distribution, cumulative probability is a key concept. It represents the area under the curve to the left of a given value. This probability tells us how likely it is for a randomly selected score from the distribution to be less than or equal to a specific value. For a standard normal distribution, the mean is 0 and the standard deviation is 1.

For example, if we have an observed value of 0.93, finding the cumulative probability involves looking at a standard normal distribution table or using a calculator. The cumulative probability for 0.93 gives us the probability that a standard normal variable is less than or equal to 0.93.

Understanding cumulative probability is crucial as it helps in assessing how an individual observation compares to the overall distribution. It's used in various statistical analyses and hypothesis testing to determine where a particular data point stands relative to the rest.
observed significance level
The observed significance level, often referred to as the P-value, plays a vital role in hypothesis testing. It indicates the probability of observing a test statistic at least as extreme as the one observed, assuming the null hypothesis is true.

To find the observed significance level for a given observed value in a standard normal distribution, we first calculate the cumulative probability. Then, subtract this cumulative probability from 1. This gives us the observed significance level.

For instance, with an observed value of 0.93, we find the cumulative probability to be 0.8238. Therefore, the observed significance level is calculated as: \(1 - P(Z \leq 0.93) = 1 - 0.8238 = 0.1762\). This means there is a 17.62% probability that a standard normal variable will be greater than 0.93.
z-score
A z-score is a measure that describes a value's position relative to the mean of a dataset, expressed in terms of standard deviations. In a standard normal distribution, the z-score indicates how many standard deviations a particular value is from the mean.

The formula for calculating a z-score is given by: \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the observed value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Since the mean and standard deviation for the standard normal distribution are 0 and 1, respectively, the z-score is equal to the observed value.

In our example, the observed value is 0.93. Hence, the z-score is also 0.93. This z-score tells us that the observed value is 0.93 standard deviations above the mean of the standard normal 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

Let \(Z\) be a Normal \((0,1)\) random variable. Find the probability that \(Z\) is in the interval. $$ [-2,2] $$

Cavities in trees are necessary for certain birds and other animals to survive. For a plot of woods in Missouri, let \(X\) represent the number of cavities per hectare. The density function for \(X\) can be approximated by \(^{24}\) $$ f(x)=\frac{a}{b}\left(\frac{x-c}{b}\right)^{a-1} e^{-(x-c)^{a} b^{a}} $$ where \(a, b\), and \(c\) are constants, and \(x \geq c\). a) Verify property 2 of a density function for \(\int(x)\). (Hint: Try substituting \(u=(x-c) / b .)\) b) For plots of trees no more than 30 yr old, it was found that \(a \approx 0.68, b \approx 0.89\), and \(c \approx 0.0\) Compute \(P(X \leq 2)\).

The probability density function of a random variable and a significance level \(\alpha\) are given. Find the critical value. $$ f(x)=0.32 e^{-x}+1.36 e^{-2 x} \text { over }[0, \infty) ; \alpha=0.01 $$

Suppose that an individual is randomly selected from a population that satisfies the Hardy-Weinberg Law. Let \(p\) and \(q\) denote the proportions of \(A\) and \(a\) alleles in the population. a) The allele inherited by the offspring may be visualized with the following multiplication tree. Fill in the probabilities of the unmarked arrows. b) In the multiplication tree, there are three ways corresponding to the transmission of the \(A\) allele. Show that the probability that the \(A\) allele is transmitted is \(p\). c) Show that the probability that the \(a\) allele is Iransmitted is \(q .\) d) Use the result of Example 9 to show that, if one generation satisfies the Hardy-Weinberg Law and survives to mate, then the offspring generation should also satisfy it.

A significance level \(\alpha\) and a tail of the standard normal distribution are given. Use the normal table to approximately determine the critical value. \(\alpha=0.05\), left tail
See all solutions

Recommended explanations on Biology Textbooks

Biological Organisms

Read Explanation

Ecological Levels

Read Explanation

Plant Biology

Read Explanation

Microbiology

Read Explanation

Biological Structures

Read Explanation

Energy Transfers

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.