/*! 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 2 Let \(z\) be a standard normal r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(z\) be a standard normal random variable with mean \(\mu=0\) and standard deviation \(\sigma=1 .\) Use Table 3 in Appendix \(I\) to find the probabilities. $$ P(z>1.16) $$

Short Answer

Expert verified
Answer: The probability is 0.1230.

Step by step solution

01

Understand the standard normal distribution

A standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is represented by the probability density function: $$ f(z) = \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} $$
02

Locate the value of \(z=1.16\) in the table

Using Table 3 in Appendix I, locate the row for \(1.1\) and the column for \(0.06\). The intersection of these corresponds to the value of \(z=1.16\).
03

Find the probability for \(z=1.16\)

The value at the intersection for \(z=1.16\) in the table represents \(P(z \le 1.16)\) (the probability that \(z\) is less than or equal to \(1.16\)). In this case, the value is \(0.8770\).
04

Calculate the probability for \(P(z > 1.16)\)

Since the total probability under the standard normal distribution is equal to 1, we have: $$ P(z > 1.16) = 1 - P(z \le 1.16) $$ Substitute the value found in Step 3: $$ P(z > 1.16) = 1 - 0.8770 $$
05

Compute the final probability

Calculate the probability: $$ P(z > 1.16) = 0.1230 $$ So the probability \(P(z > 1.16)\) is \(0.1230\).

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.

Probability Density Function
Understanding the probability density function (PDF) is crucial when studying continuous random variables. A PDF represents how the values of a random variable are distributed; it is the graph of a function that maps all possible values of the random variable to their corresponding probabilities. For the standard normal distribution, the PDF is given by the formula:
\[ f(z) = \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} \]
In simpler terms, this function tells us the probability of a random variable (in this case, the standard normal variable) falling within a particular range. The formula illustrates that the probability smoothly changes across different values of z rather than jumping from one probability to another.
Fundamentally, the PDF shows that the probabilities are highest near the mean, which is 0 for standard normal distribution, and taper off as we move away from the center. Importantly, the total area under the PDF curve equals 1, which corresponds to the certainty that a random variable will take some value within its range.
Z-Score
A z-score signifies how many standard deviations a data point is from the mean. When a random variable follows a standard normal distribution, any value of it can be converted into a z-score. The standard normal distribution's mean \(\mu = 0\)\ and standard deviation \(\sigma = 1\)\ are the benchmarks for these scores. The z-score formula is:
\[ z = \frac{X - \mu}{\sigma} \]
In practice, a z-score also serves as a tool for finding probabilities associated with the standard normal distribution. For example, to find the probability that the random variable takes a value greater than any particular z-score, one would subtract the cumulative probability up to that z-score from the total probability, 1. Moreover, z-scores help in standardizing different data sets, enabling comparisons across various scales and distributions.
Returning to the exercise, to calculate \(P(z > 1.16)\), we need to find the cumulative probability up to z-score 1.16 and deduct it from 1. This step connects the abstract idea of a z-score to a tangible method for determining probabilities.
Normal Random Variable
The term normal random variable refers to a variable that assumes values guided by the normal distribution, which is characterized by its bell-shaped curve. When the mean \(\mu\) and standard deviation \(\sigma\) of this distribution are 0 and 1, respectively, we call it a standard normal distribution. The significance of a standard normal random variable lies in its predictability; it shows us how frequently or rarely specific values occur.
Standard normal random variables are valuable in statistics because they allow us to evaluate and interpret various statistical outcomes. They are commonly used to model uncertainties and help gauge standard deviations from the mean in natural or social phenomena.
With a robust understanding of a normal random variable, one can apply the principles to more complex statistics, including hypothesis testing and confidence intervals. Moreover, since it's standardized, any normally distributed variable with a different mean or standard deviation can be transformed into a standard normal random variable using z-scores, thus simplifying calculations of probabilities and other statistical measures.

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 \(x\) have an exponential distribution with \(\lambda=1\). Find the probabilities. $$ P(1

Using Table 1 in Appendix I, find the exact values for the binomial probabilities Then approximate the probabilities using the normal approximation with the correction for continuity. Compare your answers. $$ P(x \geq 10) \text { when } n=20 \text { and } p=.4 $$

A purchaser of electric relays buys from two suppliers, \(A\) and \(B\). Supplier A supplies two of every three relays used by the company. If 75 relays are selected at random from those in use by the company, find the probability that at most 48 of these relays come from supplier A. Assume that the company uses a large number of relays.

Smartphone Shopping According to a USA Today snapshot, a large proportion of American shoppers own smartphones and use them while shopping to take pictures of items \((51 \%),\) to search for coupons \((49 \%)\) or to compare prices among retailers \((43 \%) .^{13}\) Suppose that a random sample of 50 American shoppers is selected and \(x\) is the number who use their phones to search for coupons. Assume that the \(49 \%\) figure is correct. a. What is the average number of shoppers who use their smartphones to search for coupons? b. What is the standard deviation of \(x\) ? c. If you were to observe a value of \(x\) less than or equal to \(15,\) would you consider this to be an unusual occurrence? Explain.

The discharge of suspended solids from a phosphate mine is normally distributed, with a mean daily discharge of 27 milligrams per liter \((\mathrm{mg} / \mathrm{l})\) and a standard deviation of \(14 \mathrm{mg} / \mathrm{l}\). On what proportion of days will the daily discharge exceed \(50 \mathrm{mg} / \mathrm{l} ?\)

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.