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Given that \(z\) is a standard normal random variable, compute the following probabilities. a. \(\quad P(z \leq-1.0)\) b. \(\quad P(z \geq-1)\) c. \(\quad P(z \geq-1.5)\) d. \(P(z \geq-2.5)\) e. \(\quad P(-3

Short Answer

Expert verified
a. 0.1587; b. 0.8413; c. 0.9332; d. 0.9938; e. 0.4987.

Step by step solution

01

Understanding the Standard Normal Distribution

A standard normal distribution has a mean of 0 and a standard deviation of 1. Probability calculations involving the standard normal distribution involve using Z-scores and looking up probabilities in a standard normal table or computing using software.
02

Homework Requirement Analysis

We have to find the probabilities for different conditions: - For a: Find probability that z is less than or equal to -1. - For b, c, and d: Find probabilities that z is greater than or equal to each of -1, -1.5, -2.5. - For e: Find probability that z is between -3 and 0.
03

Calculate P(z ≤ -1.0)

Referring to the standard normal distribution table or using a calculator, find the probability of Z being less than or equal to -1.0. - From the standard normal table: \[ P(z \leq -1.0) \approx 0.1587 \]
04

Calculate P(z ≥ -1)

To find P(z ≥ -1), you need the probability of the opposite condition and subtract from 1:- Since \( P(z \leq -1) \approx 0.1587 \), - \[ P(z \geq -1) = 1 - P(z \leq -1) \approx 1 - 0.1587 = 0.8413 \]
05

Calculate P(z ≥ -1.5)

Find the probability of Z being greater than or equal to -1.5:- From the standard normal table: - \( P(z \leq -1.5) \approx 0.0668 \)- \[ P(z \geq -1.5) = 1 - P(z \leq -1.5) = 1 - 0.0668 = 0.9332 \]
06

Calculate P(z ≥ -2.5)

Use the Z-table for probability less than -2.5 and subtract from 1:- \( P(z \leq -2.5) \approx 0.0062 \)- \[ P(z \geq -2.5) = 1 - P(z \leq -2.5) = 1 - 0.0062 = 0.9938 \]
07

Calculate P(-3 < z ≤ 0)

Find the probability of Z being between -3 and 0:- \( P(z \leq 0) \approx 0.5 \) since it's the mean for the standard normal distribution.- \( P(z \leq -3) \approx 0.0013 \)- \[ P(-3 < z \leq 0) = P(z \leq 0) - P(z \leq -3) = 0.5 - 0.0013 = 0.4987 \]

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Key Concepts

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

Probability Calculations
Probability calculations are essential in understanding how likely an event is to occur within a set range of values. In the context of the Standard Normal Distribution, these calculations often involve determining the probability that a Z-score falls within a certain range.

To perform these calculations, you need to understand what a Z-score is and how it relates to the normal distribution. Consider the distribution as a "bell curve." Probabilities are the areas under this curve.
  • For any given Z-score, find the probability of the score existing to the left or right of this point.
  • Use tables or computational tools to assist in these calculations.
Probability calculations often use the total area under the curve, which sums to 1, to find the desired probabilities.
Z-scores
Z-scores are a fundamental concept in statistics, particularly when working with the standard normal distribution. A Z-score shows how many standard deviations an element is from the mean. The formula for calculating a Z-score is:
\[ Z = \frac{(X - \mu)}{\sigma} \]

Where:
  • \(X\) is the value of the element,
  • \(\mu\) is the mean,
  • \(\sigma\) is the standard deviation.

For the standard normal distribution, both \(\mu\) and \(\sigma\) are 0 and 1 respectively, simplifying many calculations. The Z-score allows for comparison across different distributions by standardizing scores to a common scale. When you see a Z-score, think of it as a measure of unusualness. Higher or lower scores suggest rarity or dominance in data.
Normal Distribution Table
The Normal Distribution Table, often referred to as the Z-table, aids in probability calculations involving Z-scores. It lists the cumulative probability of a score up to and including a Z-score. This tool is invaluable for quickly determining the probability of a standard normal variable.

The table generally shows probabilities only for positive Z-scores, covering \(z > 0\). However, due to symmetry, probabilities for negative values can be found as well:
  • For any \(z\), the opposite side of the mean's probability is \(1 - P(Z \leq z)\).
  • It provides a straightforward mechanism to find probabilities and reduce computational tasks to table lookup.

Using a Z-table involves locating the row for your Z-score and the column for its decimal place, intersecting them gives the cumulative probability.
Statistical Analysis
Statistical Analysis involves the collection and interpretation of data to discover patterns and trends. In probability and statistics, the standard normal distribution is frequently used to simplify and unify analysis.

By using Z-scores, you bring a wide range of statistical inquiries under a single framework. This is crucial for making decisions based on data, as it allows analysts to:
  • relate individual observations to the overall population easily,
  • determine probabilities quickly,
  • make predictions or test hypotheses within a given confidence level.

Understanding these principles of statistical analysis empowers you to critique past performances and forecast future outcomes efficiently. It helps in turning raw data into insights and drives informed decision-making.

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Most popular questions from this chapter

Given that \(z\) is a standard normal random variable, find \(z\) for each situation. a. The area to the right of \(z\) is .01 b. The area to the right of \(z\) is .025 c. The area to the right of \(z\) is .05 d. The area to the right of \(z\) is .10 .

The time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 12 seconds. a. Sketch this exponential probability distribution. b. What is the probability that the arrival time between vehicles is 12 seconds or less? c. What is the probability that the arrival time between vehicles is 6 seconds or less? d. What is the probability of 30 or more seconds between vehicle arrivals?

Is lack of sleep causing traffic fatalities? A study conducted under the auspices of the National Highway Traffic Safety Administration found that the average number of fatal crashes caused by drowsy drivers each year was 1550 ( Business Week , January 26,2004 ). Assume the annual number of fatal crashes per year is normally distributed with a standard deviation of 300 . a. What is the probability of fewer than 1000 fatal crashes in a year? b. What is the probability the number of fatal crashes will be between 1000 and 2000 for a year? c. For a year to be in the upper \(5 \%\) with respect to the number of fatal crashes, how many fatal crashes would have to occur?

The lifetime (hours) of an electronic device is a random variable with the following exponential probability density function. \\[f(x)=\frac{1}{50} e^{-x / 50} \quad \text { for } x \geq 0\\] a. What is the mean lifetime of the device? b. What is the probability that the device will fail in the first 25 hours of operation? c. What is the probability that the device will operate 100 or more hours before failure?

Ward Doering Auto Sales is considering offering a special service contract that will cover the total cost of any service work required on leased vehicles. From experience, the company manager estimates that yearly service costs are approximately normally distributed, with a mean of \(\$ 150\) and a standard deviation of \(\$ 25\) a. If the company offers the service contract to customers for a yearly charge of \(\$ 200\), what is the probability that any one customer's service costs will exceed the contract price of \(\$ 200 ?\) b. What is Ward's expected profit per service contract?

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