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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. \(\quad P(-2.5 \leq z)\) 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 is a special normal distribution with a mean of 0 and a standard deviation of 1. Probabilities related to the standard normal distribution can be found using the standard normal table (Z-table) or statistical software.
02

Compute P(z ≤ -1.0)

To find \(P(z \leq -1.0)\), refer to the Z-table. The Z-table provides the area under the curve to the left of a given z-value. For \(z = -1.0\), the table shows that \(P(z \leq -1.0) \approx 0.1587\).
03

Compute P(z ≥ -1)

This is the complement of the probability from Step 1. Therefore, \(P(z \geq -1) = 1 - P(z \leq -1)\). From Step 1, \(P(z \leq -1) = 0.1587\), so \(P(z \geq -1) = 1 - 0.1587 = 0.8413\).
04

Compute P(z ≥ -1.5)

Use the complement rule again: \(P(z \geq -1.5) = 1 - P(z \leq -1.5)\). Using the Z-table, \(P(z \leq -1.5) \approx 0.0668\). Therefore, \(P(z \geq -1.5) = 1 - 0.0668 = 0.9332\).
05

Compute P(-2.5 ≤ z)

This probability is calculated similarly using the complement: \(P(z \geq -2.5) = 1 - P(z \leq -2.5)\). From the Z-table, \(P(z \leq -2.5) \approx 0.0062\). Thus, \(P(-2.5 \leq z) = 1 - 0.0062 = 0.9938\).
06

Compute P(-3 < z ≤ 0)

This probability can be found by calculating \(P(z \leq 0) - P(z \leq -3)\). \(P(z \leq 0) = 0.5\), since z=0 is the mean of the distribution. From the Z-table, \(P(z \leq -3) \approx 0.0013\). Hence, \(P(-3 < z \leq 0) = 0.5 - 0.0013 = 0.4987\).

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Key Concepts

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

Probability Calculation
Probability calculation in the context of a standard normal distribution involves finding the probability that a random variable falls within a specific range or above or below a certain value. The standard normal distribution is a bell-shaped curve where the mean is 0 and the standard deviation is 1.

To compute these probabilities, use tools like the Z-table, which shows the cumulative area under the curve to the left of a given z-value. This area corresponds to the probability that a standard normal random variable is less than or equal to the z-value.

For example, if you want to find the probability that a random variable is less than or equal to -1 (i.e., \(P(z \leq -1)\)), you would look up -1 in the Z-table. This lookup will provide you with the probability (or cumulative area) of this range. To calculate probabilities where the random variable is greater than a particular z-value, you can use the complement rule, which often involves subtracting the Z-table value from 1.
Z-table
The Z-table is a statistical tool that simplifies finding probabilities under the standard normal distribution curve. It provides a quick way to determine the area to the left of a specific z-value on the standard normal distribution.

When using a Z-table, you'll typically find the table divided into two main parts: one for negative z-values and another for positive z-values. You look up a z-value to find the cumulative probability (area under the curve) on the left side of the z-value.

For instance, if you check the Z-table for \(z = -1.0\), you will find that the cumulative probability (\(P(z \leq -1.0)\)) is approximately 0.1587. This means there is about a 15.87% chance that a standard normal random variable is less than or equal to -1.

If you need the probability that \(z\) is greater than a given value, you can subtract the cumulative probability from one, utilizing the complement rule.
Complement Rule
The complement rule is a useful strategy in probability calculations, particularly when dealing with scenarios where you need to find the probability that a random variable exceeds a certain number. It states that the probability of an event occurring is one minus the probability of it not occurring.

Mathematically, this can be expressed as \(P(A') = 1 - P(A)\), where \(P(A)\) is the probability of the event, and \(P(A')\) is the probability of not the event. This rule simplifies calculations when using a Z-table which provides cumulative probabilities for events that occur to the left of a z-value.

For instance, if you need to determine the probability \(P(z \geq -1)\), you would first find \(P(z \leq -1)\) from the Z-table. Then, using the complement rule, calculate \(P(z \geq -1) = 1 - P(z \leq -1)\). This reversal is not only practical but also often necessary when solving standard normal distribution problems.
Normal Distribution
The normal distribution is one of the most important concepts in statistics and is often used to represent real-valued random variables. A "standard" normal distribution is a specific form of the normal distribution that is commonly used in statistics.

The key characteristics of a normal distribution are its bell-shaped, symmetric curve with its peak at the mean. It is determined by two parameters: the mean and the standard deviation. For the standard normal distribution, these parameters are 0 and 1, respectively.

This symmetric property means that half the data under the curve falls on each side of the mean, making it very useful for probability calculations. For example, in a standard normal distribution, the probability of a random variable being less than or equal to 0 is always 0.5 because 0 is the mean.

Normal distributions can model a multitude of phenomena in the natural world, and understanding this distribution allows for profound insights into probability and statistics. It is the foundation upon which many statistical tests and methods are built.

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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 left of \(z\) is .9750 b. The area between 0 and \(z\) is .4750 c. The area to the left of \(z\) is .7291 d. The area to the right of \(z\) is .1314 e. The area to the left of \(z\) is .6700 . f. The area to the right of \(z\) is .3300 .

Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000 -unit production run in the following situations. a. The process standard deviation is \(.15,\) and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects. b. Through process design improvements, the process standard deviation can be reduced to \(.05 .\) Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects. c. What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

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?

Trading volume on the New York Stock Exchange is heaviest during the first half hour (early morning) and last half hour (late afternoon) of the trading day. The early morning trading volumes (millions of shares) for 13 days in January and February are shown here (Barron \(s\), January 23,\(2006 ;\) February 13,\(2006 ;\) and February 27,2006 ). $$\begin{array}{lllll} 214 & 163 & 265 & 194 & 180 \\ 202 & 198 & 212 & 201 & \\ 174 & 171 & 211 & 211 \end{array}$$ The probability distribution of trading volume is approximately normal. a. Compute the mean and standard deviation to use as estimates of the population mean and standard deviation. b. What is the probability that, on a randomly selected day, the early morning trading volume will be less than 180 million shares? c. What is the probability that, on a randomly selected day, the early morning trading volume will exceed 230 million shares? d. How many shares would have to be traded for the early morning trading volume on a particular day to be among the busiest \(5 \%\) of days?

Condé Nast Traveler publishes a Gold List of the top hotels all over the world. The Broadmoor Hotel in Colorado Springs contains 700 rooms and is on the 2004 Gold List (Condé Nast Traveler; January 2004 ). Suppose Broadmoor's marketing group forecasts a mean demand of 670 rooms for the coming weekend. Assume that demand for the upcoming weekend is normally distributed with a standard deviation of 30 a. What is the probability that all the hotel's rooms will be rented? b. What is the probability that 50 or more rooms will not be rented? c. Would you recommend the hotel consider offering a promotion to increase demand? What considerations would be important?

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