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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 .

Short Answer

Expert verified
a. \(z = 1.96\), b. \(z = 1.96\), c. \(z = 0.61\), d. \(z = 1.12\), e. \(z = 0.44\), f. \(z = 0.44\)

Step by step solution

01

Understanding Standard Normal Distribution

In a standard normal distribution, the variable \( z \) represents the number of standard deviations away from the mean. The distribution has a mean of 0 and a standard deviation of 1. To find \( z \) for different areas under the curve, we use the standard normal (Z) table or a calculator with normal distribution functions.
02

Situation a: Area to the Left of \( z \) is 0.9750

To find \( z \) when the area to the left is 0.9750, locate 0.9750 in the standard normal distribution table or use a calculator to find \( z \). The value of \( z \) corresponding to the area 0.9750 is 1.96.
03

Situation b: Area Between 0 and \( z \) is 0.4750

For the area between 0 and \( z \), note that this implies \( z \) is positive, and the total area to the left of \( z \) is 0.5000 + 0.4750 = 0.9750. From situation a, we know that \( z = 1.96 \).
04

Situation c: Area to the Left of \( z \) is 0.7291

Locate 0.7291 in the Z-table or use a calculator to find the corresponding \( z \)-score. The \( z \) value for 0.7291 is approximately 0.61.
05

Situation d: Area to the Right of \( z \) is 0.1314

If the area to the right is 0.1314, the area to the left is 1 - 0.1314 = 0.8686. Find \( z \) by looking up 0.8686 in the Z-table or using a calculator, which gives \( z = 1.12 \).
06

Situation e: Area to the Left of \( z \) is 0.6700

Find 0.6700 in the Z-table or use a calculator to identify \( z \). The \( z \) value corresponding to an area of 0.6700 is approximately 0.44.
07

Situation f: Area to the Right of \( z \) is 0.3300

The area to the left is 1 - 0.3300 = 0.6700. As already found in situation e, \( z \) corresponding to an area of 0.6700 is 0.44.

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Key Concepts

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

Z-table
Understanding how to use a Z-table is crucial for finding Z-scores in standard normal distribution problems. The Z-table is a tool that displays the area (probability) to the left of a given Z-score on the standard normal distribution curve. To find a Z-score using a Z-table, start by identifying the probability value you need.
  • Locate this probability value in the body of the Z-table.
  • Once you find it, look to the left margin to find the row number, and to the top margin for the column number.
  • The sum of these two values gives you the Z-score.

For example, if you are tasked with finding a Z-score where the area to the left is 0.9750, search this number on the table. In this case, you will find that the number corresponds to a Z-score of 1.96. Using a Z-table is an essential skill for anyone dealing with statistical data, allowing you to quickly find Z-scores and interpret different problems effectively.
Z-score calculations
Z-score calculations allow you to understand how far a data point is from the mean in terms of standard deviations. This standardization helps in comparing different data points from different samples or situations. To calculate a basic Z-score, use the formula: \[ Z = \frac{(X - \mu)}{\sigma} \]
  • Where \( X \) is the raw score, \( \mu \) is the mean of the population, and \( \sigma \) is the standard deviation.
However, when working with areas under the curve, as in this problem, we're more concerned with reverse calculation—finding \( z \) using known areas or probabilities.
Here are some scenarios often worked out:
  • If you know the area to the left of \( z \), you use the Z-table or a calculator to find \( z \).
  • If you have the area to the right, subtract it from 1, then find the corresponding \( z \).
  • If given the area between the mean (0) and \( z \), add 0.5 to this area to find the total area to the left, and locate \( z \).
Remember, each calculation is an essential step in interpreting data in standard normal distributions.
Area under the curve
The concept of the area under the curve in a standard normal distribution is fundamental to understanding probability. The full area under the standard normal curve equals 1, representing 100% of all possibilities. Each segment of the curve corresponds to different probability values associated with Z-scores.
  • The area to the left of a Z-score represents the cumulative probability up to that Z-score.
  • The area to the right is simply the complement of the area to the left, representing 1 minus that.
By understanding these areas, you can determine how likely it is for a data point to occur within a specified range. Understanding areas under the curve allows you to make important inferences about the likelihood or frequency of a particular Z-score occurring within a distribution. It also helps in determining confidence intervals and probabilities regarding data distribution in many practical applications.

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Most popular questions from this chapter

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?

A machine fills containers with a particular product. The standard deviation of filling weights is known from past data to be .6 ounce. If only \(2 \%\) of the containers hold less than 18 ounces, what is the mean filling weight for the machine? That is, what must \(\mu\) equal? Assume the filling weights have a normal distribution.

Collina's Italian Café in Houston, Texas, advertises that carryout orders take about 25 minutes (Collina's website, February 27,2008 ). Assume that the time required for a carryout order to be ready for customer pickup has an exponential distribution with a mean of 25 minutes. a. What is the probability than a carryout order will be ready within 20 minutes? b. If a customer arrives 30 minutes after placing an order, what is the probability that the order will not be ready? c. \(\quad\) A particular customer lives 15 minutes from Collina's Italian Café. If the customer places a telephone order at 5: 20 P.M., what is the probability that the customer can drive to the café, pick up the order, and return home by 6: 00 p.n.?

Consider the following exponential probability density function. \\[ f(x)=\frac{1}{3} e^{-x / 3} \quad \text { for } x \geq 0 \\] a. Write the formula for \(P\left(x \leq x_{0}\right)\) b. Find \(P(x \leq 2)\) c. \(\quad\) Find \(P(x \geq 3)\) d. Find \(P(x \leq 5)\) e. Find \(P(2 \leq x \leq 5)\)

Do interruptions while you are working reduce your productivity? According to a University of California-Irvine study, businesspeople are interrupted at the rate of approximately \(5^{1 / 2}\) times per hour (Fortune, March 20,2006 ). Suppose the number of interruptions follows a Poisson probability distribution. a. Show the probability distribution for the time between interruptions. b. What is the probability that a businessperson will have no interruptions during a 15 minute period? c. What is the probability that the next interruption will occur within 10 minutes for a particular businessperson?

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