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Find a z-score, say, \(z_{0}\), such that a. \(P\left(z \leq z_{0}\right)=.8708\) b. \(P\left(z \geq z_{0}\right)=.0526\) c. \(P\left(z \leq z_{0}\right)=.5\) d. \(P\left(-z_{0} \leq z \leq z_{0}\right)=.8164\) e. \(P\left(z \geq z_{0}\right)=.8023\) f. \(P\left(z \geq z_{0}\right)=.0041\)

Short Answer

Expert verified
a. z = 1.13, b. z = 1.65, c. z = 0, d. z = ±1.34, e. z = -0.84, f. z = 2.64.

Step by step solution

01

Locate z-score for P(z ≤ z0)=0.8708

To find the z-score for which the cumulative probability is 0.8708, we look up the standard normal distribution table (z-table) or use statistical software. The value corresponding to a cumulative probability of 0.8708 is approximately z = 1.13.
02

Locate z-score for P(z ≥ z0)=0.0526

If we have a probability greater than z0, we want the tail end of the standard normal distribution. Since P(z ≥ z0) = 0.0526, this is equivalent to P(z ≤ z0) = 1 - 0.0526 = 0.9474. Look for 0.9474 in the z-table; it corresponds to z = 1.65.
03

Locate z-score for P(z ≤ z0)=0.5

When the cumulative probability is 0.5, z0 represents the mean of the standard normal distribution, which is z = 0.
04

Determine z0 for symmetric interval P(-z0 ≤ z ≤ z0)=0.8164

For the symmetric interval where the probability in between is 0.8164, we find the tail probability: (1 - 0.8164) / 2 = 0.0918 on each side. Therefore, P(z ≤ z0) = 0.9082. In the z-table, this corresponds to z0 = 1.34.
05

Locate z-score for P(z ≥ z0)=0.8023

Convert the greater-than probability to a cumulative (less-than) probability: P(z ≤ z0) = 1 - 0.8023 = 0.1977. Locate 0.1977 in the z-table; it corresponds to z ≈ -0.84.
06

Locate z-score for P(z ≥ z0)=0.0041

Convert the upper tail probability to a lower tail probability: P(z ≤ z0) = 1 - 0.0041 = 0.9959. In the z-table, this cumulative probability corresponds to z ≈ 2.64.

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Key Concepts

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

Standard Normal Distribution
The standard normal distribution is a special type of normal distribution that has a mean of 0 and a standard deviation of 1. It is a very useful tool in statistics because it allows us to use z-scores for easy calculations and comparisons. This bell-shaped curve is symmetrical, meaning that it is identical on both sides of the mean. The total area under the curve is equal to 1, which represents the total probability for the distribution.

When dealing with data that follows a normal distribution, converting individual data points into z-scores enables statisticians to determine probabilities and make inferences about the population. The standard normal distribution is particularly important because it simplifies the process of working with probabilities and statistical hypothesis testing.
  • The mean of the standard normal distribution is 0.
  • The standard deviation is 1.
  • It is symmetrical around the mean.
  • The total probability under the curve is 1.
Cumulative Probability
Cumulative probability is the probability that a random variable is less than or equal to a certain value in a distribution. It is a key concept when looking up values in a z-table, as it gives the probability of falling to the left of a specified z-score.

In practice, cumulative probability helps us answer questions like, "What is the probability that a value will be less than 1.13?" For a standard normal distribution, this would mean looking up the value of 1.13 in a z-table to find the cumulative probability, which is 0.8708. In simpler terms, there is an 87.08% chance that a value will fall below 1.13.
  • Cumulative probability sums up probabilities up to a specific point.
  • It shows the likelihood of a value being less than or equal to a particular z-score.
  • Commonly used with z-tables to find probabilities to the left of a z-score.
Symmetrical Interval
A symmetrical interval in the context of a standard normal distribution refers to a range of values that are equidistant from the mean on both sides of the distribution. This characteristic is used to calculate probabilities for two-tailed tests, which consider both extremes of the curve.

For example, when we have a symmetrical interval such as \[P(-z_0 \leq z \leq z_0) = 0.8164\]we calculate the tail probability by dividing the remaining area of the distribution equally on both sides.This approach is beneficial for calculating the probability within a specific interval, ensuring accuracy when assessing variability or conducting hypothesis tests.
  • A symmetrical interval covers the same probability on both sides of the mean.
  • It allows for balanced probability calculations for both tails of the distribution.
  • Perfect for addressing two-tailed hypothesis testing.
Z-Table Lookup
The z-table is a vital tool for statisticians, allowing them to determine cumulative probabilities associated with standard normal distribution z-scores. When calculating probabilities or finding z-scores, going through this table is often necessary.

Here's how you typically use a z-table:
  • Look up the desired cumulative probability in the table (e.g., 0.8708).
  • Identify the corresponding z-score (e.g., approximately 1.13).
Remember, the z-table is crafted for the standard normal distribution, so it's specifically designed for cumulative probabilities from left to right. Learning to navigate and interpret the table efficiently can make statistical analysis much easier.
  • Z-table provides cumulative probabilities for z-scores in a standard normal distribution.
  • Essential for finding z-scores corresponding to specific probabilities.
  • A go-to tool for working with probabilities and normal distributions.

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Most popular questions from this chapter

Waiting for an elevator. The manager of a large department store with three floors reports that the time a customer on the second floor must wait for an elevator has a uniform distribution ranging from 0 to 4 minutes. If it takes the elevator 15 seconds to go from floor to floor, find the probability that a hurried customer can reach the first floor in less than 1.5 minutes after pushing the second-floor elevator button.

IQs and The Bell Curve. In their controversial book The Bell Curve (Free Press, 1994), Professors Richard J. Herrnstein (a Harvard psychologist who died while the book was in production) and Charles Murray (a political scientist at MIT) explored, as the subtitle states, "intelligence and class structure in American life." The Bell Curve employs statistical analyses heavily in an attempt to support the authors' positions. Since the book's publication, many expert statisticians have raised doubts about the authors' statistical methods and the inferences drawn from them. (See, for example, "Wringing The Bell Curve: A cautionary tale about the relationships among race, genes, and IQ," Chance, Summer \(1995 .\) ) One of the many controversies sparked by the book is the authors" tenet that level of intelligence (or lack thereof) is a cause of a wide range of intractable social problems, including constrained economic mobility. The measure of intelligence chosen by the authors is the well-known intelligent quotient (IQ). Numerous tests have been developed to measure IQ; Herrnstein and Murray use the Armed Forces Qualification Test (AFQT), originally designed to measure the cognitive ability of military recruits. Psychologists traditionally treat \(\mathrm{IQ}\) as a random variable having a normal distribution with mean \(\mu=100\) and standard deviation \(\sigma=15\). In their book, Herrnstein and Murray refer to five cognitive classes of people defined by percentiles of the normal distribution. Class I ("very bright") consists of those with IQs above the 95th percentile; Class II ("bright") are those with IQs between the 75 th and 95 th percentiles; Class III ("normal") includes IQs between the 25 th and 75 th percentiles; Class IV ("dull") are those with IQs between the 5 th and 25 th percentiles; and Class \(\mathrm{V}\) ("very dull") are IQs below the 5 th percentile. a. Assuming that the distribution of IQ is accurately represented by the normal curve, determine the proportion of people with IQs in each of the five cognitive classes defined by Herrnstein and Murray. b. Although Herrnstein and Murray define the cognitive classes in terms of percentiles, they stress that IQ scores should be compared with \(z\) -scores, not percentiles. In other words, it is more informative to give the difference in \(z\) -scores for two IQ scores than it is to give the difference in percentiles. Do you agree? c. Researchers have found that scores on many intelligence tests are decidedly nonnormal. Some distributions are skewed toward higher scores, others toward lower scores. How would the proportions in the five cognitive classes defined in part a differ for an IQ distribution that is skewed right? Skewed left?

Machine repair times. An article in IEEE Transactions (Mar. 1990 ) gave an example of a flexible manufacturing system with four machines operating independently. The repair rates for the machines (i.e., the time, in hours, it takes to repair a failed machine) are exponentially distributed with means \(\mu_{1}=1, \mu_{2}=2, \mu_{3}=.5,\) and \(\mu_{4}=.5,\) respectively. a. Find the probability that the repair time for machine 1 exceeds 1 hour. b. Repeat part a for machine 2 . c. Repeat part a for machines 3 and 4 . d. If all four machines fail simultaneously, find the probability that the repair time for the entire system exceeds 1 hour.

What is a normal probability plot and how is it used?

Tropical island temperatures. Records indicate that the daily high January temperatures on a tropical island tend to have a uniform distribution over the interval from \(75^{\circ} \mathrm{F}\) to \(90^{\circ} \mathrm{F}\). A tourist arrives on the island on a randomly selected day in January. a. What is the probability that the temperature will be above \(80^{\circ} \mathrm{F} ?\) b. What is the probability that the temperature will be between \(80^{\circ} \mathrm{F}\) and \(85^{\circ} \mathrm{F} ?\) c. What is the expected temperature?

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