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Chapter 6: Problem 20

Find the \(z\) value described and sketch the area described. Find \(z\) such that \(5 \%\) of the standard normal curve lies to the right of \(z\).

Short Answer

Expert verified
The z-score is approximately 1.645.

Step by step solution

01

Identify the Problem

We need to find the z-score such that only 5% of the standard normal curve lies to the right of this z-score. This implies that 95% of the curve lies to the left of this z-score.
02

Consult the Standard Normal Table

We need to look up the cumulative distribution function (CDF) value of 0.95 (since 95% of the curve must lie to the left) in the standard normal distribution table to find the corresponding z-score.
03

Find the Z-Score

In the z-table, locate the z-score that corresponds to a cumulative probability of 0.95. Typically, this is found to be approximately 1.645.
04

Sketch the Area

Draw the standard normal curve. Mark the mean (0) in the center. Shade the area to the right of z=1.645 because it represents 5% of the total area under the standard normal curve.

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Key Concepts

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

Z-Score Calculation
Calculating a z-score is essential when working with the standard normal distribution. The z-score tells us how many standard deviations an element is from the mean of the dataset. To calculate a z-score, you need the specific value, the mean of the dataset, and the standard deviation. The formula is: \[ z = \frac{(X - \mu)}{\sigma} \] where
  • \( X \) is the value you are analyzing,
  • \( \mu \) is the mean, and
  • \( \sigma \) is the standard deviation.
In the context of the problem, we are not given specific element data, as we are dealing with the standard normal distribution, which already assumes \( \mu = 0 \) and \( \sigma = 1 \). Thus, the z-score indicates how far and in which direction the value lies from the mean.
Cumulative Distribution Function
The cumulative distribution function (CDF) for a standard normal distribution gives us the probability that a z-score is less than or equal to a specific value. It accumulates probabilities from the left of the standard normal distribution curve. This is crucial when locating probabilities in a probability table, such as when 95% of the data lies to the left of a z-score. Understanding the CDF helps us in problems like our example, where we need to know that 95% of the data is under the curve to the left of the z-score found in a table. Thus, by reading the CDF value, which in this example is 0.95, we find the z-score corresponding to the given percentile.
Standard Normal Curve
The standard normal curve, also known as the bell curve or Gaussian distribution, is a key concept in statistics. It is symmetrical with a mean of 0 and a standard deviation of 1. The area under the curve signifies the probability of a particular range of outcomes. This curve is fundamental when visualizing the distribution of z-scores or when solving problems, such as determining the percentage of data within a certain range. In the given problem, sketching the standard normal curve helped indicate where exactly the z-score was located, specifically showing that 5% of the area lies to the right of the z = 1.645.
Probability Tables
Probability tables, or z-tables, help us find the area under the standard normal curve corresponding to z-scores. These tables display the cumulative probability of being less than a z-score, vital for problems where specific areas need to be found. In our exercise, using a z-table helped us identify a cumulative probability of 0.95 (or 95%) to find the z-score close to 1.645. Probability tables streamline the process since calculations for each point on the standard normal curve would be complex without them.

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Most popular questions from this chapter

Consider a binomial experiment with 20 trials and probability \(0.45\) of success on a single trial. (a) Use the binomial distribution to find the probability of exactly 10 successes. (b) Use the normal distribution to approximate the probability of exactly 10 successes. (c) Compare the results of parts (a) and (b).

Based on long experience, an airline found that about \(6 \%\) of the people making reservations on a flight from Miami to Denver do not show up for the flight. Suppose the airline overbooks this flight by selling 267 ticket reservations for an airplane with only 255 seats. (a) What is the probability that a person holding a reservation will show up for the flight? (b) Let \(n=267\) represent the number of ticket reservations. Let \(r\) represent the number of people with reservations who show up for the flight. Which expression represents the probability that a seat will be available for everyone who shows up holding a reservation? \(P(255 \leq r) ; \quad P(r \leq 255) ; \quad P(r \leq 267) ; \quad P(r=255)\) (c) Use the normal approximation to the binomial distribution and part (b) to answer the following question: What is the probability that a seat will be available for every person who shows up holding a reservation?

Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. $$ P(3 \leq x \leq 6) ; \mu=4 ; \sigma=2 $$

Find the \(z\) value described and sketch the area described. Find \(z\) such that \(97.5 \%\) of the standard normal curve lies to the left of \(z\).

Assume that \(x\) has a normal distribution with the specified mean and standard deviation. Find the indicated probabilities. $$ P(7 \leq x \leq 9) ; \mu=5 ; \sigma=1.2 $$
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