/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Find the \(z\) value described a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(z\) value is approximately \(1.645\).

Step by step solution

01

Understanding the Problem

We need to find the value of \(z\) such that \(5\%\) of the standard normal distribution (Z-distribution) lies to the right of \(z\). This means we are looking for a \(z\) value with a right tail probability of \(0.05\).
02

Translating the Problem into a Probability

Since the total area under the standard normal curve is \(1\), if \(5\%\) of the area is to the right of \(z\), then \(95\%\) of the area is to the left of \(z\). This translates to the statement that the cumulative probability \(P(Z \leq z) = 0.95\).
03

Using the Standard Normal Distribution Table

Consult a standard normal distribution table to find the \(z\) value that corresponds to a cumulative probability of \(0.95\). Alternatively, a statistical software or calculator with a normal distribution function can be used. The \(z\) value associated with \(P(Z \leq z) = 0.95\) is approximately \(1.645\).
04

Sketching the Area

Draw the standard normal distribution curve which is bell-shaped and symmetric around 0. Mark the \(z\) value at \(1.645\). Shade the area to the right of this \(z\) value, which visually represents the \(5\%\) probability.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Z-value
In statistics, the Z-value, or Z-score, is a measure of how many standard deviations an element is from the mean of a distribution. In a standard normal distribution, the mean is 0 and the standard deviation is 1. Therefore, the Z-score directly corresponds to a position on the standard normal curve. The formula to calculate a Z-score is:
\[Z = \frac{(X - \mu)}{\sigma}\]where:
  • \(X\) is the value being analyzed.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation of the distribution.
In the context of our exercise, finding the Z-value is crucial for determining the position on a normal distribution curve where exactly 5% of the data falls to the right. The Z-score provides the specific number needed for this alignment. When a Z-value is referenced as having a certain percentage "to the right," it ties directly to the concept of right tail probability.
Cumulative Probability
Cumulative Probability refers to the probability that a random variable will assume a value less than or equal to a specific value. In the standard normal distribution, cumulative probability helps us understand what proportion of the data falls below a certain Z-score.
For example, if a Z-score has a cumulative probability of 0.95, it suggests that 95% of the data lies below or to the left of this Z-score in the standard normal curve.
Cumulative probability is useful when translating between the probability of an event and its corresponding Z-value. In the exercise provided, identifying a Z-score where 95% of the data falls to the left helps us locate the equivalent position on the curve where only 5% remains to the right.
Normal Distribution Table
The Normal Distribution Table, also known as the Z-table, lists cumulative probabilities of the standard normal distribution. This tool helps us find the probability that a Z-value is below a particular point. The table typically provides values for cumulative probabilities up to any Z-score.
To use it, you look up the row representing the whole Z-value and the column for the hundredths place. For instance, to find the cumulative probability for a Z-score of 1.64, you would cross-reference 1.6 in the row with 0.04 in the column.
In our exercise, this table is key to identifying the Z-value that corresponds to a cumulative probability of 0.95, giving us the solution that the associated Z-value is approximately 1.645. This understanding can also be supported by using statistical software or a calculator with distribution functions as an alternative method.
Right Tail Probability
Right tail probability is the probability that a standard normal random variable is greater than a specific Z-value. It essentially represents the area under the curve from that Z-value extending infinitely rightward. This concept allows us to discern how much of the data set falls beyond a certain threshold in a distribution.
Illustratively, if we have a Z-value where the right tail probability is 0.05, this tells us that 5% of the observations are expected to be greater than that Z-score. The main point in our exercise is to find the Z-score where this tail contains precisely 5% of the total area under the standard normal curve.
Visualizing this involves sketching a standard normal distribution curve, marking the Z-value (1.645), and shading the area under the curve to the right of this point, representing the 5% right tail probability. This visual interpretation reinforces the mathematical conclusion drawn from the Z-table.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose \(5 \%\) of the area under the standard normal curve lies to the left of \(z\). Is \(z\) positive or negative?

The college physical education department offered an advanced first aid course last semester. The scores on the comprehensive final exam were normally distributed, and the \(z\) scores for some of the students are shown below: $$ \begin{array}{lcl} \text { Robert, } 1.10 & \text { Juan, } 1.70 & \text { Susan, }-2.00 \\ \text { Joel, } 0.00 & \text { Jan, }-0.80 & \text { Linda, } 1.60 \end{array} $$ (a) Which of these students scored above the mean? (b) Which of these students scored on the mean? (c) Which of these students scored below the mean? (d) If the mean score was \(\mu=150\) with standard deviation \(\sigma=20\), what was the final exam score for each student?

Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean \(\mu=75\) tons and standard deviation \(\sigma=0.8\) ton. (a) What is the probability that one car chosen at random will have less than \(74.5\) tons of coal? (b) What is the probability that 20 cars chosen at random will have a mean load weight \(\bar{x}\) of less than \(74.5\) tons of coal? (c) Suppose the weight of coal in one car was less than \(74.5\) tons. Would that fact make you suspect that the loader had slipped out of adjustment? Suppose the weight of coal in 20 cars selected at random had an average \(\bar{x}\) of less than \(74.5\) tons. Would that fact make you suspect that the loader had slipped out of adjustment? Why?

Assuming that the heights of college women are normally distributed with mean 65 inches and standard deviation \(2.5\) inches (based on information from Statistical Abstract of the United States, 112 th Edition), answer the following questions. Hint: Use Problems 5 and 6 and Figure \(6-3\). (a) What percentage of women are taller than 65 inches? (b) What percentage of women are shorter than 65 inches? (c) What percentage of women are between \(62.5\) inches and \(67.5\) inches? (d) What percentage of women are between 60 inches and 70 inches?

Let \(x\) be a random variable that represents the weights in kilograms (kg) of healthy adult female deer (does) in December in Mesa Verde National Park. Then \(x\) has a distribution that is approximately normal, with mean \(\mu=63.0 \mathrm{~kg}\) and standard deviation \(\sigma=7.1 \mathrm{~kg}\) (Source: The Mule Deer of Mesa Verde National Park, by G. W. Mierau and J. L. Schmidt, Mesa Verde Museum Association). Suppose a doe that weighs less than \(54 \mathrm{~kg}\) is considered undernourished. (a) What is the probability that a single doe captured (weighed and released) at random in December is undernourished? (b) If the park has about 2200 does, what number do you expect to be undernourished in December? (c) To estimate the health of the December doe population, park rangers use the rule that the average weight of \(n=50\) does should be more than \(60 \mathrm{~kg}\). If the average weight is less than \(60 \mathrm{~kg}\), it is thought that the entire population of does might be undernourished. What is the probability that the average weight \(\bar{x}\) for a random sample of 50 does is less than \(60 \mathrm{~kg}\) (assume a healthy population)? (d) Compute the probability that \(\bar{x}<64.2 \mathrm{~kg}\) for 50 does (assume a healthy population). Suppose park rangers captured, weighed, and released 50 does in December, and the average weight was \(\bar{x}=64.2 \mathrm{~kg}\). Do you think the doe population is undernourished or not? Explain.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.