/*! 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 23 Find the \(z\) value described a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the \(z\) value described and sketch the area described.Find the \(z\) value such that \(98 \%\) of the standard normal curve lies between \(-z\) and \(z\).

Short Answer

Expert verified
The \(z\) value is approximately \(\pm 2.33\).

Step by step solution

01

Understanding the Problem

We need to find the value of \(z\) such that 98% of the standard normal distribution is between \(-z\) and \(z\). This means that there is 1% in each tail of the distribution (since 2% is outside the interval \([-z, z]\)).
02

Identify the Percentile in a Standard Normal Distribution

Since 98% of the distribution is between \(-z\) and \(z\), look for the 1% percentile for \(-z\) and the 99% percentile for \(z\) in the standard normal distribution table, or using a calculator for a more accurate value.
03

Use a z-table or calculator

Using a z-table or calculator, find the \(z\) value that corresponds to the cumulative probability of 0.99 (since finding \(z\) for 99% will indirectly give \(z\) for the center area from -z to z). This value is approximately 2.33.
04

Determine the Final Result for both Sides

Since the normal distribution is symmetric, the \(z\) value found for 99% is the same (but positive) as the \(z\) value for 1%. Thus, \(z = \pm 2.33\) ensures 98% of the distribution lies between \(-z\) and \(z\).
05

Sketch the Distribution

Draw the standard normal curve. Mark the area between \(-2.33\) and \(2.33\) and shade it to represent the central 98% of the distribution. Indicate the tails as 1% each beyond these limits.

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Key Concepts

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

Understanding Z-Value in Standard Normal Distribution
The z-value, often called a z-score, is a measure that describes a point's position within a standard normal distribution. The standard normal distribution, also known simply as the normal distribution, is a probability distribution that is symmetric about the mean. The mean is located at zero, and the distribution has a standard deviation of one. The z-value essentially tells us how many standard deviations away a point is from the mean.

When you encounter a problem that asks for a specific z-value, it’s asking where to draw a line on a standard normal curve so that a certain amount of data falls between certain points. If you imagine this distribution as a bell-shaped curve, a z-value of 0 sits at the peak, right in the center of the curve. Positive z-values fall to the right of the mean, and negative z-values fall to the left. In the exercise, we needed to find such z-values to ensure that 98% of data is sandwiched between \(-z\) and \(+z\).

To realize this, it's crucial to grasp that the normal curve is symmetric. This means that percentages related to these z-values are mirrored across the mean, which makes calculations predictable once you understand the principle.
Exploring Percentiles
Percentiles are rankings used in statistics to indicate how a particular score compares to the overall data set. In the context of a standard normal distribution, a percentile informs us about data positioning. If you're told that a score is at the 95th percentile, it means the score is higher than or equal to 95% of the data points in the distribution.

In the exercise, we needed to focus on two specific percentiles: the 1st percentile and the 99th percentile. Why these ones specifically? Because they help determine the values of \(-z\) and \(+z\) that contain the middle 98% of the data. The 1st percentile related to the lower cutoff (left tail), and the 99th percentile to the upper cutoff (right tail) of the curve. By identifying these values, you can define the range where most (in this case, 98%) data points fall.

This approach is essential in numerous statistical analyses where one needs to set boundaries according to desired coverage or to understand data distribution characteristics.
Cumulative Probability and its Role
Cumulative probability is a concept that represents the probability of a variable falling within a certain range in a distribution. In simpler terms, it tells you the likelihood of drawing a score below a particular value in a given distribution. This becomes crucial when dealing with z-values.

When finding z-values for specific percentiles, you’re essentially tackling cumulative probabilities. The cumulative probability for a z-value is the area under the curve up to that point. For instance, if a z-value aligns with a cumulative probability of 0.99, it indicates that 99% of the data falls below that point.

In the original exercise, this understanding was instrumental. By identifying a cumulative probability of 0.99 (i.e., 99th percentile) using a z-table or calculator, we found that the corresponding z-value is approximately 2.33. Thus, \(z = \, \pm 2.33\) ensures that 98% of data lies symmetrically centered around zero, satisfying the conditions given.

Employing cumulative probabilities allows statisticians to make informed predictions and conclusions about where data points typically fall within a standard distribution.

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Most popular questions from this chapter

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) Interpretation 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 sclected 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?

Insurance: Claims Do you try to pad an insurance claim to cover your deductible? About \(40 \%\) of all U.S. adults will try to pad their insurance claims! (Source: Are You Normal?, by Bernice Kanner, St. Martin's Press.) Suppose that you are the director of an insurance adjustment office. Your office has just received 128 insurance claims to be processed in the next few days. What is the probability that (a) half or more of the claims have been padded? (b) fewer than 45 of the claims have been padded? (c) from 40 to 64 of the claims have been padded? (d) more than 80 of the claims have not been padded?

Empirical Rule What percentage of the area under the normal curve lies (a) to the left of \(\mu ?\) (b) between \(\mu-\sigma\) and \(\mu+\sigma ?\) (c) between \(\mu-3 \sigma\) and \(\mu+3 \sigma ?\)

Let \(x\) represent the dollar amount spent on supermarket impulse buying in a 10 -minute (unplanned) shopping interval. Based on a Denver Post article, the mean of the \(x\) distribution is about \(\$ 20\) and the estimated standard deviation is about \(\$ 7\) (a) Consider a random sample of \(n=100\) customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of \(\bar{x}\) the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the \(\bar{x}\) distribution? Is it necessary to make any assumption about the \(x\) distribution? Explain. (b) What is the probability that \(\bar{x}\) is between \(\$ 18\) and \(\$ 22 ?\) (c) Let us assume that \(x\) has a distribution that is approximately normal. What is the probability that \(x\) is between \(\$ 18\) and \(\$ 22 ?\) (d) Interpretation: In part (b), we used \(\bar{x},\) the average amount spent, computed for 100 customers. In part (c), we used \(x,\) the amount spent by only one customer. The answers to parts (b) and (c) are very different. Why would this happen? In this example, \(\bar{x}\) is a much more predictable or reliable statistic than \(x\). Consider that almost all marketing strategies and sales pitches are designed for the average customer and not the individual customer. How does the central limit theorem tell us that the average customer is much more predictable than the individual customer?

Does a raw score less than the mean correspond to a positive or negative standard score? What about a raw score greater than the mean?

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