/*! 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 22 Exercises 5.21 to 5.28 ask you t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Exercises 5.21 to 5.28 ask you to convert an area from one normal distribution to an equivalent area for a different normal distribution. Draw sketches of both normal distributions, find and label the endpoints, and shade the regions on both curves. The upper \(30 \%\) for a \(N(48,5)\) distribution converted to a standard normal distribution.

Short Answer

Expert verified
The Z-score equivalent to the upper 30% of the distribution \(N(48,5)\) is approximately -9.496 in the standard normal distribution \(N(0,1)\).

Step by step solution

01

Identifying the Value In Relation to the Given Distribution

First is to identify the value which represents the upper \(30 \%\) of a \(N(48,5)\) distribution. This can be found by using the Z-distribution table, and finding the Z value corresponding to cumulative probability of \(0.70\) (since upper \(30 \%\) implies lower \(70 \%)\) which gives a z-score value of \(\approx 0.52\).
02

Converting to the Z-Score for Default Distribution

Next, convert this score to normal distribution score using the formula: \(Z = \frac {(X - mu)} {\sigma}\), where \(Z\) is the Z-score, \(X\) is the score from the raw data, \(mu\) is the mean, and \(\sigma\) is the standard deviation. Substituting the given values, we get: \(Z= \frac{0.52 - 48}{5} \approx -9.496\).
03

Sketching the Normal Distributions

Finally, drawing two normal distribution curves, the first for \(N(48,5)\), with shading above the 0.52 value representing the top \(30 \%\) of the distribution, and the second for the standard normal distribution (Z - \(N(0,1)\)), with shading below the -9.496 value that also covers the equivalent top \(30 \%\) of the original distribution.

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-Score
The Z-score is a measure that describes a value's relationship to the mean of a group of values. It is expressed as the number of standard deviations a particular score is away from the mean. In practical terms, a Z-score allows you to compare different data points within the same dataset or in different datasets entirely.

To calculate the Z-score, you use the formula:
  • \[ Z = \frac{X - \mu}{\sigma} \]
Here:
  • \(X\) is the raw score,
  • \(\mu\) is the mean of the dataset,
  • \(\sigma\) is the standard deviation.
By assigning a Z-score to a value, you are identifying how far and in what direction the value deviates from the group's mean. A positive Z-score indicates a value above the mean, while a negative Z-score indicates a value below the mean.

Z-scores are particularly useful because they allow researchers and statisticians to compare scores from different normal distributions. As in the original exercise, when converting the upper 30% of a \(N(48,5)\) distribution to its Z-score equivalent, you transform raw data into a standard format, enabling easier comparison and interpretation.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution. Instead of being defined by any mean and standard deviation, it has a mean of 0 and a standard deviation of 1.

This makes it a key concept in statistics because it allows for the uniform comparison of different datasets, no matter their original mean and standard deviation. When data is converted into Z-scores and thus into the standard normal distribution, it facilitates a deeper understanding of data phenomena and data comparison across different contexts.

This conversion allows for the simplicity of using standardized Z-tables, which provide cumulative probabilities for standard normal distributions. These tables are universally applied in statistical analysis, fundamentally aiding in finding probabilities for ranges of data. Then the original values can be interpreted within these constructs. The beauty of the standard normal distribution is that it transforms unique data into standardized forms simplifying complex processes into digestible parts for clearer insights.
Cumulative Probability
Cumulative probability in the context of the normal distribution refers to the likelihood that a variable will take a value less than or equal to a certain point. It is calculated using the area under the curve to the left of the specified value.

In practice, cumulative probability is crucial when determining where a specific value falls within a given distribution. It gives a proportion between 0 and 1, where 1 (or 100%) represents the certainty of a value falling below a given point.

For instance, when identifying the Z-value that corresponds to cumulative probability for the upper 30% of a distribution, you are effectively calculating the value below which 70% of the data lies. This shift from a direct value to its cumulative probability allows for the determination and integration of probability boundaries essential in statistical analysis. The connection between cumulative probability and Z-scores empowers the transformation of data into a digestible, visual form, making interpretation more intuitive and insightful.

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

Average Age for ICU Patients The ICUAdmissions dataset includes a variable indicating the age of the patient. Find and interpret a \(95 \%\) confidence interval for mean age of ICU patients using the facts that, in the sample, the mean is 57.55 years and the standard error for such means is \(S E=1.42\). The sample size of 200 is large enough to use a normal distribution.

Smoke-Free Legislation and Asthma Hospital admissions for asthma in children younger than 15 years was studied \(^{11}\) in Scotland both before and after comprehensive smoke-free legislation was passed in March \(2006 .\) Monthly records were kept of the annualized percent change in asthma admissions, both before and after the legislation was passed. For the sample studied, before the legislation, admissions for asthma were increasing at a mean rate of 5.2\% per year. The standard error for this estimate is \(0.7 \%\) per year. After the legislation, admissions were decreasing at a mean rate of \(18.2 \%\) per year, with a standard error for this mean of \(1.79 \% . \mathrm{In}\) both cases, the sample size is large enough to use a normal distribution. (a) Find and interpret a \(95 \%\) confidence interval for the mean annual percent rate of change in childhood asthma hospital admissions in Scotland before the smoke-free legislation. (b) Find a \(95 \%\) confidence interval for the same quantity after the legislation. (c) Is this an experiment or an observational study? (d) The evidence is quite compelling. Can we conclude cause and effect?

In Exercises 5.44 to 5.49 , find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A \(90 \%\) confidence interval for a mean \(\mu\) if the sample has \(n=30\) with \(\bar{x}=23.1\) and \(s=5.7,\) and the standard error is \(S E=1.04\)

Exercises 5.50 to 5.55 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\) when the sample has \(n=50\) and \(\hat{p}=0.41,\) with \(S E=0.07\).

Penalty Shots in World Cup Soccer A study \(^{13}\) of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only \(41 \%\) of the time. The article notes that this is "slightly worse than random chance." We use these data as a sample of all World Cup penalty shots ever. Test at a \(5 \%\) significance level to see whether there is evidence that the percent guessed correctly is less than \(50 \%\). The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is \(S E=0.043 .\)

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.