/*! 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 41 Empirical Rule for Normal Distri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Empirical Rule for Normal Distributions Pick any positive values for the mean and the standard deviation of a normal distribution. Use your selection of a normal distribution to answer the questions below. The results of parts (a) to (c) form what is often called the Empirical Rule for the standard deviation in a normal distribution. (a) Verify that about \(95 \%\) of the values fall within two standard deviations of the mean. (b) What proportion of values fall within one standard deviation of the mean? (c) What proportion of values fall within three standard deviations of the mean? (d) Will the answers to (b) and (c) be the same for any normal distribution? Explain why or why not.

Short Answer

Expert verified
According to the empirical rule, about \(95 \%\) of the values fall within two standard deviations of the mean, about \(68 \%\) of the values fall within one standard deviation of the mean, and about \(99.7 \%\) fall within three standard deviations of the mean. These percentages hold true for all normal distributions.

Step by step solution

01

Select Mean and Standard Deviation

Choose any positive values for the mean and the standard deviation of a normal distribution. For instance, let's take the mean (\( \mu \)) as 100 and the standard deviation (\( \sigma \)) as 15.
02

Apply Empirical Rule for two standard deviations

The empirical rule states that 95% of the data falls within two standard deviations of the mean. Therefore, this means that 95% of the values will fall between \( \mu - 2\sigma \) and \( \mu + 2\sigma \). In our case, this is between (100 - 2*15) and (100 + 2*15), which is between 70 and 130. So, \(95 \%\) of the values fall within two standard deviations of the mean.
03

Empirical Rule for one standard deviation

According to empirical rule, 68% of the data falls within one standard deviation of the mean. That means 68% of the values will fall between \( \mu - \sigma \) and \( \mu + \sigma \). With our values, this is between (100 - 15) and (100 + 15), which is between 85 and 115.
04

Empirical Rule for three standard deviations

For three standard deviations from the mean, the empirical rule states that 99.7% of the values in a normal distribution fall within this range. Thus, 99.7% of the data will be between \( \mu - 3\sigma \) and \( \mu + 3\sigma \). In our case, this is between (100 - 3*15) and (100 + 3*15) which is between 55 and 145.
05

Explanation

The empirical rule applies to all normal distributions, so the answers to (b) and (c) will be the same regardless of the chosen mean and standard deviation. But for distributions that are not normal, the empirical rule may not hold.

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!

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

Heights of Men in the US Heights of adult males in the US are approximately normally distributed with mean 70 inches \((5 \mathrm{ft} 10 \mathrm{in})\) and standard deviation 3 inches. (a) What proportion of US men are between \(5 \mathrm{ft}\) 8 in and \(6 \mathrm{ft}\) tall \((68\) and 72 inches, respectively)? (b) If a man is at the 10 th percentile in height, how tall is he?

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 \(H_{0}: \mu=80\) vs \(H_{a}: \mu>80\) when the sample has \(n=20, \bar{x}=82.4\), and \(s=3.5,\) with \(S E=0.8\).

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 \(H_{0}: \mu=10\) vs \(H_{a}: \mu \neq 10\) when the sample has \(n=75, \bar{x}=11.3,\) and \(s=0.85,\) with \(S E=0.10\)

Hearing Loss in Teenagers A recent study" found that, of the 1771 participants aged 12 to 19 in the National Health and Nutrition Examination Survey, \(19.5 \%\) had some hearing loss (defined as a loss of 15 decibels in at least one ear). This is a dramatic increase from a decade ago. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error for the proportion is \(S E=0.009 .\) Find and interpret a \(90 \%\) confidence interval for the proportion of teenagers with some hearing loss.

Curving Grades on an Exam A statistics instructor designed an exam so that the grades would be roughly normally distributed with mean \(\mu=75\) and standard deviation \(\sigma=10 .\) Unfortunately, a fire alarm with ten minutes to go in the exam made it difficult for some students to finish. When the instructor graded the exams, he found they were roughly normally distributed, but the mean grade was 62 and the standard deviation was 18\. To be fair, he decides to "curve" the scores to match the desired \(N(75,10)\) distribution. To do this, he standardizes the actual scores to \(z\) -scores using the \(N(62,18)\) distribution and then "unstandardizes" those \(z\) -scores to shift to \(N(75,10)\). What is the new grade assigned for a student whose original score was \(47 ?\) How about a student who originally scores a \(90 ?\)

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.