/*! 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 The distribution of hours of sle... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The distribution of hours of sleep per weeknight among college students is found to be Normally distributed, with a mean of \(6.5\) hours and a standard deviation of 1 hour. The percentage of college students that sleep at least eight hours per weeknight is about a. \(95 \%\) b. \(6.7 \%\) c. \(2.5 \%\)

Short Answer

Expert verified
Approximately 6.7% of college students sleep at least 8 hours per weeknight.

Step by step solution

01

Understand the Problem

We have a normal distribution of college students' sleep hours with a mean \( \mu = 6.5 \) hours and a standard deviation \( \sigma = 1 \) hour. We need to find the percentage of students that sleep at least 8 hours per weeknight.
02

Calculate Z-score

The Z-score formula is \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value we are interested in. Substituting \( X = 8 \), \( \mu = 6.5 \), and \( \sigma = 1 \), we have:\[Z = \frac{8 - 6.5}{1} = 1.5\]
03

Use Z-score to Find Probability

Using a standard normal distribution table, we find the probability that a Z-score is less than 1.5. The cumulative probability for \( Z = 1.5 \) is approximately 0.9332, meaning 93.32% of students sleep less than 8 hours.
04

Find the Complementary Probability

Since we want the percentage of students that sleep at least 8 hours, we calculate the complementary probability:\[1 - 0.9332 = 0.0668\]This corresponds to about 6.68%.

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 calculation
To tackle problems involving normal distribution, calculating the Z-score is an essential step. The Z-score allows us to determine how many standard deviations a data point is away from the mean of the distribution. This is crucial in understanding the position of a value within a dataset.
The formula for the Z-score is given by:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here, \( X \) represents the data point we're interested in, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
In this specific problem, we're looking to find out how 8 hours compares to the average of 6.5 hours with a standard deviation of 1 hour.
Plugging in these numbers:
  • \( Z = \frac{8 - 6.5}{1} = 1.5 \)
This tells us that 8 hours of sleep is 1.5 standard deviations above the mean sleep duration for the students.
Cumulative probability
Once we have the Z-score, the next step is to find the cumulative probability. This will tell us the probability of a random data point falling below a particular value.
Using the Z-score we calculated for 8 hours of sleep, 1.5, we refer to standard normal distribution tables or use statistical software for precise values.
For a Z-score of 1.5, the cumulative probability (area under the curve to the left of the Z-score) is approximately 0.9332. This means that 93.32% of the students sleep less than 8 hours.
Cumulative probability gives us a powerful overview of where a particular observation stands in comparison to the rest of the data. It is particularly useful in assessing how typical or unusual a certain observation is within the data distribution.
Complementary probability
Understanding complementary probability is key when dealing with probability distributions. After determining the cumulative probability, which shows us how many students sleep less than 8 hours, we can find the complementary probability to answer the question of interest: How many sleep at least 8 hours?
Since cumulative probability informs us of the percentage of occurrences up to a point, complementary probability looks at the opposite side.
To calculate it, we simply subtract the cumulative probability from 1:
  • Complementary Probability = \( 1 - 0.9332 = 0.0668 \)
This calculation tells us that approximately 6.68% of students manage to get 8 or more hours of sleep.
Complementary probability highlights the significance of the rare outcomes which are often equally important in understanding the full picture of data distribution.

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

Weights Aren't Normal. The heights of people of the same sex and similar ages follow a Normal distribution reasonably closely. Weights, on the other hand, are not Normally distributed. The weights of men aged 20-29 in the United States have mean 186.8 pounds and median \(177.8\) pounds. The first and third quartiles are \(152.9\) pounds and \(208.5\) pounds, respectively. In addition, the bottom \(10 \%\) have weights less than or equal to \(137.6\) pounds while the top \(10 \%\) have weights greater than or equal to 247.2. What can you say about the shape of the weight distribution? Why?

The scores of adults on an IQ test are approximately Normally distributed, with mean 100 and standard deviation 15. Alysha scores 135 on such a test. Her \(z\)-score is about a. \(1.33 .\) b. \(2.33\) c. \(6.33 .\)

Runners. In a study of exercise, a large group of male runners walk on a treadmill for 6 minutes. Their heart rates in beats per minute at the end vary from runner to runner according to the \(N(104,12.5)\) distribution. The heart rates for male nonrunners after the same exercise have the \(N(130,17)\) distribution. a. What percentage of the runners have heart rates above 140 ? b. What percentage of the nonrunners have heart rates above 140 ?

The proportion of observations from a standard Normal distribution that take values greater than \(1.78\) is about a. \(0.9554 .\) b. \(0.0446 .\) c. \(0.0375 .\)

Standard Normal Drill. a. Find the number \(z\) such that the proportion of observations that are less than \(z\) in a standard Normal distribution is \(0.2\). b. Find the number \(z\) such that \(40 \%\) of all observations from a standard Normal distribution are greater than \(z\).

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.