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According to the Sleep Foundation, the average night's sleep is 6.8 hours (Fortune, March 20,2006 ). Assume the standard deviation is .6 hour and that the probability distribution is normal a. What is the probability that a randomly selected person sleeps more than 8 hours? b. What is the probability that a randomly selected person sleeps 6 hours or less? c. Doctors suggest getting between 7 and 9 hour of sleep each night. What percentage of the population gets this much sleep?

Short Answer

Expert verified
a) 2.28% b) 9.18% c) 37.06%

Step by step solution

01

Identify Given Values

We are given that the average night's sleep (mean \( \mu \)) is 6.8 hours and the standard deviation (\( \sigma \)) is 0.6 hours. We assume a normal distribution for sleep hours.
02

Compute Z-score for Part (a)

For part (a), we need to find the probability that a person sleeps more than 8 hours. First, calculate the Z-score using the formula: \[Z = \frac{X - \mu}{\sigma}\]where \(X = 8\) hours: \[Z = \frac{8 - 6.8}{0.6} = \frac{1.2}{0.6} = 2\].
03

Find Probability for Part (a)

Use standard normal distribution tables or a calculator to find the probability that \(Z > 2\). The Z-table value for \(Z = 2\) is approximately 0.9772. Thus, the probability is \[P(Z > 2) = 1 - 0.9772 = 0.0228\].
04

Compute Z-score for Part (b)

For part (b), find the probability that a person sleeps 6 hours or less by calculating the Z-score where \(X = 6\):\[Z = \frac{6 - 6.8}{0.6} = \frac{-0.8}{0.6} = -1.3333\].
05

Find Probability for Part (b)

Using the Z-table, find \(P(Z \leq -1.3333)\). The value for \(Z \approx -1.33\) is about 0.0918, thus \[P(Z \leq -1.3333) = 0.0918\].
06

Set Range for Part (c) and Find Z-scores

For part (c), the range is between 7 and 9 hours. Calculate the Z-scores:- For \(X = 7\): \[Z_1 = \frac{7 - 6.8}{0.6} = \frac{0.2}{0.6} = 0.3333\] - For \(X = 9\): \[Z_2 = \frac{9 - 6.8}{0.6} = \frac{2.2}{0.6} = 3.6667\].
07

Calculate Probability for Part (c)

Find the probability that \( 0.3333 \leq Z \leq 3.6667 \). Using the Z-table:- \(P(Z \leq 0.3333) = 0.6293\)- \(P(Z \leq 3.6667) \approx 0.9999\)So, \[P(0.3333 \leq Z \leq 3.6667) = 0.9999 - 0.6293 = 0.3706\].Thus, 37.06% of the population sleeps between 7 and 9 hours.

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Key Concepts

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

Understanding Z-score
A Z-score is a way to measure how far away a particular value is from the mean, in terms of standard deviations. It helps us understand where a specific data point lies in a standard normal distribution.
Imagine the Z-score as a translator. It takes your raw data point and tells you how unusual it is based on the known average and the spread of the data.
Here's the formula to calculate it: \[ Z = \frac{X - \mu}{\sigma} \]- **\(X\)** is the value of interest.- **\(\mu\)** is the mean of the distribution.- **\(\sigma\)** is the standard deviation.
In our sleep study, to determine how often people sleep more than 8 hours, you first compute the Z-score. This calculation tells you if 8 hours is far from the average sleep duration, which is 6.8 hours, given a spread of 0.6 hours. Understanding Z-scores is crucial because it standardizes different datasets, allowing for easier comparison and probability assessment across datasets.
Probability Calculation in Normal Distribution
When we talk about probability in the context of a normal distribution, we're usually interested in finding out how likely it is that a data point falls in a certain range.
Probabilities in a normal distribution are calculated using Z-scores to find the area under the curve. This area tells you how much of the data falls below or above a particular data point. You can use Z-tables or statistical software to identify this area easily.
  • For example, if we want the probability that a person sleeps more than 8 hours, we first compute the Z-score (which is 2 in this case) and then use the Z-table to find the probability of a Z-score greater than 2.
  • Likewise, to find the probability of sleeping 6 hours or less, you calculate the Z-score for 6 and find the corresponding probability from the Z-table.
Because the normal distribution is symmetrical, once the Z-score is known, you can quickly determine probabilities by referencing Z-tables. These calculations are like finding areas on a map that tell you what's normal (or common) in your dataset.
Application in Sleep Study
In a practical setting like a sleep study, understanding the normal distribution and related probabilities can help us analyze how people sleep.
Sleep studies focus on patterns and how they compare to the general population. By knowing the mean and standard deviation, we can predict the sleeping habits of the wider population.
- **Example Scenarios:** - Are you getting more than the average sleep of 6.8 hours? - How unusual is it to sleep only 6 hours or less? - What percentage of people manage to sleep between 7 to 9 hours as recommended by doctors? In our exercise, we could see that a significant percentage, specifically 37.06%, sleeps within the optimal range of 7 to 9 hours.
This information helps health professionals understand general sleep habits and design better sleep-related health interventions to help promote well-being.

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