/*! 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 18 Assume that a quantitative chara... [FREE SOLUTION] | 91Ó°ÊÓ

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Assume that a quantitative character is normally distributed with mean \(\mu\) and standard deviation \(\sigma .\) Determine what fraction of the population falls into the given interval. \([\mu+\sigma, \mu+2 \sigma]\)

Short Answer

Expert verified
Approximately 13.59% of the population is in the interval \\( [\mu + \sigma, \mu + 2\sigma] \\).

Step by step solution

01

Understanding the Normal Distribution

In a normal distribution, the mean \( \mu \) defines the center of the distribution, and the standard deviation \( \sigma \) determines the spread. The area under the normal distribution curve represents probabilities or fractions of the population. We want to find the fraction of the population that falls between \([\mu + \sigma, \mu + 2\sigma]\).
02

Standardizing the Interval

To find probabilities in a normal distribution, we convert the interval bounds to z-scores using the formula \((x - \mu)/\sigma\). For the lower bound, \( z = (\mu + \sigma - \mu) / \sigma = 1 \). For the upper bound, \( z = (\mu + 2\sigma - \mu) / \sigma = 2 \). Thus, we need to find the probability between \( z = 1 \) and \( z = 2 \).
03

Using the Standard Normal Distribution Table

A standard normal distribution table gives the probability that a standard normal random variable is less than or equal to a given z-score. For \( z = 1 \,\) the probability is approximately 0.8413; for \( z = 2 \,\) the probability is approximately 0.9772. We find the area between these two z-scores by subtracting the probabilities: \( 0.9772 - 0.8413 = 0.1359 \).
04

Conclusion

The fraction of the population that falls between \( \mu + \sigma \) and \( \mu + 2\sigma \) is approximately 0.1359, or 13.59%. This means that about 13.59% of the distribution lies within this interval.

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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 of expressing a data point in relation to the mean, measured in terms of standard deviations. It's a very useful statistic in the world of normal distributions because it allows us to easily understand and compare different values from various distributions. The formula to calculate a z-score is:
  • \( z = \frac{x - \mu}{\sigma} \)
Where:
  • \( x \) is the value you're looking at,
  • \( \mu \) is the mean of the distribution,
  • \( \sigma \) is the standard deviation.
To standardize or find the z-score for a range, such as from \( \mu + \sigma \) to \( \mu + 2\sigma \), you substitute these expressions into the formula to compute the respective z-scores. By doing this, values from different distributions can be directly compared using their z-scores. The z-score helps in determining how far away a particular data point is from the mean in terms of standard deviations.
Decoding Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. When we speak of a normal distribution, the standard deviation tells us how spread out the values are from the mean. The smaller the standard deviation, the closer the data points are to the mean, resulting in a steeper bell-curve shape. Conversely, a larger standard deviation indicates a wider spread, creating a flatter shape on the graph. The standard deviation is pivotal when working with normal distributions because it tells us the 'spread' or 'width' of the data. In statistics, it allows precise statements about where particular fractions of the population, or probabilities, lie in relation to the mean. For instance, within one standard deviation of the mean, approximately 68% of values lie in a normal distribution. When applied to a range, it helps us define intervals such as \( [\mu + \sigma, \mu + 2\sigma] \), quantifying the fraction of data within this space.
Interpreting the Standard Normal Distribution Table
The standard normal distribution table, often referred to as a z-table, helps determine the percentage of values below a certain z-score in standard normal distribution. This table is essential for converting z-scores into probabilities.When you calculate a z-score, you can look it up in the z-table to find the cumulative probability up to that z-score. This value tells you the proportion of data below that z-score in a standard normal distribution. Here's how it works:
  • Find the z-score in the table (most tables showcase the first digit and first decimal together, with the second decimal being isolated on a second column).
  • Locate the cumulative probability corresponding to the z-score.
For intervals, such as between \( z = 1 \) and \( z = 2 \), you find the cumulative probabilities for both values and subtract them: \( P(z = 2) - P(z = 1) \). This result gives you the probability of a data point falling within this specific range. Such tables are an invaluable tool in statistics, particularly when analyzing data and making probability statements within normally distributed data sets.

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Most popular questions from this chapter

A number of human traits are caused by a single pair of recessive genes and thus manifest themselves only in individuals who are homozygous for the mutant gene. An individual with one normal and one mutant gene is a carrier, but does not exhibit the trait.Calculate each of the probabilites. Tay-Sachs disease is caused by a single pair of recessive genes. If both parents are carriers of the mutant gene, what is the likelihood that none of their four children will be affected?

An urn contains 12 green and 24 blue balls. (a) You take 10 balls out of the urn. Find the probability that 6 of the 10 balls are blue. (b) You take a ball out of the urn, note its color, and replace it. You repeat these steps 10 times. Find the probability that 6 of the 10 balls are blue.

Suppose that the probability mass function of a discrete random variable \(X\) is given by the following table: $$\begin{array}{rc} \hline \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) \\ \hline-3 & 0.2 \\ -1 & 0.3 \\ 1.5 & 0.4 \\ 2 & 0.1 \\ \hline \end{array}$$ Find the mean, the variance, and the standard deviation of \(X\).

Suppose \(X\) is Poisson distributed with parameter \(\lambda=1.5\). Find the probability that \(X\) exceeds \(3 .\)

Suppose that the probability mass function of a discrete random variable \(X\) is given by the following table: $$\begin{array}{cc} \hline \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) \\ \hline-2 & 0.1 \\ -1 & 0.4 \\ 0 & 0.3 \\ 1 & 0.2 \\ \hline \end{array}$$ (a) Find \(E(X)\). (b) Find \(E\left(X^{2}\right)\). (c) Find \(E[X(X-1)]\).

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