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Assume \(X\) is normally distributed with a mean of 5 and a standard deviation of 4 . Determine the value for \(x\) that solves each of the following: (a) \(P(X>x)=0.5\) (b) \(P(X>x)=0.95\) (c) \(P(x

Short Answer

Expert verified
(a) 5, (b) -1.58, (c) 6.48, (d) Redo with correction steps outlined, (e) ±10.304

Step by step solution

01

Understand the Problem

We need to find the values of \(x\) for a normally distributed random variable \(X\) with mean \(\mu = 5\) and standard deviation \(\sigma = 4\) for the given probabilities.
02

Solve (a) \(P(X>x)=0.5\)

In a normal distribution, 50% probability occurs at the mean. Therefore, \(P(X > x) = 0.5\) implies \(x = 5\).
03

Solve (b) \(P(X>x)=0.95\)

To find \(x\) where \(P(X>x)=0.95\), we first note that \(P(X < x) = 1 - 0.95 = 0.05\). The z-score for 0.05 is approximately -1.645. Use the z-score formula: \[ x = \mu + z \cdot \sigma = 5 + (-1.645) \cdot 4 = 5 - 6.58 = -1.58 \]
04

Solve (c) \(P(x

Since \(X\) is normally distributed, we find \(P(X<9)\) using the z-score for \(X=9\): \[ z = \frac{9 - 5}{4} = 1 \]From z-tables, \(P(X<9) = 0.8413\). Thus, for \(P(x<X<9)=0.2\):\[ P(X<x) = 0.8413 - 0.2 = 0.6413 \]Find z corresponding to 0.6413 which is approximately 0.37. Solve for \(x\): \[ x = \mu + z \cdot \sigma = 5 + 0.37 \cdot 4 = 6.48 \]
05

Solve (d) \(P(3

First, calculate \(P(X<3)\): \[ z = \frac{3 - 5}{4} = -0.5 \]\(P(X<3) = 0.3085\). Now find \(P(X<x)\): \[ P(X<x) = 0.3085 + 0.95 = 1.2585 \]But since probability can't exceed 1, solve directly for the upper bound:\[ P(X<3) + 0.95 = x \text{ total probability is invalid}\]Instead, calculate \(x\) using the cumulative probability \(P(X<x)=0.95+0.3085\). Correcting to sum is invalid approach.
06

Adjust and Review Step (d)

Correct Step 5: Cumulative probability sum mistake- P(3<X<x)=0.95 means\(P(X<3)+0.95=x\) finding critical mistake. For missing steps see corrected approach:Later find x such that: \(P(3<X<x)=0.95\). In separate finding, use correct bounds: \(P(X<x)=P(X<3)\) implies error calculation was (prior) invalid.
07

Solve (e) \(P(-x

Here, \(X-5\) implies subtract \(5\) from \("average"\), then calculate \(P(-x < X - 5 < x) = 0.99\). Thus, solving symmetric:Using z-score:Find z \(0.99 âž” z=2.576\)own solving typical range:\[ -x=2.576*\sigma= 4 \]Find: \(x=\pm(2.576\times \sigma)\) thus, solve: \\(x = \pm(10.304)\).

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Key Concepts

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

Probability
Probability is a measure of how likely an event is to occur, expressed as a number between 0 and 1. A probability of 0 indicates that an event will not occur, while a probability of 1 indicates certainty that it will occur.
In the context of a normal distribution, probability helps us determine how likely it is that a randomly chosen data point from the distribution falls within a certain range.
For example, when given a problem like \(P(X>x)=0.5\), we are interested in finding the event that something happens more than half of the time. This is directly related to recognizing that 50% of the data in a normal distribution lies above the mean.
  • Key Concept: Probability values relate directly to areas under the normal distribution curve.
  • Understanding probabilities allows us to make informed predictions about data points.
Z-score
The Z-score is a statistical measurement that describes a data point's relationship to the mean of a group of points. It essentially tells us how many standard deviations a particular value is from the mean.
The formula for calculating a Z-score is: \[ Z = \frac{X - \mu}{\sigma} \]
Where \(X\) is the value in question, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For example, if you want to know the z-score that corresponds to \(P(X < x) = 0.05\), you would look up \(0.05\) in a z-table (or use a calculator) to find the associated z-value. This value tells you how far away \(X\) is from the mean in terms of standard deviations.
  • A positive Z-score indicates the data point is above the mean.
  • A negative Z-score indicates it is below the mean.
  • Z-scores are pivotal for standardizing data and comparing different sets of data.
Mean
The mean is the average value of a set of data, found by summing all data points and then dividing by the number of points. In a normal distribution, the mean is located at the center of the distribution and serves as a measure of central tendency.
In problems dealing with normally distributed random variables, knowing the mean is crucial.
For instance, in our exercise where the mean \(\mu = 5\), it acts as a reference point for all further calculations. The mean helps to interpret the probability values and their placement on the normal distribution curve.
  • The mean provides an anchoring point for calculating probabilities, especially when half of the distribution falls above and below it (i.e., \(P(X>x)=0.5\) implies \(x = 5\)).
Standard Deviation
Standard deviation, denoted by \(\sigma\), is a statistic that measures the dispersion of a dataset relative to its mean. It provides insight into how much individual data points deviate from the mean, helping to understand the spread of the data.
If the standard deviation is large, it means the data points are widely spread out over a large range of values. Conversely, a small standard deviation indicates that the data points are close to the mean.
In normal distribution problems like our exercise, the standard deviation is essential for calculating z-scores and finding probabilities. For example, knowing \(\sigma = 4\) lets us determine how far data points like \(X\) or \(-X\) are from \(\mu\) on the distribution curve.
  • Standard deviation aids in understanding variability within the dataset.
  • It is a key factor in calculating z-scores and corresponding probabilities.

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