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Assume that \(X\) is normally distributed with a mean of 10 and a standard deviation of \(2 .\) 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) x = 10; b) x = 6.71; c) x = 10; d) x = 3.92; e) x = 5.15.

Step by step solution

01

Understand the Distribution

First, we note that the random variable \(X\) is normally distributed with a mean \(\mu = 10\) and a standard deviation \(\sigma = 2\). This is denoted as \(X \sim N(10, 4)\).
02

Problem a - Find x for P(X > x) = 0.5

Since the normal distribution is symmetric about its mean, the point where \(P(X > x) = 0.5\) is the mean \(X = 10\). So, for part (a), the value of \(x\) is 10.
03

Problem b - Find x for P(X > x) = 0.95

Using a standard normal distribution table or calculator, we need to find \(z\) such that \(P(Z > z) = 0.95\). Then convert this z-score to the actual value using the formula \(x = \mu + z\sigma\), where \(\mu = 10\) and \(\sigma = 2\). The z-score for 0.05 (since \(1 - 0.95 = 0.05\) for left-tail probability) is approximately -1.645. Thus, \(x = 10 + (-1.645)\times 2 = 6.71\).
04

Problem c - Find x for P(x < X < 10) = 0

This probability is zero when \(x \geq 10\). Since no values are less than 10 if \(X\) is always 10 or more, \(x = 10\).
05

Problem d - Find x for P(-x < X-10 < x) = 0.95

Convert to the standard normal variable, \(P(-x/2 < Z < x/2) = 0.95\). Find \(z\) such that \(P(-z < Z < z) = 0.95\), which gives \(z \approx 1.96\). Solve \(-x/2 = -1.96\) and \(x/2 = 1.96\) to get \(x = 3.92\).
06

Problem e - Find x for P(-x < X-10 < x) = 0.99

Similarly, \(P(-x/2 < Z < x/2) = 0.99\). The z-score that leaves 0.01 tails in the total probability is approximately \(z \approx 2.576\). Solve \(-x/2 = -2.576\) and \(x/2 = 2.576\) to get \(x = 5.15\).

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Key Concepts

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

Probability
When we talk about probability in the context of normal distribution, we're interested in the likelihood of a certain event occurring. In the normal distribution, this often relates to the chances that a random variable falls within certain ranges of values.
For example, when you see expressions like \(P(X > x)\), it simply means the probability that the random variable \(X\) is greater than some value \(x\). This can help answer questions such as "What are the chances a test score is greater than 85?" for a normally distributed test.

Probability in a normal distribution is fully defined by the area under the bell curve. The total area under the curve is always equal to 1 (or 100%). This is because the probability of all possible outcomes of a random variable occurring is always equal to 1.
  • If the area to the left of \(x\) is 0.5, it indicates that there's a 50% chance.
  • The right half is also 0.5, showing another 50% chance.
Thus, probabilities allow us to make quantitative assessments about the likelihood of certain outcomes.
Mean and Standard Deviation
The mean and the standard deviation are key components of a normal distribution. Understanding these concepts helps in comprehending how data is spread across the distribution.

The **mean** (denoted \(\mu\)) is the average of all the values in a data set. It's the center point of a normal distribution. For instance, in our case, the mean is 10, which indicates most of the values cluster around this point.
The **standard deviation** (denoted \(\sigma\)) measures how spread out the numbers are across the distribution. A small standard deviation indicates that the data points tend to be close to the mean, while a large standard deviation shows a wide range of values.
  • In the context of a normal distribution problem, the standard deviation tells us how much the curve is spread out from the mean.
  • For our example, a standard deviation of 2 implies that most values fall within 2 units of the mean (either direction).
These metrics help shape the curve of the distribution, determining both its center and spread.
Z-score
The Z-score is a critical component in working with normal distributions, as it helps translate raw data into a standard form that is easy to use. The Z-score tells you how many standard deviations an element is from the mean.

To find a Z-score, you use the formula: \[Z = \frac{(X - \mu)}{\sigma}\]where \(X\) is the value in question, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
  • A Z-score of 0 indicates the value is exactly at the mean.
  • A positive Z-score means the value is above the mean.
  • A negative Z-score means it is below the mean.
In exercises, calculating the Z-score allows us to use standard normal distribution tables to find probabilities and percentiles. For instance, if you want to know how likely it is for \(X\) to be a certain value or less, you'd calculate its Z-score first and use standard tables.
Standard Normal Distribution
The standard normal distribution is a special instance of the normal distribution. It's a distribution with a mean of 0 and a standard deviation of 1. It is often used for calculating probabilities and understanding distributions because it's a 'universal' form of the normal distribution.

Why do we use the standard normal distribution? Primarily, it simplifies calculations. Once a value \(X\) is converted to a Z-score, you can easily look up cumulative probabilities using standard Z-tables.
  • When we mention the 'table' in statistics, we are referring to these Z-tables that map every Z-score to a percentile.
  • This transformation makes any normal distribution easier to work with since any normal distribution can be related back to this standardized form.
Thus, understanding how to use the standard normal distribution, and convert normal variables into standard normals using the Z-score, is critical for applying statistical concepts effectively.

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