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Magazine Chapter 4: Problem 68
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 鈮� 6.71; (c) x = 10; (d) x 鈮� 3.92; (e) x 鈮� 5.152.
Step by step solution
01
Understanding the Normal Distribution
The normal distribution is symmetric around the mean. Here, the mean \( \mu = 10 \) and the standard deviation \( \sigma = 2 \). The probability \( P(X>x) \) represents the area to the right of \( x \) on the normal distribution curve.
02
Solving (a) 鈥� Finding x for P(X>x)=0.5
Since the distribution is symmetric about the mean, \( P(X > 10) = 0.5 \) because 10 is the mean. Therefore, the value of \( x \) that satisfies \( P(X > x) = 0.5 \) is \( x = 10 \).
03
Solving (b) 鈥� Finding x for P(X>x)=0.95
For \( P(X > x) = 0.95 \), we need the 5th percentile of the distribution because 0.95 probability to the right corresponds to 0.05 probability to the left. This can be calculated using standard normal distribution tables or a calculator. The 5th percentile corresponds to approximately 1.645 standard deviations below the mean: \[ x = 10 - 1.645 imes 2 = 10 - 3.29 = 6.71. \]
04
Solving (c) 鈥� Finding x for P(x
Since \( P(x < X < 10) = 0 \), and the distribution is normal and continuous, this only happens when \( x = 10 \). Hence, there is no area under the curve if \( x \) is equal to the upper bound, \( 10 \). Thus, \( x = 10 \).
05
Solving (d) 鈥� Finding x for P(-x
The expression \( P(-x < X-10 < x) = 0.95 \) can be rewritten as \( P(10-x < X < 10+x) = 0.95 \). This implies that \( x \) corresponds to the distance such that it covers 95% of the distribution around the mean. For a normal distribution, 95% corresponds approximately to \( \pm 1.96 \sigma \):\[ x = 1.96 imes 2 = 3.92. \]
06
Solving (e) 鈥� Finding x for P(-x
The expression \( P(-x < X-10 < x) = 0.99 \) corresponds to \( P(10-x < X < 10+x) = 0.99 \). For a normal distribution, 99% is approximately under \( \pm 2.576 \sigma \):\[ x = 2.576 imes 2 = 5.152. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability
Probability is a way to quantify the likelihood of a particular event occurring. In the context of a normal distribution, we often answer questions about finding the probability that a value falls above, below, or between specific points in the distribution.
The normal distribution is symmetric, meaning probabilities can be calculated accurately using the standard normal distribution table or software. For any given random variable, like our exercise with mean \( \mu = 10 \) and standard deviation \( \sigma = 2 \), we calculate probability to find the area under the curve.
The normal distribution is symmetric, meaning probabilities can be calculated accurately using the standard normal distribution table or software. For any given random variable, like our exercise with mean \( \mu = 10 \) and standard deviation \( \sigma = 2 \), we calculate probability to find the area under the curve.
- For example, \( P(X>x)=0.5 \) implies that the point \( x \) is the mean, since exactly half of the data falls to the right.
- The probability \( P(X>x) \) can also find values corresponding to certain cumulative percentages.
Standard Deviation
Standard deviation is a critical measure in statistics that quantifies how spread out the data are in a distribution. Within a normal distribution, it signifies the average distance of each data point from the mean.
In our problem, the standard deviation \( \sigma \) is 2, which means most values fall within 2 units of the mean \( \mu = 10 \).
Understanding standard deviation is key for estimating the likelihood of events and interpreting statistical data.
In our problem, the standard deviation \( \sigma \) is 2, which means most values fall within 2 units of the mean \( \mu = 10 \).
- This indicates that any value within this range is considered close to the average.
- Standard deviations are used to identify percentiles and probabilities within the distribution, such as in question \((b)\) where we use the concept to find values corresponding to the 5th percentile.
Understanding standard deviation is key for estimating the likelihood of events and interpreting statistical data.
Percentile
Percentiles in a distribution indicate the value below which a given percentage of observations fall. For example, the 5th percentile will be the point below which 5% of observations are found.
Percentiles are widely used in normal distributions to find critical values. In the exercise, determining \( P(X>x)=0.95 \) requires knowledge of the 5th percentile because 0.95 probability to the right translates to 0.05 to the left, making it similar to positioning at the 5th percentile.
By using percentiles, we effectively identify critical points in the distribution that have specific probabilities associated with them.
Percentiles are widely used in normal distributions to find critical values. In the exercise, determining \( P(X>x)=0.95 \) requires knowledge of the 5th percentile because 0.95 probability to the right translates to 0.05 to the left, making it similar to positioning at the 5th percentile.
- Percentile ranks allow comparisons to be made about the relative standing of a value within a data set.
- They are crucial for boundary cases, like when 95% or 99% of observations are within specific limits.
By using percentiles, we effectively identify critical points in the distribution that have specific probabilities associated with them.
Continuous Distribution
A continuous distribution is one in which data points fall within a continuous range. Unlike discrete distributions, where data points are distinct and separate, continuous distributions cover every value within an interval.
The normal distribution is a classic example of a continuous distribution. It is characterized by its bell-shaped curve, symmetric about the mean. In our problem, this property allows for evaluating frequencies of any possible outcome.
A deep understanding of continuous distributions allows for better data handling and more informed predictions.
The normal distribution is a classic example of a continuous distribution. It is characterized by its bell-shaped curve, symmetric about the mean. In our problem, this property allows for evaluating frequencies of any possible outcome.
- Solutions often involve calculating areas under the curve, which are infinite because of its continuous nature.
- Using properties of continuity, we see that expressions such as \( P(x < X < 10) = 0 \) imply that x must equate to the upper boundary.
A deep understanding of continuous distributions allows for better data handling and more informed predictions.
