Chapter 3: Problem 4
Let \(X\) be \(N\left(\mu, \sigma^{2}\right)\) so that \(P(X<89)=0.90\) and \(P(X<94)=0.95 .\) Find \(\mu\) and \(\sigma^{2}\).
Short Answer
Expert verified
\(\mu \approx 86.34\) and \(\sigma^{2}\approx 19.31\)
Step by step solution
01
Calculate Z-scores
You begin by calculating the Z-score for each probability value using the Z-tables or a Z-score calculator. The Z-score is calculated as follows: the Z score for \(P(X<89)=0.90\) is approximately \(1.28\) and the Z-score for \(P(X<94) = 0.95\) is approximately \(1.64\). You find these numbers in the Z-table that corresponds to the probabilities given.
02
Set up equations using Z-score formula
Next, you set up two equations using the formula for the Z score, which is \(Z = \frac{X - \mu}{\sigma}\). From the equations \(Z = \frac{89 - \mu}{\sigma} = 1.28\) and \(Z = \frac{94 - \mu}{\sigma} = 1.64\), we can now form two equations: \(89 - \mu = 1.28\sigma\) and \(94 - \mu = 1.64\sigma\).
03
Solve the system of equations
Now we have a system of two equations with two unknowns \(\mu\) and \(\sigma\). To find the values of \(\mu\) and \(\sigma\), subtract the two equations: \(94 -89 = 1.64\sigma - 1.28\sigma\), which simplifies into \(5=0.36\sigma\). Solving for \(\sigma\), gives us \(\sigma = \frac{5}{0.36}\). Plug the \(\sigma\) value obtained into one of the equations to find \(\mu = 89 - 1.28\sigma\).
04
Check our solution
You check to make sure that the solutions obtained truly fit the conditions given in the original problem. That is \(P(X<89)=0.90\) and \(P(X<94)=0.95\) in our case. This confirms the correctness of the result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Z-Score
A z-score is a statistical measure that tells us how many standard deviations an element is from the mean of its distribution. Imagine it as a way to put a specific number within a data set into perspective with the entire set. It's commonly used in statistics to determine how normal an observation is, compared to a standard distribution like the normal distribution.
In our exercise, we used z-scores to find probabilities under a normal distribution. You calculate a z-score with the formula: \[ Z = \frac{X - \mu}{\sigma} \] where:
In our exercise, we used z-scores to find probabilities under a normal distribution. You calculate a z-score with the formula: \[ Z = \frac{X - \mu}{\sigma} \] where:
- \(X\) is the value in the distribution,
- \(\mu\) is the mean, and
- \(\sigma\) is the standard deviation.
Mean and Variance Calculation
The mean and variance are fundamental concepts in statistics that describe the central tendency and spread of a data set, respectively. The mean, often referred to as the average, is a measure that summarizes an entire dataset with a single value indicating the center of the data.
When it comes to variance, it's a measure of how much the values in a dataset differ from the mean. The standard deviation is simply the square root of the variance and provides a direct measure of how much the values deviate from the mean.
In the problem above, to find the mean \(\mu\) and variance \(\sigma^2\), we set up equations using z-scores. For example,
When it comes to variance, it's a measure of how much the values in a dataset differ from the mean. The standard deviation is simply the square root of the variance and provides a direct measure of how much the values deviate from the mean.
In the problem above, to find the mean \(\mu\) and variance \(\sigma^2\), we set up equations using z-scores. For example,
- Given \(Z = \frac{89 - \mu}{\sigma} = 1.28\), we can rearrange it to find \(\mu\), the mean.
- Similarly, \(Z = \frac{94 - \mu}{\sigma} = 1.64\) helps us establish another equation for \(\mu\) and \(\sigma\).
Probability
Probability is the measure of likelihood that a particular event will occur. It's the foundation upon which many statistical concepts are built, including the normal distribution, which shows how probabilities are distributed under a symmetric bell curve.
In the context of a normal distribution, probability tells us how likely it is for a value or a range of values to appear in our dataset. For example, in the exercise above, given probabilities are \(P(X < 89) = 0.90\) and \(P(X < 94) = 0.95\), which indicate that 90% and 95% of data fall below 89 and 94, respectively.
This is crucial because:
In the context of a normal distribution, probability tells us how likely it is for a value or a range of values to appear in our dataset. For example, in the exercise above, given probabilities are \(P(X < 89) = 0.90\) and \(P(X < 94) = 0.95\), which indicate that 90% and 95% of data fall below 89 and 94, respectively.
This is crucial because:
- It allows statisticians to make predictions about data.
- It helps in decision-making processes based on likely outcomes.
- It informs us about the relative standing of a value within a dataset.