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Chapter 17: Problem 17

\(X \sim N\left(\mu, \sigma^{2}\right), P(x>162)=0.122\) and \(P(x<56)=0.0276 .\) Find \(\mu\) and \(\sigma\)

Short Answer

Expert verified
\(\mu = 121.94\), \(\sigma \approx 34.53\)

Step by step solution

01

Understand the Problem

We are given a normally distributed random variable \(X\) with an unknown mean \(\mu\) and standard deviation \(\sigma\). We know that \(P(X > 162) = 0.122\) and \(P(X < 56) = 0.0276\). We need to find \(\mu\) and \(\sigma\).
02

Calculate Z-scores

Using a standard normal distribution table, we find the Z-scores for the given probabilities. For \(P(X > 162) = 0.122\), the corresponding Z-score is the value such that \(P(Z > z) = 0.122\). We find \(z_1 \approx 1.16\). Similarly, \(P(X < 56) = 0.0276\) corresponds to \(z_2 \approx -1.91\).
03

Set up Equations

The given Z-scores for a normal distribution relate to the mean and standard deviation by the formulas: \(z_1 = \frac{162 - \mu}{\sigma}\) and \(z_2 = \frac{56 - \mu}{\sigma}\). Substitute the Z-scores found: \(1.16 = \frac{162 - \mu}{\sigma}\) and \(-1.91 = \frac{56 - \mu}{\sigma}\).
04

Solve Simultaneous Equations

Take the equations: \(1.16\sigma = 162 - \mu\) and \(-1.91\sigma = 56 - \mu\). Solve for \(\mu\) and \(\sigma\): \(\mu = 162 - 1.16\sigma\) and \(\mu = 56 + 1.91\sigma\). Set the equations equal: \(162 - 1.16\sigma = 56 + 1.91\sigma\). Simplify and solve for \(\sigma\): \(106 = 3.07\sigma\) yields \(\sigma \approx 34.53\).
05

Solve for \(\mu\)

Substitute \(\sigma = 34.53\) back into any of the expressions for \(\mu\), for example: \(\mu = 162 - 1.16 \times 34.53\). Calculate \(\mu\): \(\mu \approx 121.94\).
06

Verify Solutions

Verify that \(\mu = 121.94\) and \(\sigma = 34.53\) satisfy both probability conditions. Substitute back into the equations: \(z_1 = \frac{162 - 121.94}{34.53} \approx 1.16\) and \(z_2 = \frac{56 - 121.94}{34.53} \approx -1.91\). This verifies the correctness.

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Key Concepts

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

Z-scores
A Z-score is a way to describe a data point's location within a normal distribution. It tells us how many standard deviations away a data point is from the mean. Z-scores are vital because they standardize different data points, giving a common frame for comparison against a standard normal distribution with a mean of 0 and a standard deviation of 1.
  • To find a Z-score, you subtract the mean from the data point and then divide by the standard deviation.
  • The formula is: \( z = \frac{X - \mu}{\sigma} \).
Z-scores can be positive or negative. A positive Z-score means the data point is above the mean, while a negative one indicates it is below. By transforming data into Z-scores, it's easier to determine probabilities and make meaningful comparisons.
Mean
The mean is the average value in a data set and a crucial component of the normal distribution. It signifies the center of the distribution and provides insight into where data values typically cluster. Calculating the mean involves summing all data points and dividing by the number of points.
  • The formula for the mean of a continuous random variable in a normal distribution is \( \mu \).
  • In a normal distribution curve, the mean is located at the peak.
Understand that the mean is not always a reliable measure of central tendency if the data is heavily skewed. However, in normal distribution, the mean is equal to the median and mode, making it particularly meaningful.
Standard Deviation
Standard deviation measures the spread or variation of the data around the mean. In a normal distribution, standard deviation determines the width of the curve. A larger standard deviation means the data points are more spread out, resulting in a flatter curve, whereas a smaller standard deviation indicates the data points are closely packed around the mean.
  • The formula for standard deviation in a normal distribution is \( \sigma \).
  • It's calculated as the square root of the variance, \( \sigma = \sqrt{\sigma^2} \).
Standard deviation helps us understand the variability in the data. In context, it allows prediction of where data points are likely to fall and contributes to forming Z-scores, which standardize differences across different datasets. Understanding its role is vital for accurately interpreting any normal distribution.
Probability
Probability in the context of normal distribution often involves calculating the likelihood of a data point falling within a certain range. For a normal distribution, probabilities are typically looked up using a Z-table which helps decipher the area under the curve of a standard normal distribution.
  • Knowing probabilities associated with Z-scores helps assess the likelihood of different outcomes.
  • In this exercise, probabilities guide the identification of Z-scores for extreme values.
Visualizing probability helps in interpreting data in real-world contexts, like determining how likely a particular event is. The areas under different sections of the normal curve represent cumulative probabilities, which are essential for statistical interpretation. Mastery of these concepts allows for meaningful applications in data sciences and beyond.

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