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Chapter 7: Problem 55

Explain why \(P(X \leq 220)\) should be reported as \(>0.9999\) if \(X\) is a normal random variable with mean 100 and standard deviation \(15 .\)

Short Answer

Expert verified
A Z-score of 8 indicates that 220 is far right in a normal distribution, making \(P(X \leq 220)\) essentially 1, reported as \(>0.9999\).

Step by step solution

01

Identify the Distribution Parameters

The given problem states that the random variable \(X\) is normally distributed with a mean (\(\mu\)) of 100 and a standard deviation (\(\sigma\)) of 15.
02

Determine the Z-Score for the Given Value

To find \(P(X \leq 220)\), first convert the value 220 to a standard normal variable (Z) using the formula: \[ Z = \frac{X - \mu}{\sigma} \] Substituting the given values, we get: \[ Z = \frac{220 - 100}{15} = \frac{120}{15} = 8 \]
03

Interpret the Z-Score

A Z-score of 8 is extremely high. In the standard normal distribution, this suggests that the value 220 is very far out in the right tail of the distribution.
04

Find the Probability from Z-Table

Using the Z-table, or standard normal distribution tables, probabilities for Z-scores greater than 3.9 are very close to 1.
05

Report the Probability

Since \(Z = 8\) is much larger than 3.9, the probability \(P(X \leq 220)\) must be extremely close to 1. Therefore, it is appropriate to report \(P(X \leq 220)\) as \(>0.9999\).

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Key Concepts

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

z-score calculation
To understand why we need to use the z-score in calculating probabilities for a normal distribution, we must start with the basics. The z-score, also known as the standard score, tells us how many standard deviations an element is from the mean. This metric helps to standardize different data points, making it easier to compare them.

We use the formula: \[ Z = \frac{X - \text{mean}}{\text{standard deviation}} \] Here, - \(X\) is the value from the dataset (in our problem, it's 220), - The mean is 100, - The standard deviation is 15.

By plugging these values into the formula, we get:\[ Z = \frac{220 - 100}{15} = 8 \] This means that 220 is 8 standard deviations above the mean.

Calculating the z-score is crucial because it allows us to interpret the normal distribution in a standardized way.
standard normal distribution
The standard normal distribution is a special type of normal distribution that has a mean of 0 and a standard deviation of 1. When we calculate z-scores, we transform our data so that it fits into this standard normal distribution.

This transformation helps because: - All z-scores can be compared regardless of the original meaning or units of the data.- It simplifies the process of finding probabilities using standard normal distribution tables or Z-tables.

In the context of our problem, transforming the value 220 to a z-score of 8 situates it on this standard normal curve. This standardization makes it easier to find associated probabilities.
interpreting z-scores
Interpreting z-scores effectively is crucial for understanding and explaining probability in a normal distribution. A z-score tells us how rare or common a value is when compared to the mean of a normal distribution.

In our exercise, a z-score of 8 is extraordinarily high. Standard normal distribution tables (Z-tables) typically only go up to around 3.9, where the probability is very close to 1. Therefore, a z-score of 8 implies that the probability of finding a value below 220 is far over 99.99%, hence why we report it as >0.9999.

Understanding z-scores and their significance is essential for accurately interpreting data in the realm of statistics.

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