Problem 24 - Calculus and its applications

Question

Let \(x\) be a continuous random variable with a standard normal distribution. Using Table A, find each of the following. a) \(P(-1 \leq x \leq 1)\) b) What percentage of the area is from -1 to \(1 ?\)

Step-by-Step Solution

Answer

a) 0.6826, b) 68.26%

Step 1: Understand the Distribution

A standard normal distribution means that the random variable \(x\) follows a normal distribution with a mean of \(0\) and a standard deviation of \(1\). This is also referred to as the \(Z\)-distribution.

Step 2: Use the Z-Table for Cumulative Probability

Find the cumulative probability for \(Z = 1\) and \(Z = -1\) using a standard normal (\(Z\)) table. The table generally provides \(P(Z  - \text{value})\).- For \(Z = 1\), the table gives \(P(Z \leq 1) \approx 0.8413\).- For \(Z = -1\), it gives \(P(Z \leq -1) \approx 0.1587\).

Step 3: Calculate Probability Between Two Points

The problem \(P(-1 \leq x \leq 1)\) requires finding the probability that \(x\) is between \(-1\) and \(1\). This can be calculated as the difference between the cumulative probabilities:\[ P(-1 \leq x \leq 1) = P(x \leq 1) - P(x \leq -1) = 0.8413 - 0.1587 = 0.6826 \]

Step 4: Convert Probability to Percentage

To express this probability as a percentage, multiply by 100%. Therefore:\[ 0.6826 \times 100\% = 68.26\% \]Thus, 68.26% of the area under the standard normal curve is between \(Z = -1\) and \(Z = 1\).

Key Concepts

Cumulative Probability

Cumulative probability is a key concept when dealing with the standard normal distribution. It refers to the probability that a random variable, such as \(Z\), is less than or equal to a specific value. This type of probability helps us to understand how much of the total area under the probability curve is covered up to a certain point.
For example, when we use the \(Z\)-table to find \(P(Z \leq 1)\), we are calculating the cumulative probability up to \(Z = 1\). According to the table, this cumulative probability is approximately \(0.8413\) or \(84.13\%\), indicating that 84.13% of the area under the curve is to the left of \(Z = 1\).
Experienced statisticians often use cumulative probability to determine the likelihood of events occurring within certain boundaries, especially when analyzing data that follows a normal distribution.

Z-Distribution

The Z-distribution, also known as the standard normal distribution, plays a vital role in statistics, particularly in the context of hypothesis testing and standardization of data. A key feature of the Z-distribution is its symmetry around the mean, which is 0. Furthermore, it has a standard deviation of 1.
What makes the Z-distribution unique is its ability to convert raw scores into standardized scores (or z-scores). This transformation allows us to compare scores from different normal distributions with varied means and standard deviations by expressing them on a common scale.
Additionally, the bell-shaped curve of the Z-distribution represents the density of the variables, where approximately \(68\%\) of the data falls within one standard deviation from the mean, \(95\%\) within two, and \(99.7\%\) within three. Through this, the Z-distribution standardizes data, enabling effective comparisons and analyses.

Z-Table

The Z-table is an essential tool used to find cumulative probabilities associated with the standard normal distribution. It provides a quick reference to understand the cumulative probability for any z-score, allowing users to easily determine probabilities over intervals.
When you look up a value in the Z-table, you will typically find the cumulative probability that \(Z\) is less than or equal to your chosen value. For example, if you want to find \(P(Z \leq -1)\), the Z-table shows approximately \(0.1587\). Similarly, for \(Z = 1\), it shows \(P(Z \leq 1) \approx 0.8413\).
Utilizing the Z-table helps in answering questions regarding the likelihood of a random variable falling within a certain range, as seen in the textbook problem where the probability \(P(-1 \leq x \leq 1)\) was calculated by accessing the cumulative probabilities from the Z-table and subtracting them. This resource is crucial for efficient problem-solving in statistics.