Problem 15 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

If the random variable \(X\) is normally distributed with mean \(\mu=7\) and standard deviation \(\sigma=2\), what is \(P(X \geq 9)\) ?

Step-by-Step Solution

Answer

P(X \geq 9) = 0.1587.

Step 1: Convert to Standard Normal Variable

First, convert the normal variable to a standard normal variable (Z) using the formula: \[ Z = \frac{X - \mu}{\sigma} \] Here, \(X = 9\), \(\mu = 7\) and \(\sigma = 2\). Substitute these values into the formula to get: \[ Z = \frac{9 - 7}{2} = 1 \]

Step 2: Find Z-Score Probability

Next, find the probability corresponding to \(Z=1\) using standard normal distribution tables or a calculator. Typically, \(P(Z < 1)\) is found to be 0.8413.

Step 3: Calculate the Desired Probability

Since we need \(P(X \geq 9)\), which in terms of Z is \(P(Z \geq 1)\), recognize that \(P(Z \geq 1) = 1 - P(Z < 1)\). Using the value from Step 2, \[ P(Z \geq 1) = 1 - 0.8413 = 0.1587 \]

Key Concepts

standard normal variable

In statistics, a standard normal variable (denoted as Z) is essential for simplifying calculations involving normally distributed data. Normal distributions have variables with a specific mean and standard deviation. By converting these variables to standard normal variables, we can use just one standard normal distribution for all cases. This process is called standardization. Using the formula: \[ Z = \frac{X - \mu}{\sigma} \]we can convert any normal variable X into Z. Here, X is the original variable value, \mu is the mean, and \sigma is the standard deviation. This conversion helps when comparing different datasets or determining probabilities involving normal distributions.

Z-score

The Z-score represents how many standard deviations a data point is from its mean. It's a critical concept in statistics for comparing data. By determining Z, you can easily understand where a specific value lies within a normal distribution. The formula is simple: \[ Z = \frac{X - \mu}{\sigma} \].
For example, to find the Z-score for X = 9 when the mean \mu is 7 and \sigma is 2, you get: \[ Z = \frac{9 - 7}{2} = 1 \].
This Z-score tells you that 9 is exactly one standard deviation above the mean. Understanding Z-scores simplifies working with probability tables and helps in comparing different datasets.

probability calculation

Once you've found the Z-score, you can calculate probabilities. For any given Z value, standard normal distribution tables or online calculators give the probability of a variable being less than the Z value. This is denoted as \( P(Z < z) \).
Looking at standard normal distribution tables, the probability for \( Z = 1 \) is 0.8413. This means there's an 84.13% chance that a value is less than 1 standard deviation above the mean.
To find the probability of a value being greater than a Z value (\(P(Z \geq z) \)), subtract the table value from 1:
\[ P(Z \geq 1) = 1 - P(Z < 1) = 1 - 0.8413 = 0.1587 \].
Thus, for our example, there's a 15.87% chance that a value is greater than one standard deviation above the mean.

mean and standard deviation

The mean \mu and standard deviation \sigma are fundamental concepts in statistics that describe a dataset. The mean is the average value of a dataset and gives a central location of the data.
Standard deviation describes the spread of data around the mean, indicating how dispersed the values are. A smaller standard deviation means the data points are close to the mean, while a larger standard deviation indicates more variability.
In our example, the mean \mu is 7, and the standard deviation \sigma is 2. To find the probability of \( X \geq 9 \) where \mu = 7 and \sigma = 2, convert X to a Z-score and find the corresponding probability. This conversion and probability calculation are essential in understanding where the data point lies within the distribution.