Question

A normal random variable \(x\) has mean \(\mu=5\) and \(\sigma=2\). Find the following probabilities of these \(x\) -values: a. \(1.27.5\) c. \(x \leq 0\)

Short answer

Based on the given normal distribution with a mean of 5 and a standard deviation of 2, calculate the probabilities for the following ranges of x-values: 1. \(1.2 < x < 10\) 2. \(x > 7.5\) 3. \(x \leq 0\) The probabilities for each range are as follows: 1. The probability that \(1.2 < x < 10\) is approximately \(0.9651\). 2. The probability that \(x > 7.5\) is approximately \(0.2119\). 3. The probability that \(x \leq 0\) is approximately \(0.0062\).

Step-by-step solution

Convert x-values to z-scores

To find the corresponding z-scores for each range of x-values, use the z-score formula: \(z = \frac{x-\mu}{\sigma}\). For each part, we'll plug in the given x-values, mean (\(\mu = 5\)), and standard deviation (\(\sigma = 2\)) into the formula. a. \(1.2 < x < 10\) For the lower limit, \(x_L = 1.2\):\ \(z_L = \frac{1.2-5}{2} = -1.9\) For the upper limit, \(x_U = 10\):\ \(z_U = \frac{10-5}{2} = 2.5\) b. \(x > 7.5\) \(x_L = 7.5\):\ \(z_L = \frac{7.5-5}{2} = 1.25\) c. \(x \leq 0\) \(x_U = 0\):\ \(z_U = \frac{0-5}{2} = -2.5\)

Find probabilities using the Z-table

Now, using the Z-table, we will find the probabilities corresponding to the calculated z-scores. a. \(1.2 < x < 10\) \(P(1.2 < x < 10) = P(-1.9 < z < 2.5) = P(z < 2.5) - P(z < -1.9)\)\ Approximate values from the Z-table are:\ \(P(z < 2.5) = 0.9938\)\ \(P(z < -1.9) = 0.0287\) So, the probability is:\ \(P(1.2 < x < 10) = 0.9938 - 0.0287 = 0.9651\) b. \(x > 7.5\) \(P(x > 7.5) = 1 - P(x \leq 7.5) = 1 - P(z \leq 1.25)\)\ Approximate value from the Z-table is:\ \(P(z \leq 1.25) = 0.7881\) So, the probability is:\ \(P(x > 7.5) = 1 - 0.7881 = 0.2119\) c. \(x \leq 0\) \(P(x \leq 0) = P(z \leq -2.5)\)\ Approximate value from the Z-table is:\ \(P(z \leq -2.5) = 0.0062\) So, the probability is:\ \(P(x \leq 0) = 0.0062\)

Final Results

The probabilities that we found for each part are: a. \(P(1.2 < x < 10) = 0.9651\) b. \(P(x > 7.5) = 0.2119\) c. \(P(x \leq 0) = 0.0062\)

Key concepts

Z-Score Calculation

Understanding Z-score calculation is crucial when working with normal distribution probabilities. A Z-score represents the number of standard deviations a data point is from the mean. The formula for calculating a Z-score is given by:
\[ z = \frac{x - \mu}{\sigma} \]
Here, x is the value from the dataset, \(\mu\) is the mean of the dataset, and \(\sigma\) is the standard deviation. In our exercise where the mean is 5 and the standard deviation is 2, converting a raw score of 7.5 to a Z-score involves substituting into the formula: \[ z = \frac{7.5 - 5}{2} = 1.25 \]
This Z-score tells us that 7.5 is 1.25 standard deviations above the mean. To enhance understanding, visualize the normal distribution curve. The mean (\(\mu\)) lies at the center, and each Z-score specifies how far away from the center a score lies, either to the left (negative Z-score) or to the right (positive Z-score). Understanding Z-scores helps in comparing different data points from different normal distributions.
To assist in grasping this concept, imagine different values of x and calculate their corresponding Z-scores. This hands-on approach makes it easier to appreciate how the Z-score reflects the position of a value in relation to the mean.

Z-Table

Once we have the Z-scores, we proceed to use a Z-table to find probabilities. A Z-table, also known as the standard normal distribution table, lists the cumulative probabilities of Z-scores. The table typically shows the probability of a Z-score being less than or equal to a certain value.
For example, to find the probability that x is more than 7.5 in our exercise, we first calculated the Z-score for 7.5 which is 1.25. Using the Z-table: \[ P(z \leq 1.25) = 0.7881 \]
This probability suggests that approximately 78.81% of the values lie below a Z-score of 1.25. To find the probability of x being more than 7.5, we calculate: \[ P(x > 7.5) = 1 - P(z \leq 1.25) = 1 - 0.7881 = 0.2119 \]
Understanding how to read the Z-table and apply it directly to Z-score results is a valuable skill. It enables students to calculate the likelihood of events occurring within a given normal distribution. Always remember, for greater than scenarios, subtract the table value from one, and for less than scenarios, use the table value directly. Practice by looking up different Z-scores in the Z-table to solidify this critical skill.

Standard Deviation

Now, let's illuminate the concept of standard deviation (\(\sigma\)). It's a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.
In the context of our problem, the standard deviation is 2, which means that most values lie within 2 units of the mean on either side. It's important to remember that the standard deviation can significantly alter the Z-score because it denominates the difference between a value and the mean. Thus, if the standard deviation is large, even a substantial difference from the mean could result in a small Z-score, suggesting that such a difference is not uncommon. Conversely, a small standard deviation signals that the same difference is more noteworthy.
To cement this concept in practice, examine different datasets with varying standard deviations and observe how they affect the distribution of data. Also, try calculating the standard deviation for various sets of data to appreciate how it summarizes the dataset's spread. This understanding underpins many aspects of statistical analysis and applications in real-world scenarios.