Question

A normal random variable \(x\) has mean 35 and standard deviation \(10 .\) Find a value of \(x\) that has area .01 to its right. This is the 99 th percentile of this normal distribution.

Short answer

Answer: The value of x corresponding to the 99th percentile is approximately 58.3.

Step-by-step solution

Identify given values

We are given the mean (μ) as 35 and the standard deviation (σ) as 10. We need to find the value of x that corresponds to the 99th percentile.

Use a standard normal distribution table

We need to use a standard normal distribution table (z-table) to find the z-score that corresponds to the 99th percentile (0.99). Looking up 0.99 in the z-table, we find a z-score of 2.33.

Decode the z-score to find x

To find the value of x corresponding to the z-score we found, we can use the z-score formula which is: \(z = \frac{x - \mu}{\sigma}\) We already know the mean (μ) is 35, the standard deviation (σ) is 10, and the z-score is 2.33. Now we can plug in the values and solve for x: \(2.33 = \frac{x - 35}{10}\)

Solve for x

Finally, we can solve for x: \(x - 35 = 2.33 \cdot 10\) \(x - 35 = 23.3\) \(x = 58.3\) So, the value of x that has an area of .01 to its right (the 99th percentile) is approximately 58.3.

Key concepts

Percentile

A percentile is a measure used in statistics to indicate the value below which a given percentage of observations in a group falls. For example, the 99th percentile refers to the value below which 99% of the observations lie. It is a way of ranking and comparing data. In our exercise, we were asked to find the 99th percentile, or the value of the random variable where only 1% of the distribution lies to the right. This concept is particularly useful in data analysis because it helps in understanding the relative standing of an individual value within a distribution. It is often used in standardized testing to determine how a score compares to others. To find a percentile in a normal distribution, we typically use a z-table, which associates percentile ranks with z-scores. These z-scores indicate how many standard deviations a value is from the mean.

Z-score

A z-score is a statistical measurement that describes a value's relation to the mean of a group of values, measured in terms of standard deviations. If a z-score is 0, it indicates that the value is exactly average, while a positive or negative z-score reflects above or below average, respectively.The formula for calculating a z-score is: \[ z = \frac{x - \mu}{\sigma} \] where \( x \) is the value of the variable, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. In the exercise, we sought the z-score for the 99th percentile, which was found to be 2.33 using the z-table. This means that a value at the 99th percentile is 2.33 standard deviations above the mean.Understanding z-scores is vital because they allow for the comparison of scores from different distributions, providing a method to standardize scores. This is especially useful in assessing probabilities within a normal distribution.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation points to data that is close to the mean, whereas a high standard deviation indicates data that is spread out over a wider range.The formula for standard deviation is derived as follows: \[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2} \] where \( \sigma \) is the standard deviation, \( N \) is the number of data points, \( x_i \) is each individual data point, and \( \mu \) is the mean of the data set.In the context of our exercise, the standard deviation is 10. This tells us how much the values in our normal distribution tend to deviate from the mean, which is 35 in this case. By knowing the standard deviation, we can calculate other statistics, such as z-scores, and predict the likelihood of certain results within our data set. Understanding standard deviation enhances our ability to conclude how data behaves and forms the foundation for further statistical analysis.