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(a) The 90 th percentile of a normal distribution is how many standard deviations above the mean? (b) The 10 th percentile of a normal distribution is how many standard deviations below the mean?

Short Answer

Expert verified
The 90th percentile corresponds to approximately 1.28 standard deviations above the mean, while the 10th percentile corresponds to approximately 1.28 standard deviations below the mean.

Step by step solution

01

- Understand percentiles in a normal distribution

In a standard normal distribution (mean = 0, standard deviation = 1), percentiles correspond to the number of standard deviations from the mean. The 90th percentile is the value below which 90% of the data can be found. Similarly, the 10th percentile is the value below which 10% of the data lies.
02

- Find the z-score for the 90th percentile

Consult a standard normal distribution (z-table) or use a statistical software to find the z-score that corresponds to the 90th percentile. The z-score is the number of standard deviations the percentile is away from the mean.
03

- Find the z-score for the 10th percentile

Similarly, consult the z-table or software to find the z-score for the 10th percentile. This z-score is negative, indicating that it is below the mean.

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Key Concepts

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

Z-score
The z-score is a powerful tool that tells you how many standard deviations a data point is from the mean. It's a measure of position that allows comparisons across different data sets, which could have different means and standard deviations. By converting a score from a normal distribution to a z-score, we create a standard normal distribution.

Imagine you've taken an exam and you want to know how well you did in comparison to your classmates. Your score is turned into a z-score and now you can see if you're above or below average, and by how much. For example, a z-score of 1.96 indicates a value that is 1.96 standard deviations above the mean. In the context of percentile ranks, finding the z-score associated with a certain percentile helps to understand the relative standing of that percentile within the distribution.
Standard Normal Distribution
When we talk about the standard normal distribution, we're referring to a special type of normal distribution that has a mean (average) of 0 and a standard deviation of 1. It's a bell-shaped curve where the majority of values cluster around the mean and it tapers off symmetrically on both sides.

Why is this so important? Because it allows us to standardize scores from different normal distributions, making them comparable. It transforms our tests, measurements, and data into a universal scale—the 'standard' scale—where we can find probabilities and percentiles systematically. This standardization is done using z-scores, and it's what makes tables or software necessary for finding exact positions within a distribution possible.
Percentile Rank
Let's dive into percentile ranks. These ranks tell you the percentage of data points that fall below a certain value in a distribution. For instance, if you're at the 90th percentile, you've scored better than 90% of all the data points.

It's like being in a race; if you're in the top 10%, you've outrun 90% of the competitors. Percentiles are not about how much better you are, but rather about your position in the crowd. For standardized tests or measurements, understanding percentile ranks is crucial, because they help pinpoint your relative performance or standing.
Standard Deviations
The standard deviation is a statistic that measures the amount of variation or dispersion in a set of values. In simpler terms, it shows how much the individual data points deviate from the mean of the set. A small standard deviation means the values are close to the mean, while a large standard deviation indicates a wide spread of values.

In the context of a normal distribution, the standard deviation is key because it defines the curve's width. Knowing the number of standard deviations from the mean—a value's z-score—is like knowing how far you are from the center of a city. The further you are, the less common and more extreme the position is. Thus, when you're looking at percentiles, you're actually measuring the distance from the mean in units of standard deviation.

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Most popular questions from this chapter

The brain weights of a certain population of adult Swedish males follow approximately a normal distribution with mean \(1,400 \mathrm{gm}\) and standard deviation \(100 \mathrm{gm} .\) What percentage of the brain weights are (a) \(1,500 \mathrm{gm}\) or less? (b) between 1,325 and \(1,500 \mathrm{gm} ?\) (c) 1,325 gm or more? (d) 1,475 gm or more? (e) between 1,475 and \(1,600 \mathrm{gm} ?\) (f) between 1,200 and \(1,325 \mathrm{gm} ?\)

A certain assay for serum alanine aminotransferase (ALT) is rather imprecise. The results of repeated assays of a single specimen follow a normal distribution with mean equal to the true ALT concentration for that specimen and standard deviation equal to \(4 \mathrm{U} / \mathrm{l}\) (see Example 2.2 .12 ). Suppose that a certain hospital lab measures many specimens every day, performing one assay for each specimen, and that specimens with ALT readings of \(40 \mathrm{U} / \mathrm{l}\) or more are flagged as "unusually high." If a patient's true ALT concentration is \(35 \mathrm{U} / \mathrm{l}\), what is the probability that his specimen will be flagged as "unusually high"?

In the nerve-cell activity of a certain individual fly, the time intervals between "spike" discharges follow approximately a normal distribution with mean \(15.6 \mathrm{~ms}\) and standard deviation \(0.4 \mathrm{~ms}\) (as in Example 4.1.3). Let \(Y\) denote a randomly selected interspike interval. Find (a) \(\operatorname{Pr}\\{Y>15\\}\) (b) \(\operatorname{Pr}\\{Y>16.5\\}\) (c) \(\operatorname{Pr}\\{15

The bill lengths of a population of male Blue Jays follow approximately a normal distribution with mean equal to \(25.4 \mathrm{~mm}\) and standard deviation equal to \(0.8 \mathrm{~mm}\) (as in Example 4.1.2). Find the 95 th percentile of the bill length distribution.

Find the \(z\) -values corresponding to the following percentiles of the standard normal distribution. (a) the 75 th percentile (b) the 90 th percentile (c) the 95 th percentile (d) the 99 th percentile

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