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Chapter 6: Problem 11

Find the following percentiles for the standard normal random variable \(z\) : a. 90 th percentile b. 95 th percentile c. 98 th percentile d. 99 th percentile

Short Answer

Expert verified
Answer: The z-values for the 90th, 95th, 98th, and 99th percentiles of a standard normal distribution are as follows: a. The 90th percentile has a z-score of 1.28 b. The 95th percentile has a z-score of 1.645 c. The 98th percentile has a z-score of 2.05 d. The 99th percentile has a z-score of 2.33

Step by step solution

01

Understanding What a Percentile Represents

A percentile measures the relative standing of a value within a distribution. For example, if a test score is at the 90th percentile, this means that 90% of the other scores are below this value, and 10% are above it. In this exercise, we are given the following percentiles to find: 90th, 95th, 98th, and 99th.
02

Using the Z-Score Table to Find the Z Values

The first step is to use a standard normal table to find the z values corresponding to each percentile. This table shows the area (probability) to the LEFT of a z-score, which represents the percentile we are interested in. a. To find the 90th percentile, we need to find the z-value for which 0.900 of the data lies to the left. Looking up 0.900 in the z-score table, we get a value of z = 1.28. b. To find the 95th percentile, we need to find the z-value for which 0.950 of the data lies to the left. Looking up 0.950 in the z-score table, we get a value of z = 1.645. c. To find the 98th percentile, we need to find the z-value for which 0.980 of the data lies to the left. Looking up 0.980 in the z-score table, we get a value of z = 2.05. d. To find the 99th percentile, we need to find the z-value for which 0.990 of the data lies to the left. Looking up 0.990 in the z-score table, we get a value of z = 2.33.
03

Writing the Results

We have now found the z-values corresponding to the given percentiles: a. The 90th percentile has a z-score of 1.28 b. The 95th percentile has a z-score of 1.645 c. The 98th percentile has a z-score of 2.05 d. The 99th percentile has a z-score of 2.33

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Key Concepts

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

Percentiles
Percentiles are a measure used to indicate the relative position of a value in a dataset. Understanding percentiles can give you a heads-up on how data points compare to each other. For example, if you score in the 90th percentile, 90% of the data points are below your score, and only 10% are above it. This is particularly useful in determining how extreme or typical a particular observation is within a dataset.

Percentiles are widely used in statistics and are instrumental in evaluating standard normal distributions. They allow us to understand where a particular value lies in relation to the entire dataset. To find which value corresponds to a particular percentile in a standard normal distribution, we often use a Z-score table. The percentile helps determine the probability or the area under the curve to the left of that score.
Z-Score Calculation
Z-scores are used to determine how many standard deviations an element is from the mean. This is particularly useful in standard normal distribution because it allows for easy calculation and comparison of different values in the dataset.

To calculate a Z-score, you use the formula: \[ Z = \frac{(X - \mu)}{\sigma} \] where:
  • \( X \) is the data point you are evaluating,
  • \( \mu \) is the mean of the dataset,
  • \( \sigma \) is the standard deviation.

Since the standard normal distribution already has a mean (\( \mu \)) of 0 and a standard deviation (\( \sigma \)) of 1, calculating Z-scores becomes straightforward. When dealing with percentiles, we often look for the Z-score that corresponds directly to the percentage of data that falls below a given point in the Z-score table.
Normal Distribution Table
A normal distribution table, also known as a Z-table, is a mathematical tool that provides the probability that a standard normal random variable will be less than or equal to a given value. In simpler terms, it shows the area (probability) under the curve to the left of a given Z-score.

This table is crucial because it allows us to determine probabilities and percentiles for normal distributions without needing to perform complex calculations. The table is divided into rows and columns. Each cell contains a probability that corresponds to a Z-score.

To use the Z-table effectively:
  • First, identify the row corresponding to the cellular value for your Z-score.
  • Next, locate the column that matches the second decimal place of your Z-score.
  • The intersection of the row and column provides the area under the curve to the left of the Z-score.


To find percentiles like the 90th or 95th, you look for Z-scores that correspond to an area like 0.900 or 0.950 in the table, indicating that 90% or 95% of the dataset falls below that score.

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Most popular questions from this chapter

Your pulse rate is a measure of the number of heartbeats per minute. It can be measured in several places on your body, where an artery passes close to the skin. Once you find the pulse, count the number of beats per minute, or, count for 30 seconds and multiply by two. What's a normal pulse rate? That depends on a variety of factors. Pulse rates between 60 and 100 beats per minute are considered normal for children over 10 and adults. \({ }^{4}\) Suppose that these pulse rates are approximately normally distributed with a mean of 78 and a standard deviation of 12 . a. What proportion of adults will have pulse rates between 60 and \(100 ?\) b. What is the 95 th percentile for the pulse rates of adults? c. Would a pulse rate of 110 be considered unusual? Explain.

A stringer of tennis rackets has found that the actual string tension achieved for any individual racket stringing will vary as much as 6 pounds per square inch from the desired tension set on the stringing machine. If the stringer wishes to string at a tension lower than that specified by a customer only \(5 \%\) of the time, how much above or below the customer's specified tension should the stringer set the stringing machine? (NOTE: Assume that the distribution of string tensions produced by the stringing machine is normally distributed, with a mean equal to the tension set on the machine and a standard deviation equal to 2 pounds per square inch.)

a. Find the probability that \(z\) is greater than \(-.75 .\) b. Find the probability that \(z\) is less than 1.35 .

Calculate the area under the standard normal curve to the left of these values: a. \(z=1.6\) b. \(z=1.83\) c. \(z=.90\) d. \(z=4.18\)

The average length of time required to complete a college achievement test was found to equal 70 minutes with a standard deviation of 12 minutes. When should the test be terminated if you wish to allow sufficient time for \(90 \%\) of the students to complete the test? (Assume that the time required to complete the test is normally distributed.)
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