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Use the following information to answer the next four exercises: X ~ N(54, 8) Find the \(80^{\text { th }}\) percentile.

Short Answer

Expert verified
The 80th percentile is approximately 60.73.

Step by step solution

01

Identify the Distribution

The variable X follows a normal distribution with a mean (μ) of 54 and a standard deviation (σ) of 8. This is denoted as \(X \sim N(54, 8)\).
02

Understand the Percentile Concept

The 80th percentile of a distribution is the value below which 80% of the observations may be found. In a normal distribution, you use the standard normal distribution to find this value.
03

Convert to Z-Score

To find the percentile, convert the percentile value to a z-score using the standard normal distribution tables. For the 80th percentile, find the z-score (approximately 0.8416) where the cumulative distribution function equals 0.80.
04

Use Z-Score Formula

Apply the z-score formula to convert from the standard normal distribution to the original distribution: \[x = \, \mu \,+ z \times \sigma\], replacing \(\mu\) with 54, \(z\) with 0.8416, and \(\sigma\) with 8.
05

Calculate the Percentile

Calculate the value using the formula: \[x = 54 + 0.8416 \times 8 = 54 + 6.7328 = 60.7328\]. Round if necessary.

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Key Concepts

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

Percentile
In statistics, a percentile is a measure used to indicate the relative standing of a value within a data set. The 80th percentile, for example, means that 80% of the data falls below this particular value, while 20% is above it. This is especially useful for understanding how a particular score compares to the rest of a distribution.

To find a percentile in a normal distribution, we often look at the standard normal distribution. This involves using the cumulative distribution function (CDF) to determine the point at which the specified percentage of observations fall below a certain value. For the normal distribution, you can achieve this by converting the percentile into a z-score, which indicates how many standard deviations away a value is from the mean.
Z-Score
The z-score is a statistical measurement that tells you how far away a data point is from the mean in terms of the number of standard deviations. It is a way to standardize scores on the same scale, making it easier to compare different data points from different normal distributions.

Given a normal distribution, finding the z-score that corresponds to a specific percentile involves determining the score that has that percentage of data to its left on the standard normal curve. For instance, the 80th percentile corresponds to a z-score of approximately 0.8416. To find this value, one might consult standard normal distribution tables or use software that calculates z-scores. Once the z-score is determined, it can be translated back to a specific value in the original distribution using the formula:
  • \[x = \, \mu \,+ z \times \sigma\]
Standard Deviation
Standard deviation is a key concept in statistics, showing the amount of variation or dispersion in a data set. A low standard deviation means that the data points tend to be close to the mean of the data set. Conversely, a high standard deviation indicates a wide spread around the mean.

In the context of a normal distribution, standard deviation serves as a scaling factor that directly influences the computation of z-scores and percentiles. For example, in our initial setup, the standard deviation (σ) is 8. This value is essential when transforming a z-score back into the context of the original data set.

Knowing the standard deviation enables you to understand how spread out the values are and predict where most data points lie. Approximately 68% of data within a normal distribution lies within one standard deviation of the mean. This profound insight grants clarity to the spread and predictability of the dataset.

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Most popular questions from this chapter

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. In 2005, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean ? = 520 and standard deviation ? = 115. a. Calculate the z-score for an SAT score of 720. Interpret it using a complete sentence. b. What math SAT score is 1.5 standard deviations above the mean? What can you say about this SAT score? c. For 2012, the SAT math test had a mean of 514 and standard deviation 117. The ACT math test is an alternate to the SAT and is approximately normally distributed with mean 21 and standard deviation 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 30, who did better with respect to the test they took?

Use the following information to answer the next two exercises: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. The heights of the 430 National Basketball Association players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with mean, ? = 79 inches and a standard deviation, ? = 3.89 inches. For each of the following heights, calculate the z-score and interpret it using complete sentences. a. 77 inches b. 85 inches c. If an NBA player reported his height had a z-score of 3.5, would you believe him? Explain your answer.

What is the z-score of \(x=12,\) if it is two standard deviations to the right of the mean?

If the area to the left of x in a normal distribution is 0.123, what is the area to the right of x?

Use the following information to answer the next three exercises: The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. Based upon the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space? a. Yes b. No c. Unable to determine

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