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A normal distribution has a mean of 80 and a standard deviation of 14. Determine the value above which 80 percent of the values will occur.

Short Answer

Expert verified
Values above 68.24 comprise 80% of the data in this distribution.

Step by step solution

01

Understand the Problem

We need to determine the value above which 80% of the data falls in a normal distribution with a mean (\(\mu\)) of 80 and a standard deviation (\(\sigma\)) of 14.
02

Convert Percentage to Z-Score

Find the Z-score for the 20th percentile (since we want 80% of values above this point) using a Z-table or calculator. The Z-score corresponding to the 20th percentile is approximately -0.84.
03

Apply the Z-Score Formula

Use the Z-score formula \[ Z = \frac{X - \mu}{\sigma} \]where \(Z\) is the Z-score, \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Rearrange to solve for \(X\):\[ X = Z \cdot \sigma + \mu \]
04

Calculate the Value of X

Insert the values into the rearranged formula: \\(X = -0.84 \cdot 14 + 80\).Calculate to find \\(X = 68.24\).

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Key Concepts

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

Percentile
A percentile is a measure that tells us what percentage of the data lies below a certain value in a distribution. This is particularly useful in understanding how a specific score compares to the rest of the data. For instance, if you're in the 20th percentile, 20% of the data falls below you.

In the context of a normal distribution, percentiles can help us find where most of the data lies. For example, saying a value is at the 20th percentile means that 20% of the data values are below this point, and consequently, 80% of the data values are above this point.
  • Useful for comparisons
  • Gives an idea of data distribution
Understanding percentiles allows you to quickly gauge position among a set of values.
Z-score
The Z-score is a statistical measure that tells you how many standard deviations a data point is from the mean. It helps standardize different data points from different normal distributions, enabling comparison.

To find a Z-score, you can use the formula:\[ Z = \frac{X - \mu}{\sigma} \]Where:
  • \(X\) is the value in question
  • \(\mu\) is the mean of the distribution
  • \(\sigma\) is the standard deviation
The Z-score can be used alongside tables or calculators to determine the percentile or probability associated with a particular Z-score. For example, a Z-score of -0.84 corresponds to the 20th percentile in a standard normal distribution. In other words, a Z-score tells you how far, in terms of standard deviations, a particular score is from the mean.
Standard Deviation
Standard deviation is a measure of how much the values in a data set vary from the mean. It tells us the spread or dispersion of the data points. A low standard deviation means data points are close to the mean, whereas a high standard deviation indicates data points are spread out over a wider range of values.

In our normal distribution example, the standard deviation is 14, meaning most of the data points fall within 14 units above or below the mean. Here's why standard deviation is important:
  • It reveals the extent of variation within the data.
  • Helps to understand how representative the mean is.
  • Informs decisions based on data distribution.
Understanding this concept is crucial as it forms the foundation for calculating Z-scores and consequently determining percentiles.
Mean
The mean is one of the most common measures of central tendency, representing the average of a data set. It is calculated by adding all the numbers in a data set and dividing by the number of values. The mean gives us an idea of what a "typical" value in the data set looks like.

For example, in the normal distribution exercise, the mean value is 80. This implies that, on average, the data points revolve around 80. The mean helps us understand the center point of a normal distribution, acting as a critical anchor in calculations involving Z-scores and standard deviation.
  • Central reference point in the data
  • Influences the symmetry of the normal distribution
The mean is crucial for interpreting the data's overall trend and serves as a baseline from which deviations are measured.

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Most popular questions from this chapter

Shaver Manufacturing, Inc., offers dental insurance to its employees. A recent study by the human resource director shows the annual cost per employee per year followed the normal probability distribution, with a mean of \(\$ 1,280\) and a standard deviation of \(\$ 420\) per year. a. What fraction of the employees cost more than \(\$ 1,500\) per year for dental expenses? b. What fraction of the employees cost between \(\$ 1,500\) and \(\$ 2,000\) per year? c. Estimate the percent that did not have any dental expense. d. What was the cost for the 10 percent of employees who incurred the highest dental expense?

The April rainfall in Flagstaff, Arizona, follows a uniform distribution between 0.5 and 3.00 inches. a. What are the values for \(a\) and \(b\) ? b. What is the mean amount of rainfall for the month? What is the standard deviation? C. What is the probability of less than an inch of rain for the month? d. What is the probability of exactly 1.00 inch of rain? e. What is the probability of more than 1.50 inches of rain for the month?

A study of long-distance phone calls made from the corporate offices of Pepsi Bottling Group, Inc., in Somers, New York, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.2 minutes and the standard deviation was 0.60 minutes. a. What fraction of the calls last between 4.2 and 5 minutes? b. What fraction of the calls last more than 5 minutes? c. What fraction of the calls last between 5 and 6 minutes? d. What fraction of the calls last between 4 and 6 minutes? e. As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4 percent of the calls. What is this time?

A uniform distribution is defined over the interval from 6 to 10 a. What are the values for \(a\) and \(b\) ? b. What is the mean of this uniform distribution? c. What is the standard deviation? d. Show that the total area is 1.00 . e. Find the probability of a value more than 7 . f. Find the probability of a value between 7 and \(9 .\)

The funds dispensed at the ATM machine located near the checkout line at the Kroger's in Union, Kentucky, follows a normal probability distribution with a mean of \(\$ 4,200\) per day and a standard deviation of \(\$ 720\) per day. The machine is programmed to notify the nearby bank if the amount dispensed is very low (less than \(\$ 2,500)\) or very high (more than \(\$ 6,000)\) a. What percent of the days will the bank be notified because the amount dispensed is very low? b. What percent of the time will the bank be notified because the amount dispensed is high? c. What percent of the time will the bank not be notified regarding the amount of funds dispersed?

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