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Chapter 7: Problem 23

A normal distribution has a mean of 50 and a standard deviation of 4 . Determine the value below which \(95 \%\) of the observations will occur.

Short Answer

Expert verified
The value is approximately 56.58.

Step by step solution

01

Understand the Problem

You need to determine the value below which 95% of the observations in a normal distribution will occur, given that the distribution has a mean of 50 and a standard deviation of 4.
02

Identify the Corresponding Z-Score

For a normal distribution, 95% of the observations lie to the left of the point. This corresponds to the 95th percentile. For standard normal distributions, the Z-score for the 95th percentile is approximately 1.645.
03

Use the Z-Score Formula

The Z-score formula is given by \( Z = \frac{X - \mu}{\sigma} \), where \(X\) is the unknown value, \(\mu = 50\) is the mean, and \(\sigma = 4\) is the standard deviation. Substitute \(Z = 1.645\) into the formula: \[ 1.645 = \frac{X - 50}{4} \]
04

Solve for X

Isolate \(X\) in the formula:\[ 1.645 \times 4 = X - 50 \]\[ 6.58 = X - 50 \]\[ X = 6.58 + 50 \]\[ X = 56.58 \]

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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 statistical measure used to describe a data point's relation to the mean of a group of data. It tells us how many standard deviations an element is from the mean. Here's what you need to know to understand z-scores effectively:
  • The z-score formula is: \( Z = \frac{X - \mu}{\sigma} \), where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
  • If the z-score is 0, it means the data point's score is identical to the mean score.
  • A positive z-score indicates the data point is above the mean, while a negative z-score indicates it is below.
  • Z-scores allow us to compare scores across different distributions, as they standardize scores regardless of the original mean and standard deviation.
The specific z-score can help identify percentiles in a normal distribution, revealing the relative standing of a data point.
Percentile
A percentile is a measure that indicates the value below which a given percentage of observations in a group falls. Moving from z-scores to percentiles involves understanding:
  • The 95th percentile means that 95% of the data falls below this value.
  • In a normal distribution, certain z-scores correspond to percentile ranks. For instance, the z-score of approximately 1.645 equates to the 95th percentile.
  • This concept is crucial in determining where a value lies in relation to the entire data set.
Percentiles can be visualized through a cumulative distribution function (CDF), which provides comprehensive insights into the distribution of the data points across the whole range of values.
Mean and Standard Deviation
Mean and standard deviation are fundamental concepts in the description of data sets within a normal distribution. They serve as the primary descriptors for the data's central tendency and spread:
  • The **mean**, often denoted as \(\mu\), is the average of all data points in the distribution.
  • The **standard deviation**, denoted as \(\sigma\), measures the extent of variation or dispersion from the mean.
  • Together, these descriptors tell us not only where the center of the data is situated but also how spread out the data points are.
In the context of a normal distribution, knowing the mean and standard deviation allows us to utilize the z-score formula effectively, enabling us to identify specific data points such as those meeting a particular percentile.

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Most popular questions from this chapter

A recent Gallup study (http://www.gallup.com/poll/175286/hour-workweek- actually-Ionger-seven- hours.aspx?g_source=polls+work+hours\(\&\)g_medium=search\(\&\)g_campaign=tiles) found the typical American works an average of 46.7 hour per week. The study did not report the shape of the distribution of hours worked or the standard deviation. It did, however, indicate that \(40 \%\) of the workers worked less than 40 hours a week and that 18 percent worked more than 60 hours. a. If we assume that the distribution of hours worked is normally distributed, and knowing \(40 \%\) of the workers worked less than 40 hours, find the standard deviation of the distribution. b. If we assume that the distribution of hours worked is normally distributed and \(18 \%\) of the workers worked more than 60 hours, find the standard deviation of the distribution. c. Compare the standard deviations computed in parts \(a\) and \(b\). Is the assumption that the distribution of hours worked is approximately normal reasonable? Why?

The accounting department at Weston Materials Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution. a. Determine the \(z\) values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect? b. What percent of the garages take between 29 hours and 34 hours to erect? c. What percent of the garages take 28.7 hours or less to erect? d. Of the garages, \(5 \%\) take how many hours or more to erect?

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, 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 minute. a. What is the probability that calls last between 4.2 and 5 minutes? b. What is the probability that calls last more than 5 minutes? c. What is the probability that calls last between 5 and 6 minutes? d. What is the probability that 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) \(4 \%\) of the calls. What is this time?

The weights of canned hams processed at Henline Ham Company follow the normal distribution, with a mean of 9.20 pounds and a standard deviation of 0.25 pound. The label weight is given as 9.00 pounds. a. What proportion of the hams actually weigh less than the amount claimed on the label? b. The owner, Glen Henline, is considering two proposals to reduce the proportion of hams below label weight. He can increase the mean weight to 9.25 and leave the standard deviation the same, or he can leave the mean weight at 9.20 and reduce the standard deviation from 0.25 pound to \(0.15 .\) Which change would you recommend?

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?
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