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A normal population has a mean of 80.0 and a standard deviation of \(14.0 .\) a. Compute the probability of a value between 75.0 and 90.0 . b. Compute the probability of a value of 75.0 or less. c. Compute the probability of a value between 55.0 and 70.0 .

Short Answer

Expert verified
a. 0.4019, b. 0.3602, c. 0.2000

Step by step solution

01

Understanding the Normal Distribution

The given problem describes a normal distribution with a mean \( \mu = 80.0 \) and a standard deviation \( \sigma = 14.0 \). We will use the standard normal distribution (z-distribution) to calculate probabilities. The standard normal distribution has a mean of 0 and a standard deviation of 1.
02

Calculating Z-Scores for Part (a)

For part (a), we want the probability that a value is between 75.0 and 90.0. We convert these values to z-scores using the formula \( z = \frac{x - \mu}{\sigma} \). For 75.0, \( z = \frac{75.0 - 80.0}{14.0} = -0.3571 \). For 90.0, \( z = \frac{90.0 - 80.0}{14.0} = 0.7143 \).
03

Finding Probability for Part (a)

Using a standard normal z-table or calculator, find the probabilities for \( z = -0.3571 \) and \( z = 0.7143 \). The probability for \( z = -0.3571 \) is approximately 0.3602, and for \( z = 0.7143 \) it is approximately 0.7621. Subtract these probabilities to find the probability between them: \( P(75.0 < X < 90.0) = 0.7621 - 0.3602 = 0.4019 \).
04

Calculating Z-Score for Part (b)

For part (b), we need the probability of a value of 75.0 or less. The z-score for 75.0 is \( z = -0.3571 \).
05

Finding Probability for Part (b)

Using the z-table, the probability for \( z = -0.3571 \) is approximately 0.3602. Therefore, \( P(X \leq 75.0) = 0.3602 \).
06

Calculating Z-Scores for Part (c)

For part (c), we calculate the z-scores for values 55.0 and 70.0. For 55.0, \( z = \frac{55.0 - 80.0}{14.0} = -1.7857 \). For 70.0, \( z = \frac{70.0 - 80.0}{14.0} = -0.7143 \).
07

Finding Probability for Part (c)

Using the z-table or calculator, find probabilities for \( z = -1.7857 \) and \( z = -0.7143 \). The probability for \( z = -1.7857 \) is approximately 0.0375 and for \( z = -0.7143 \) is approximately 0.2375. Subtract these to get the probability: \( P(55.0 < X < 70.0) = 0.2375 - 0.0375 = 0.2000 \).

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Key Concepts

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

Mean and Standard Deviation
In any data set, understanding the mean and standard deviation is crucial since these elements define the normal distribution. The **mean** is the average value of all data points. In this exercise, the mean is 80.0. It represents the central value around which the data is distributed.

The **standard deviation** tells us how much the data varies from the mean. A smaller standard deviation means the data points are closer to the mean, whereas a larger standard deviation indicates they are spread out over a wider range of values. Here, the standard deviation is 14.0. So, most data points fall within 14 units of the mean. These concepts help you understand where most values lie in a normal distribution.
Z-Score
A **z-score** helps you understand how far away a specific data point is from the mean in terms of standard deviations. To calculate the z-score, use the formula:

\( z = \frac{x - \mu}{\sigma} \)

Where \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. By converting data points to z-scores, you're able to compare different data points directly, even if they come from different distributions, by seeing how many standard deviations away they are from the mean.

In the given problem, converting values like 75.0 and 90.0 to z-scores lets you analyze them using the standard normal distribution.
Probability Calculation
Probability calculation in a normal distribution gives us the likelihood or chance that a particular outcome will happen. This probability is found by determining the area under the bell curve of the normal distribution. When you know the z-scores, as in this exercise, you can use them to find probabilities.

For a range like between 75.0 and 90.0, calculate the probability by finding the difference between the probabilities of each z-score. It's essentially measuring the proportion of the area of the distribution that falls between these two z-scores.

This way of calculating probabilities makes it easier to understand how likely it is for a data point to fall within a specific range.
Z-Table
A **z-table** is a critical tool for finding probabilities in a standard normal distribution. The table provides the probabilities that correspond to different z-scores.

The z-table lists standardized z-scores and their cumulative probabilities from the left of the standard normal distribution up to that point.

For example, a z-score of -0.3571 corresponds to a probability of approximately 0.3602, based on the z-table. Similarly, each z-score calculated in the exercise corresponds to a specific probability found in the z-table, which helps to compute the probability over a range of values in the normal distribution.

This method streamlines probability calculations and makes working with normal distributions much more manageable.
Standard Normal Distribution
The **standard normal distribution** is a special form of the normal distribution where the mean is 0, and the standard deviation is 1. This standard form makes it universally applicable because any normal distribution can be transformed to the standard normal distribution using z-scores.

This transformation simplifies probability calculations and allows us to use z-scores universally for any normal distribution's analysis. In essence, it acts as a universal template for analyzing different datasets normalized to the same scale.

By understanding and using the standard normal distribution, you can apply findings from one data set to another, given how they scale with respect to their respective means and standard deviations.

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Most popular questions from this chapter

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?

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