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Chapter 6: Problem 28

Determine the \(z\) value for each of the following \(x\) values for a normal distribution with \(\mu=16\) and \(\sigma=3 .\) a. \(x=12\) b. \(x=22\) c. \(x=19\) d. \(x=13\)

Short Answer

Expert verified
The z scores for each of the x values are: For a. \(z=-1.33\), b. \(z=2.00\), c. \(z=1.00\) and d. \(z=-1.00\).

Step by step solution

01

Understanding Z-Score formula

The formula to calculate the z-score from raw scores is \(z= \frac{x - \mu}{\sigma}\), where \(x\) is the raw score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
02

Solution for a. \(x=12\)

To get the z score for \(x=12\), apply the formula as \(z= \frac{12 - 16}{3}\). This calculates to a z score of \(-1.33\).
03

Solution for b. \(x=22\)

To get the z score for \(x=22\), apply the formula as \(z= \frac{22 - 16}{3}\). This calculates to a z score of \(2.00\).
04

Solution for c. \(x=19\)

To get the z score for \(x=19\), apply the formula as \(z= \frac{19 - 16}{3}\). This calculates to a z score of \(1.00\).
05

Solution for d. \(x=13\)

To get the z score for \(x=13\), apply the formula as \(z= \frac{13 - 16}{3}\). This calculates to a z score of \(-1.00\).

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Key Concepts

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

Understanding Normal Distribution
Normal distribution, often referred to as a bell curve, is a probability distribution that is symmetric about the mean. Imagine it as a graph where most values cluster around a central peak.
This peak represents the mean, and the probabilities for values gradually decrease as they move away from this mean.
  • The total area under the curve is 1, meaning all possibilities are covered.
  • In a perfectly normal distribution, about 68% of values lie within one standard deviation of the mean.
  • 95% are within two standard deviations, and 99.7% are within three.
This makes the normal distribution a powerful tool in statistics, as it allows for predictions about data patterns and variability.
Mean and Standard Deviation Explained
The mean is like the average of a set of numbers. It's the central value of your data set. You calculate it by adding all numbers together and dividing by the number of values.
In our example, the mean \( \mu \) is 16. This is the central point around which our normal distribution is built.
The standard deviation, denoted as \( \sigma \), measures how spread out numbers are in your data. It's about the average distance each data point is from the mean. In our scenario, \( \sigma \) is 3.
  • A small standard deviation means data points are close to the mean.
  • A larger standard deviation indicates more widespread data.
Understanding these concepts is key to calculating z-scores and interpreting data distributions.
Raw Score Conversion to Z-Score
Converting a raw score to a z-score involves a simple formula: \( z = \frac{x - \mu}{\sigma} \).
This equation helps you understand where a specific raw score (\( x \)) lies in relation to the mean (\( \mu \)) of the distribution.
Here's a breakdown:
  • Subtract the mean from the raw score: This tells you how far away, and in which direction, the raw score is from the mean.
  • Divide this difference by the standard deviation: This scales the difference according to the typical distribution spread.
Each z-score tells you how many standard deviations away \( x \) is from the mean. For example, a z-score of 1 means the raw score is one standard deviation above the mean. It helps standardize scores from different data sets, making them comparable across different contexts.

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Most popular questions from this chapter

Find the following binomial probabilities using the normal approximation. a. \(n=140, \quad p=.45, \quad P(x=67)\) b. \(n=100, \quad p=.55, \quad P(52 \leq x \leq 60)\) c. \(n=90, \quad p=.42, \quad P(x \geq 40)\) d. \(n=104, p=.75, \quad P(x \leq 72)\)

The print on the package of 100 -watt General Electric soft-white lightbulbs states that these bulbs have an average life of 750 hours. Assume that the lives of all such bulbs have a normal distribution with a mean of 750 hours and a standard deviation of 50 hours. a. Let \(x\) be the life of such a lightbulb. Find \(x\) so that only \(2.5 \%\) of such lightbulbs have lives longer than this value. b. Let \(x\) be the life of such a lightbulb. Find \(x\) so that about \(80 \%\) of such lightbulbs have lives shorter than this value.

Do the width and/or height of a normal distribution change when its standard deviation remains the same but its mean increases?

For a binomial probability distribution, \(n=120\) and \(p=.60\). Let \(x\) be the number of successes in 120 trials. a. Find the mean and standard deviation of this binomial distribution. b. Find \(P(x \leq 69)\) using the normal approximation. c. Find \(P(67 \leq x \leq 73)\) using the normal approximation.

A psychologist has devised a stress test for dental patients sitting in the waiting rooms. According to this test, the stress scores (on a scale of 1 to 10 ) for patients waiting for root canal treatments are found to be approximately normally distributed with a mean of \(7.59\) and a standard deviation of \(.73\). a. What percentage of such patients have a stress score lower than \(6.0\) ? b. What is the probability that a randomly selected root canal patient sitting in the waiting room has a stress score between \(7.0\) and \(8.0\) ? c. The psychologist suggests that any patient with a stress score of \(9.0\) or higher should be given a sedative prior to treatment. What percentage of patients waiting for root canal treatments would need a sedative if this suggestion is accepted?
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