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Chapter 6: Problem 6

A normal distribution has \(\mu=10\) and \(\sigma=2\). (a) Find the \(z\) score corresponding to \(x=12\). (b) Find the \(z\) score corresponding to \(x=4\). (c) Find the raw score corresponding to \(z=1.5\). (d) Find the raw score corresponding to \(z=-1.2\).

Short Answer

Expert verified
(a) 1; (b) -3; (c) 13; (d) 7.6

Step by step solution

01

Understanding the Z-score Formula

The Z-score formula is used to find how many standard deviations a data point is from the mean. The formula is given by: \( z = \frac{x - \mu}{\sigma} \), where \( x \) is the raw score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
02

Calculate Z-score for Part (a)

Given that \( \mu = 10 \), \( \sigma = 2 \), and \( x = 12 \). Substitute these values into the Z-score formula: \[ z = \frac{12 - 10}{2} = \frac{2}{2} = 1 \]. So, the Z-score for \( x = 12 \) is 1.
03

Calculate Z-score for Part (b)

Using the same formula, for \( x = 4 \): \[ z = \frac{4 - 10}{2} = \frac{-6}{2} = -3 \]. Thus, the Z-score for \( x = 4 \) is -3.
04

Understanding the Raw Score Formula

To find the raw score from a Z-score, we use the formula: \( x = z \cdot \sigma + \mu \).
05

Calculate Raw Score for Part (c)

Given \( z = 1.5 \), substitute the values \( \mu = 10 \) and \( \sigma = 2 \) in the formula: \[ x = 1.5 \cdot 2 + 10 = 3 + 10 = 13 \]. Therefore, the raw score corresponding to \( z = 1.5 \) is 13.
06

Calculate Raw Score for Part (d)

For \( z = -1.2 \): substitute \( \mu = 10 \) and \( \sigma = 2 \) in the formula: \[ x = -1.2 \cdot 2 + 10 = -2.4 + 10 = 7.6 \]. Thus, the raw score corresponding to \( z = -1.2 \) is 7.6.

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Key Concepts

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

Understanding Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a fundamental concept in statistics. It describes how data is spread out, following a bell-shaped curve. This curve is symmetric around the mean, showing that data near the mean are more frequent in occurrence.

Here are some key features:
  • The total area under the curve is 1, representing the whole probability.
  • The mean, median, and mode of a normal distribution are all equal, located at the center of the distribution.
  • The curve is defined by two main parameters: the mean (8...) and standard deviation (9...).
The mean (8...) determines the location of the center of the curve, while the standard deviation (9...) controls the spread of the distribution. A smaller standard deviation means the data is tightly clustered around the mean, whereas a larger one indicates that the data points are spread out over a wider range.
Explaining the Z-score
A Z-score, also known as a standard score, measures how many standard deviations an element is from the mean of the dataset. It is a valuable tool for understanding the position of data points within a normal distribution.

The formula for the Z-score is:\[ z = \frac{x - \mu}{\sigma} \]Where:
  • 8... is the individual data point.
  • 9... is the mean of the data set.
  • 9... represents the standard deviation.
A Z-score of 0 indicates that the data point's score is identical to the mean. A positive Z-score signifies that the data point is above the mean, while a negative Z-score shows it falls below the mean. This standardized way of measuring allows for comparability across different data sets and scales.
Demystifying Standard Deviation
The standard deviation is a measure that is used to quantify the amount of variation or dispersion in a set of data values. It provides context to what is 'normal' in a normal distribution.

Here's why it's important:
  • A small standard deviation indicates that the data points tend to be close to the mean.
  • A large standard deviation suggests that the data are spread out over a wide range.
  • It is always a positive number or zero but never negative.
Standard deviation helps in understanding data variability. For instance, in a quality control setting, a small standard deviation would imply consistent quality, while a large one could point to issues that need addressing.
Comprehending the Mean
The mean, often referred to as the average, is one of the most common measures of central tendency. It provides a single value that summarizes the entire data set.

To calculate the mean:
  • Add up all data values.
  • Divide the sum by the number of values in the data set.
The mean is useful because it provides a quick snapshot of the overall trend of the data set. However, it can be sensitive to outliers, which are unusually high or low values that may skew the mean .

While a great tool for summarization, it's often used alongside other measures like median and mode for a more comprehensive understanding of the data set's distribution.

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Most popular questions from this chapter

Suppose \(x\) has a distribution with a mean of 20 and a standard deviation of \(3 .\) Random samples of size \(n=36\) are drawn. (a) Describe the \(\bar{x}\) distribution and compute the mean and standard deviation of the distribution. (b) Find the \(z\) value corresponding to \(\bar{x}=19\). (c) Find \(P(\bar{x}<19)\). (d) Would it be unusual for a random sample of size 36 from the \(x\) distribution to have a sample mean less than 19? Explain.

Let \(z\) be a random variable with a standard normal distribution. Find the indicated probability, and shade the corresponding area under the standard normal curve. $$ P(z \leq-2.15) $$

Let \(z\) be a random variable with a standard normal distribution. Find the indicated probability, and shade the corresponding area under the standard normal curve. $$ P(-0.73 \leq z \leq 3.12) $$

More than a decade ago, high levels of lead in the blood put \(88 \%\) of children at risk. A concerted effort was made to remove lead from the environment. Now, according to the Third National Health and Nutrition Examination Survey (NHANES III) conducted by the Centers for Disease Control, only \(9 \%\) of children in the United States are at risk of high blood-lead levels. (a) In a random sample of 200 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (b) In a random sample of 200 children taken now, what is the probability that 50 or more have high blood-lead levels?

Sketch the areas under the standard normal curve over the indicated intervals, and find the specified areas. Between \(z=-2.42\) and \(z=-1.77\)
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