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Chapter 6: Problem 93

Let \(x\) denote the IQ of an individual selected at random from a certain population. The value of \(x\) must be a whole number. Suppose that the distribution of \(x\) can be approximated by a normal distribution with mean value 100 and standard deviation 15. Approximate the following probabilities. a. \(P(x=100)\) b. \(P(x \leq 110)\) c. \(P(x<110)\) (Hint: \(x<110\) is the same as \(x \leq 109 .\) ) d. \(P(75 \leq x \leq 125)\)

Short Answer

Expert verified
a. \(P(x=100) \approx 0.0268\) b. \(P(x\leq110) \approx 0.7486\) c. \(P(x<110) \approx P(x\leq 109) \approx 0.7357\) d. \(P(75\leq x \leq 125) \approx 0.9103\)

Step by step solution

01

Understand the provided information

We are given that the IQ scores (\(x\)) are a whole number and follow a normal distribution with an average (\(\mu\)) of 100 and a standard deviation (\(\sigma\)) of 15. We will use the provided parameters to find specific probabilities.
02

Convert \(x\) to \(z\)-score

In order to use the standard normal table, we must convert the scores into \(z\)-scores, using the formula: \[z = \frac{x - \mu}{\sigma}\]
03

Approximate the required probabilities

Using the formula from Step 2 and the standard normal table, approximate each of the requested probabilities: a. Approximate \(P(x=100)\): Since the IQ scores are a whole number, \(P(x=100)\) can be rewritten as \(P(99.5 \le x \le 100.5)\). Calculate the \(z\)-scores corresponding to 99.5 and 100.5 and find the difference of their probabilities in the standard normal table. b. Approximate \(P(x\leq 110)\): Calculate the \(z\)-score corresponding to \(x=110\) and find its probability in the standard normal table. c. Approximate \(P(x<110)\): As the hint suggests, we can rewrite this as \(P(x\leq109)\). Calculate the \(z\)-score corresponding to \(x=109\) and find its probability in the standard normal table. d. Approximate \(P(75\leq x \leq 125)\): Calculate the \(z\)-scores corresponding to \(x=75\) and \(x=125\), and find their probabilities in the standard normal table. Subtract the probabilities to get the approximate probability for \(75\leq x \leq 125\).

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Key Concepts

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

Z-Score Calculation
Understanding the concept of a z-score is crucial when working with normal distributions. A z-score indicates how many standard deviations an element is from the mean. To calculate a z-score, you use the formula:
\[z = \frac{x - \mu}{\sigma}\]
where \(x\) is the value in the distribution, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
When you calculate a z-score, you're essentially standardizing different variables so that you can compare them on the same scale.
  • If the z-score is 0, it means the value is exactly at the mean.
  • A positive z-score indicates a value above the mean.
  • A negative z-score signifies a value below the mean.
Calculating z-scores allows us to find the probability of a certain value occurring within a normal distribution, and facilitates comparison between different measurements.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution with a mean of 0 and a standard deviation of 1. It's also known as the Z-distribution. When we convert any normal distribution to z-scores, we essentially use the properties of the standard normal distribution to make probability calculations.
We can use standardized tables, known as Z-tables, to find the probability of a z-score falling within a certain range. These tables are built on the standard normal distribution's characteristics and provide areas under the curve from the mean to the z-score value. The total area under the curve in a standard normal distribution equals 1, or 100% probability. This simplification allows you to find probabilities associated with any normal distribution once you've calculated the corresponding z-scores.
Normal Distribution IQ
IQ scores are often represented by a normal distribution, with the average population IQ set at a mean value, which is traditionally 100. The spread, or dispersion, of these scores is described by the standard deviation, often assumed to be 15. This means that most people have IQs within one standard deviation (15 points) above or below the mean (100).
In practice, the probability of an exact IQ score, such as \(P(x=100)\), is calculated over a range (e.g., \(99.5 \leq x \leq 100.5\)), due to the continuous nature of the normal distribution. When looking at the range of IQs, such as finding the probability of an IQ between 75 and 125, we're interpreting the area under the normal curve between these values. This area represents the likelihood of an individual having an IQ in that range. In educational settings, understanding the distribution of IQ scores can help educators tailor their teaching to the varying abilities of their students.

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Most popular questions from this chapter

A grocery store has an express line for customers purchasing at most five items. Consider the random variable \(x=\) the number of items purchased by a randomly selected customer using this line. Make two tables that represent two different possible probability distributions for \(x\) that have the same standard deviation but different means.

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each question. Suppose that you have not studied and so must guess (randomly select one of the five answers) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Calculate the variance and standard deviation of \(x\). d. Based on your answers to Parts \((\mathrm{b})\) and \((\mathrm{c}),\) is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

Suppose that the \(\mathrm{pH}\) of soil samples taken from a certain geographic region is normally distributed with a mean of 6.00 and a standard deviation of \(0.10 .\) Suppose the \(\mathrm{pH}\) of a randomly selected soil sample from this region will be determined. a. What is the probability that the resulting \(\mathrm{pH}\) is between 5.90 and \(6.15 ?\) b. What is the probability that the resulting \(\mathrm{pH}\) exceeds \(6.10 ?\) c. What is the probability that the resulting \(\mathrm{pH}\) is at most \(5.95 ?\) d. What value will be exceeded by only \(5 \%\) of all such \(\mathrm{pH}\) values?

The probability distribution of \(x\), the number of tires needing replacement on a randomly selected automobile checked at a certain inspection station, is given in the following table: 0 \(\begin{array}{cccc}1 & 2 & 3 & 4 \\ 0.16 & 0.06 & 0.04 & 0.20\end{array}\) $$ p(x) \quad 0.54 $$ The mean value of \(x\) is \(\mu_{x}=1.2\). Calculate the values of \(\sigma_{x}^{2}\) and \(\sigma\).

The distribution of the number of items produced by an assembly line during an 8 -hour shift can be approximated by a normal distribution with mean value 150 and standard deviation 10 . a. What is the approximate probability that the number of items produced is at most \(120 ?\) b. What is the approximate probability that at least 125 items are produced? c. What is the approximate probability that between 135 and 160 (inclusive) items are produced?
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