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Chapter 6: Problem 48

Let \(z\) denote a random variable having a normal distribution with \(\mu=0\) and \(\sigma=1\). Determine each of the following probabilities: a. \(P(z<0.10)\) b. \(P(z<-0.10)\) c. \(P(0.40-1.25)\) g. \(P(z<-1.50\) or \(z>2.50)\)

Short Answer

Expert verified
The probabilities for each part are as follows: a. \(P(z < 0.10) = 0.5398\) b. \(P(z < -0.10) = 0.4602\) c. \(P(0.40 < z < 0.85) = 0.1469\) d. \(P(-0.85 < z < -0.40) = 0.1469\) e. \(P(-0.40 < z < 0.85) = 0.4577\) f. \(P(z > -1.25) = 0.7890\) g. \(P(z < -1.50 \text{ or } z > 2.50) = 0.0730\)

Step by step solution

01

(1. Standard normal distribution)

The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1.
02

(2. Using the z-table or calculator to find the probabilities)

When finding probabilities for the standard normal distribution, you can either use a standard normal distribution table or a calculator. We will use a z-table for our example.
03

(a. Find P(z < 0.10))

Look up the z-score 0.10 in the z-table, which gives the probability P(z < 0.10) = 0.5398.
04

(b. Find P(z < -0.10))

Look up the z-score -0.10 in the z-table, which gives the probability P(z < -0.10) = 0.4602.
05

(c. Find P(0.40 < z < 0.85))

To find P(0.40 < z < 0.85), look up 0.40 and 0.85 in the z-table. P(z < 0.85) = 0.8023 and P(z < 0.40) = 0.6554. Subtract the two probabilities: P(0.40 < z < 0.85) = 0.8023 - 0.6554 = 0.1469.
06

(d. Find P(-0.85 < z < -0.40))

To find P(-0.85 < z < -0.40), look up -0.85 and -0.40 in the z-table. P(z < -0.40) = 0.3446 and P(z < -0.85) = 0.1977. Subtract the two probabilities: P(-0.85 < z < -0.40) = 0.3446 - 0.1977 = 0.1469.
07

(e. Find P(-0.40 < z < 0.85))

To find P(-0.40 < z < 0.85), look up 0.85 and -0.40 in the z-table. P(z < 0.85) = 0.8023 and P(z < -0.40) = 0.3446. Subtract the two probabilities: P(-0.40 < z < 0.85) = 0.8023 - 0.3446 = 0.4577.
08

(f. Find P(z > -1.25))

To find P(z > -1.25), we can find the complement, P(z < -1.25) and subtract from 1. Look up -1.25 in the z-table, and we get P(z < -1.25) = 0.2110. So, P(z > -1.25) = 1 - 0.2110 = 0.7890.
09

(g. Find P(z < -1.50 or z > 2.50))

To find P(z < -1.50 or z > 2.50), we will find the complement probabilities and subtract from 1. Look up -1.50 and 2.50 in the z-table, and we get P(z < -1.50) = 0.0668 and P(z < 2.50) = 0.9938. So, P(z > 2.50) = 1 - 0.9938 = 0.0062. Finally, we find P(z < -1.50 or z > 2.50) = 0.0668 + 0.0062 = 0.0730.

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Key Concepts

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

The Standard Normal Distribution
Understanding the standard normal distribution is crucial for students grappling with statistics. This is a specific instance of a normal distribution where the mean (average) is zero, denoted as \(\mu=0\), and the standard deviation (the measure of spread) is one, \(\sigma=1\). Conceptually, this means that the data is centered around the middle value of zero, with most outcomes falling within a certain distance (standard deviation) from the center.

What makes the standard normal distribution so important is its role as a reference point. Due to its standardized nature, researchers and students alike can relate real-world data to this model, converting any normal distribution to the standard normal distribution through a process called standardization. This is done by subtracting the mean and dividing by the standard deviation of the distribution in question, yielding a 'z-score'.

The beauty of the standard normal distribution lies in its universality - all possible z-scores and corresponding probabilities are well-charted out and can be applied to any normally distributed variable.
Navigating the Z-table
The z-table is a vital tool for statisticians and students working with normal distributions. It鈥檚 essentially a chart that outlines the probability that a statistic is less than a given z-value. The z-table helps us navigate the intricate landscape of probabilities without having to perform complex calculus integrations each time.

To effectively use a z-table, one must be able to find the corresponding z-score for the data point in question. This z-score tells us how many standard deviations away from the mean our data point is. With this score in hand, the z-table provides the cumulative probability up to that point in the standard normal distribution.

For instance, in the given exercise, when asked to calculate probabilities like \(P(z < 0.10)\), it鈥檚 the z-table that students turn to. It鈥檚 worth noting that z-tables may give the area to the left of a z-value only, so for probabilities greater than a certain z-value, we鈥檇 take the complement by subtracting the table value from one.
Probability Calculations for Normal Distributions
Calculating probabilities for normal distributions can initially seem daunting, but with understanding of the standard normal distribution and the z-table, it becomes a structured process. Students learn to convert scores from any normal distribution to z-scores, as this allows for standardized calculations.

When we perform probability calculations for normal distributions, we are typically finding the likelihood of a value falling within a certain range. This can involve looking for the probability that a value is less than, greater than, or between certain points, as seen in the given exercise. After converting to z-scores, probabilities for 'less than' can be read directly from the z-table. For 'greater than' scenarios, as mentioned earlier, we use the complement rule.

Differential probabilities, like \(P(0.40

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Most popular questions from this chapter

Consider the population of all one-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean \(\mu=5 \mathrm{ml}\) and standard deviation \(\sigma=0.2 \mathrm{ml}\) is a reasonable model for the distribution of the variable \(x=\) amount of red dye in the paint mixture Use the normal distribution to calculate the following probabilities. (Hint: See Example \(6.21 .\) ) a. \(P(x<5.0)\) b. \(P(x<5.4)\) c. \(P(x \leq 5.4)\) d. \(P(4.64.5)\) f. \(P(x>4.0)\)

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 \(\sigma\).

6.99 A symptom validity test (SVT) is sometimes used to confirm diagnosis of psychiatric disorders. The paper "Developing a Symptom Validity Test for Posttraumatic Stress Disorder: Application of the Binomial Distribution" (Journal of Anxiety Disorders [2008]: 1297-1302) investigated the use of SVTs in the diagnosis of post-traumatic stress disorder. One SVT proposed is a 60 -item test (called the MENT test), where each item has only a correct or incorrect response. The MENT test is designed so that responses to the individual questions can be considered independent of one another. For this reason, the authors of the paper believe that the score on the MENT test can be viewed as a binomial random variable with \(n=60 .\) The MENT test is designed to help in distinguishing fictitious claims of post-traumatic stress disorder. The items on the test are written so that the correct response to an item should be relatively obvious, even to people suffering from stress disorders. Researchers have found that a patient with a fictitious claim of stress disorder will try to "fake" the test, and that the probability of a correct response to an item for these patients is 0.7 (compared to 0.96 for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=0.7\) to calculate various probabilities of interest, where \(x=\) number of correct responses on the MENT test for a patient who is trying to fake the test. a. Verify that it is appropriate to use a normal approximation to the binomial distribution in this situation. b. Approximate the following probabilities: $$ \text { i. } \quad P(x=42) $$ ii. \(P(x<42)\) $$ \text { iii. } P(x \leq 42) $$ c. Explain why the probabilities calculated in Part (b) are not all equal. d. The authors calculated the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=0.96,\) they found $$ P(x \leq 42)=.000000000013 $$ Explain why the authors calculated this probability using the binomial formula rather than using a normal approximation. e. The authors propose that someone who scores 42 or less on the MENT exam is faking the test. Explain why this is reasonable, using some of the probabilities from Parts (b) and (d) as justification.

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 mean but different standard deviations.

Consider the following sample of 25 observations on \(x=\) diameter (in centimeters) of CD disks produced by a particular manufacturer: $$ \begin{array}{lccccccc} 15.66 & 15.78 & 15.82 & 15.84 & 15.89 & 15.92 & 15.94 & 15.95 \\ 15.99 & 16.01 & 16.04 & 16.05 & 16.06 & 16.07 & 16.08 & 16.10 \\ 16.11 & 16.13 & 16.13 & 16.15 & 16.15 & 16.19 & 16.22 & 16.27 \end{array} $$ 16.11 16.29 The 13 largest normal scores for a sample of size 25 are 1.964,1.519,1.259,1.064,0.903,0.763,0.636,0.519,0.408 \(0.302,0.200,0.099,\) and \(0 .\) The 12 smallest scores result from placing a negative sign in front of each of the given nonzero scores. Construct a normal probability plot. Is it reasonable to think that the disk diameter distribution is approximately normal? Explain.
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