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Let \(Z\) be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. a. \(P(0 \leq Z \leq 2.17)\) b. \(P(0 \leq Z \leq 1)\) c. \(P(-2.50 \leq Z \leq 0)\) d. \(P(-2.50 \leq Z \leq 2.50)\) e. \(P(Z \leq 1.37)\) f. \(P(-1.75 \leq Z)\) g. \(P(-1.50 \leq Z \leq 2.00)\) h. \(P(1.37 \leq Z \leq 2.50)\) i. \(P(1.50 \leq Z)\) j. \(P(|Z| \leq 2.50)\)

Short Answer

Expert verified
a. 0.4850, b. 0.3413, c. 0.4938, d. 0.9876, e. 0.9147, f. 0.9599, g. 0.9104, h. 0.0791, i. 0.0668, j. 0.9876.

Step by step solution

01

Understanding the Standard Normal Distribution

The standard normal distribution, denoted as \( Z \), has a mean of 0 and a standard deviation of 1. Probabilities involving \( Z \) can be found using the standard normal distribution table (z-table) or a calculator that provides cumulative probabilities. Area under the standard normal curve represents probabilities.
02

Solving Probability a

To find \( P(0 \leq Z \leq 2.17) \), we use the cumulative distribution function (CDF). This is the probability that \( Z \) is less than or equal to 2.17 minus the probability that \( Z \) is less than 0, e.g., \( P(0 \leq Z \leq 2.17) = P(Z \leq 2.17) - P(Z \leq 0) \). According to the standard normal table, \( P(Z \leq 2.17) = 0.9850 \) and \( P(Z \leq 0) = 0.5 \). Therefore, \( 0.9850 - 0.5 = 0.4850 \).
03

Solving Probability b

For \( P(0 \leq Z \leq 1) \), calculate \( P(Z \leq 1) - P(Z \leq 0) \). From the z-table, \( P(Z \leq 1) = 0.8413 \) and \( P(Z \leq 0) = 0.5 \), thus \( P(0 \leq Z \leq 1) = 0.8413 - 0.5 = 0.3413 \).
04

Solving Probability c

To find \( P(-2.50 \leq Z \leq 0) \), use \( P(Z \leq 0) - P(Z \leq -2.50) \). With \( P(Z \leq 0) = 0.5 \) and \( P(Z \leq -2.50) = 0.0062 \), the probability is \( 0.5 - 0.0062 = 0.4938 \).
05

Solving Probability d

For \( P(-2.50 \leq Z \leq 2.50) \), add \( P(0 \leq Z \leq 2.50) \) and \( P(-2.50 \leq Z \leq 0) \). Since \( P(Z \leq 2.50) = 0.9938 \) and \( P(Z \leq -2.50) = 0.0062 \), the result is \( 0.9938 - 0.0062 = 0.9876 \).
06

Solving Probability e

For \( P(Z \leq 1.37) \), lookup directly from the z-table, which provides \( P(Z \leq 1.37) = 0.9147 \).
07

Solving Probability f

For \( P(-1.75 \leq Z) \), this is 1 minus \( P(Z \leq -1.75) \). With \( P(Z \leq -1.75) = 0.0401 \), the probability is \( 1 - 0.0401 = 0.9599 \).
08

Solving Probability g

To calculate \( P(-1.50 \leq Z \leq 2.00) \), use \( P(Z \leq 2.00) - P(Z \leq -1.50) \). From the z-table, \( P(Z \leq 2.00) = 0.9772 \), and \( P(Z \leq -1.50) = 0.0668 \), giving \( 0.9772 - 0.0668 = 0.9104 \).
09

Solving Probability h

For \( P(1.37 \leq Z \leq 2.50) \), calculate \( P(Z \leq 2.50) - P(Z \leq 1.37) \). With \( P(Z \leq 2.50) = 0.9938 \) and \( P(Z \leq 1.37) = 0.9147 \), the result is \( 0.9938 - 0.9147 = 0.0791 \).
10

Solving Probability i

For \( P(1.50 \leq Z) \), this is 1 minus \( P(Z \leq 1.50) \). Lookup gives \( P(Z \leq 1.50) = 0.9332 \), hence \( P(1.50 \leq Z) = 1 - 0.9332 = 0.0668 \).
11

Solving Probability j

For \( P(|Z| \leq 2.50) \), we need \( P(-2.50 \leq Z \leq 2.50) \), which is found as \( P(Z \leq 2.50) - P(Z \leq -2.50) \). Since it's symmetric, use earlier result: \( 0.9938 - 0.0062 = 0.9876 \).

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Key Concepts

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

Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is crucial when working with the standard normal distribution. Essentially, it helps us find the probability that a random variable is less than or equal to a certain value. For a standard normal distribution, the CDF, denoted as \( F(Z) \), gives the probability that \( Z \) will be less than or equal to a given z-value.
For example, \( F(Z = 2.17) = 0.9850 \) tells us that there is a 98.50% chance that the standard normal random variable \( Z \) is less than or equal to 2.17. This function is very helpful because it allows us to convert a complex area under the normal curve into an easily interpretable percentage or probability. Understanding the CDF is key to determining probabilities in various scenarios, as it simplifies the process significantly.
Z-Table
The Z-Table, also known as the standard normal distribution table, is a helpful tool in statistical analysis involving normal distributions. It lists the cumulative probabilities associated with each z-value, which are essentially distances from the mean measured in terms of standard deviation. When you look up a value in the Z-Table, you're finding the area under the normal curve to the left of that z-score.
Suppose we want \( P(Z \leq 1) \). By looking up 1 in the Z-Table, we find 0.8413, indicating an 84.13% probability that \( Z \) is less than or equal to 1. It simplifies the task of finding probabilities without needing extensive calculations by hand, making it a staple in solving standard normal distribution problems.
Probability Calculation
Probability calculations in the context of the standard normal distribution involve determining the likelihood of events occurring within certain bounds. This often uses CDF and Z-Table data.
For instance, to determine \( P(-1.50 \leq Z \leq 2.00) \), the process involves two steps: first, find \( P(Z \leq 2.00) \), which is 0.9772 from the Z-Table, and second, find \( P(Z \leq -1.50) \), which is 0.0668. Subtract these probabilities to get the final result (\( 0.9104 \)).
By breaking down probability calculations into simple subtractions or additions of CDF values, we can tackle complex probability questions with ease.
Normal Curve
The normal curve, represented by a bell-shaped graph, shows the distribution of data around a mean (in our case, 0 for the standard normal distribution) with a standard deviation (1 in our case). This curve is symmetrical, meaning the probability of being in a specific range is the same on both sides of the mean.
Key characteristics include its symmetry, single peak at the mean, and the property that about 68% of data falls within one standard deviation of the mean, about 95% within two, and about 99.7% within three.
Visualizing problems on the curve, drawing shaded areas for distributions can help better grasp concepts and probabilities. Understanding the normal curve helps us comprehend how distributions work, providing valuable intuition for estimating probabilities and solving related problems.

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Most popular questions from this chapter

The error involved in making a certain measurement is a continuous rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cc} .09375\left(4-x^{2}\right) & -2 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Compute \(P(X>0)\). c. Compute \(P(-1.5)\).

The mode of a continuous distribution is the value \(x^{*}\) that maximizes \(f(x)\). a. What is the mode of a normal distribution with parameters \(\mu\) and \(\sigma\) ? b. Does the uniform distribution with parameters \(A\) and \(B\) have a single mode? Why or why not? c. What is the mode of an exponential distribution with parameter \(\lambda\) ? (Draw a picture.) d. If \(X\) has a gamma distribution with parameters \(\alpha\) and \(\beta\), and \(\alpha>1\), find the mode. [Hint: \(\ln [f(x)]\) will be maximized iff \(f(x)\) is, and it may be simpler to take the derivative of \(\ln [f(x)]\).] e. What is the mode of a chi-squared distribution having \(v\) degrees of freedom?

Consider an rv \(X\) with mean \(\mu\) and standard deviation \(\sigma\), and let \(g(X)\) be a specified function of \(X\). The first-order Taylor series approximation to \(g(X)\) in the neighborhood of \(\mu\) is $$ g(X) \Rightarrow g(\mu)+g^{\prime}(\mu) \cdot(X-\mu) $$ The right-hand side of this equation is a linear function of \(X\). If the distribution of \(X\) is concentrated in an interval over which \(g(\cdot)\) is approximately linear [e.g., \(\sqrt{x}\) is approximately linear in \((1,2)\) ], then the equation yields approximations to \(E(g(X))\) and \(V(g(X))\). a. Give expressions for these approximations. [Hint: Use rules of expected value and variance for a linear function \(a X+b .]\) b. If the voltage \(v\) across a medium is fixed but current \(l\) is random, then resistance will also be a random variable related to \(I\) by \(R=v / I\). If \(\mu_{l}=20\) and \(\sigma_{l}=.5\), calculate approximations to \(\mu_{R}\) and \(\sigma_{R}\).

When circuit boards used in the manufacture of compact dise players are tested, the long-run percentage of defectives is \(5 \%\). Suppose that a batch of 250 boards has been received and that the condition of any particular board is independent of that of any other board. a. What is the approximate probability that at least \(10 \%\) of the boards in the batch are defective? b. What is the approximate probability that there are exactly 10 defectives in the batch?

Let \(X\) have a Weibull distribution with the pdf from Expression (4.11). Verify that \(\mu=\beta \Gamma(1+1 / \alpha)\). [Hint: In the integral for \(E(X)\), make the change of variable \(y=(x / \beta)^{\alpha}\), so that \(x=\beta y^{1 / \alpha}\).]

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