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Chapter 3: Problem 3

If \(X\) is \(N\left(\mu, \sigma^{2}\right)\), find \(b\) so that \(P[-b<(X-\mu) / \sigma

Short Answer

Expert verified
The value of b, i.e., the number of standard deviations away from the mean where approximately 90% of the data is located, can be determined via z-table or the pnorm function in R or S-PLUS.

Step by step solution

01

Transform the problem into standard normal form

The standard normal distribution is often represented by \(Z \sim N(0,1)\) where \(Z = \frac{(X - \mu)}{\sigma}\). The given problem can be transformed in terms of Z as: \(P(-b<Z<b) = 0.90\)
02

Understand Z table or pnorm function

The standard normal distribution table (also known as Z table) gives the probability that a standard normal random variable Z is less than or equal to z. Similarly, the pnorm function in R or S-PLUS provides the same functionality, it gives the probability that a standard normal variable is less than or equal to a certain value.
03

Find the Z score corresponding to the 90% probability

As the standard normal distribution is symmetrical around zero, for an area of 0.90 to be present between -b and b, there would be 0.05 in each tail beyond these points (because 0.90 + 2*0.05 = 1.00 or 100% of the distribution). By looking up 0.05 in the Z table or using the pnorm function in R or S-PLUS in 'lower.tail = FALSE' mode, the corresponding value b can be found. Let's use R in this case: b = abs(qnorm(0.05, lower.tail = FALSE))
04

Interpret the result

The value obtained for 'b' refers to the number of standard deviations away from the mean between which approximately 90% of the data in a normal distribution lies.

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Key Concepts

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

Understanding the Z Score
The Z score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If you have a Z score of 0, it indicates that the value is exactly at the mean, while a Z score of 1.0 signifies that the value is one standard deviation above the mean. Conversely, a Z score of -1.0 signifies that the value is one standard deviation below the mean.

When applied to the exercise, finding a Z score involves standardizing the given normal random variable so that the resulting value can be compared against the standard normal distribution. In our case, the transform \(Z = \frac{\(X - \mu\)}{\sigma}\) is used to convert the variable into a Z score, allowing us to compare it to the standard normal distribution which has a mean of 0 and a standard deviation of 1.

In practical terms, calculating the Z score for a given probability in the context of a standard normal distribution allows us to understand how extreme or unusual an event is. For instance, a high Z score represents an event that is very unusual, which has occurred far away from the mean. This concept is critical for performing statistical tests and is widely used in various fields such as finance, quality control, and psychology.
The Normal Random Variable
A normal random variable is a variable that has a probability distribution which is normal, also known as a Gaussian distribution. The normal distribution is symmetric about its mean, and its shape is defined by its mean (\(\mu\)) and its variance (\(\sigma^2\)). It has the famous bell-shaped curve where the majority of the observations cluster around the central peak and the probabilities for values further from the mean taper off equally in both directions.

In the context of our exercise, the variable \(X\) is normally distributed with mean \(\mu\) and variance \(\sigma^2\). This means that if we pick a random value for \(X\), it is most likely to be close to the mean \(\mu\), and the likelihood decreases as we move away from the mean. To find the specific bounds within which a certain percentage of the values lie, we convert this variable into a standard normal variable \(Z\) by subtracting the mean and dividing by the standard deviation. This conversion is essential when we aim to use the standard normal tables or computational tools like R for determining probabilities associated to \(X\).

Understanding the properties of a normal random variable is crucial because it serves as the foundation for many statistical methods and analyses. The techniques derived from it aid in making predictions and conclusions about data that is assumed to be normally distributed.

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Most popular questions from this chapter

Consider a random variable \(X\) of the continuous type with cdf \(F(x)\) and pdf \(f(x)\). The hazard rate (or failure rate or force of mortality) is defined by $$ r(x)=\lim _{\Delta \rightarrow 0} \frac{P(x \leq X0\). (a) Show that \(r(x)=f(x) /(1-F(x))\). (b) If \(r(x)=c\), where \(c\) is a positive constant, show that the underlying distribution is exponential. Hence, exponential distributions have constant failure rate over all time. (c) If \(r(x)=c x^{b}\), where \(c\) and \(b\) are positive constants, show that \(X\) has a Weibull distribution, i.e., $$ f(x)=\left\\{\begin{array}{ll} c x^{b} \exp \left\\{-\frac{c x^{b+1}}{b+1}\right\\} & 0

Let the mutually independent random variables \(X_{1}, X_{2}\), and \(X_{3}\) be \(N(0,1)\), \(N(2,4)\), and \(N(-1,1)\), respectively. Compute the probability that exactly two of these three variables are less than zero.

Determine the constant \(c\) in each of the following so that each \(f(x)\) is a \(\beta\) pdf: (a) \(f(x)=c x(1-x)^{3}, 0

If \(M\left(t_{1}, t_{2}\right)\) is the mgf of a bivariate normal distribution, compute the covariance by using the formula $$ \frac{\partial^{2} M(0,0)}{\partial t_{1} \partial t_{2}}-\frac{\partial M(0,0)}{\partial t_{1}} \frac{\partial M(0,0)}{\partial t_{2}} $$ Now let \(\psi\left(t_{1}, t_{2}\right)=\log M\left(t_{1}, t_{2}\right) .\) Show that \(\partial^{2} \psi(0,0) / \partial t_{1} \partial t_{2}\) gives this covariance directly.

On the average a grocer sells 3 of a certain article per week. How many of these should he have in stock so that the chance of his running out within a week will be less than 0.01? Assume a Poisson distribution.
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