91Ó°ÊÓ

Let \(X_{1}, X_{2}\), and \(X_{3}\) represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv's with expected values \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) and variances \(\sigma_{1}^{2}, \sigma_{2}^{2}\), and \(\sigma_{3}^{2}\), respectively. a. If \(\mu=\mu_{2}=\mu_{3}=60\) and \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=15\), calculate \(P\left(T_{o} \leq 200\right)\) and \(P\left(150 \leq T_{o} \leq 200\right)\) ? b. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate both \(P(55 \leq \bar{X})\) and \(P(58 \leq \bar{X} \leq 62)\). c. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate and interpret \(P\left(-10 \leq X_{1}-.5 X_{2}-.5 X_{3} \leq 5\right)\). d. If \(\mu_{1}=40, \mu_{2}=50, \mu_{3}=60, \sigma_{1}^{2}=10, \sigma_{2}^{2}=12\), and \(\sigma_{3}^{2}=14\), calculate \(P\left(X_{1}+X_{2}+X_{3} \leq 160\right)\) and also \(P\left(X_{1}+X_{2} \geq 2 X_{3}\right)\).

Step by step solution

01

Calculate Expected Value and Variance of Total Repair Time (a)

The total time is given by \(T_o = X_1 + X_2 + X_3\). The expected value of \(T_o\) is \(\mu_1 + \mu_2 + \mu_3 = 40 + 60 + 60 = 160\). Since the variances add for independent variables, the variance is \(\sigma_1^2 + \sigma_2^2 + \sigma_3^2 = 15 + 15 + 15 = 45\). Thus, \(T_o\) is normally distributed with mean 160 and variance 45.

02

Calculate Probability (Part a)

To find \(P(T_o \leq 200)\), we standardize the variable: \(Z = \frac{T_o - 160}{\sqrt{45}}\). Thus, \(P(T_o \leq 200) = P\left(Z \leq \frac{200 - 160}{\sqrt{45}}\right) = P(Z \leq 5.961)\). Use the standard normal distribution table or calculator to find this probability, which is essentially 1.

03

Calculate Probability Range (Part a)

For \(P(150 \leq T_o \leq 200)\), calculate for both bounds: \(P(150 \leq T_o \leq 200) = P\left(\frac{150 - 160}{\sqrt{45}} \leq Z \leq \frac{200 - 160}{\sqrt{45}}\right) = P(-1.494 \leq Z \leq 5.961)\). From standard normal tables, \(P(-1.494 \leq Z \leq 5.961) = 0.9319\).

04

Calculate Expected Value and Variance of Average Time (b)

The average time is \(\bar{X} = \frac{X_1 + X_2 + X_3}{3}\), so \(E(\bar{X}) = \frac{160}{3} = 53.3\), and \(Var(\bar{X}) = \frac{45}{3^2} = 5\).

05

Calculate Probability for Mean (b)

For \(P(55 \leq \bar{X})\), standardize using \(Z = \frac{\bar{X} - 53.3}{\sqrt{5}}\), then \(P(55 \leq \bar{X}) = P(Z \geq 0.758)\), which gives roughly 0.2249 from the table or calculator.

06

Calculate Probability Range for Mean (b)

For \(P(58 \leq \bar{X} \leq 62)\), standardize both bounds: \(-P(Z \leq \frac{58 - 53.3}{\sqrt{5}}) + P(Z \leq \frac{62 - 53.3}{\sqrt{5}})\), resulting in probabilities yielding 0.0548.

07

Transformation of Variables (c)

For \(Y = X_1 - 0.5X_2 - 0.5X_3\), find \(E[Y] = 40 - 0.5(60) - 0.5(60) = -10\). Calculate variance: \(Var(Y) = 10 + 0.25(15) + 0.25(15) = 17.5\). \(Y\) is normally distributed.

08

Calculate Probability Range (c)

Standardize the range: \(P(-10 \leq Y \leq 5) = P\left(\frac{-10 + 10}{\sqrt{17.5}} \leq Z \leq \frac{5 + 10}{\sqrt{17.5}}\right) = P(0 \leq Z \leq 3.58)\), yielding approximately 0.5 from standard normal.

09

Calculate Probability for New Parameters (d)

Compute \(E[X_1 + X_2 + X_3] = 40 + 50 + 60 = 150\), and find variance \(Var(X_1 + X_2 + X_3) = 10 + 12 + 14 = 36\). For \(P(X_1 + X_2 + X_3 \leq 160)\), \(P(Z \leq 1.67)\) gives 0.9525.

10

Calculate Inequality (d)

For \(P(X_1 + X_2 \geq 2X_3)\), transform to \(P(X_1 + X_2 - 2X_3 \geq 0)\), with mean \(-30\) and variance \(48\). Calculate \(P(Z \geq 2.17)\), which is 0.0150.

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Key Concepts

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

Normal Distribution

A normal distribution, often known as a bell curve, is a probability distribution that is symmetric about the mean. It means the data around the mean value are distributed in a way where most occurrences take place close to the mean. The characteristic bell shape reflects how numbers are distributed; there are fewer occurrences the further away you move from the center.
For math enthusiasts, the probability density for a normal distribution is given by the formula \[ f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2} \left(\frac{x - \mu}{\sigma}\right)^2} \]. Here, \( \mu \) is the mean, and \( \sigma^2 \) is the variance.
This distribution is crucial in statistics because it describes many phenomena naturally occurring in life, from heights of people to errors in measurements.

• When dealing with normally distributed variables, you are often interested in probabilities involving the range of these variables.
• Normal distributions are completely described by two parameters: the mean (\(\mu\)) and the standard deviation (\(\sigma\)).

Expected Value

The expected value of a random variable is a fundamental concept in probability theory. It provides a measure of the center of the distribution of the variable and can be thought of as the "average" value you would expect if you were to repeat an experiment many times.
Mathematically, for a discrete variable, the expected value is calculated using \( E[X] = \sum x_i P(x_i) \) where \( x_i \) are the possible values of the random variable and \( P(x_i) \) is the probability of each value.

• The expected value can also be applied to continuous distributions using an integral form.
• In the exercise, each task's expected time indicates how long, on average, each repair is predicted to take.

Understanding expected values helps make better predictions and decisions based on probabilistic outcomes.

Variance

Variance is a measure of how spread out numbers are in a distribution. It tells you the degree to which each number differs from the mean in a dataset.
The formula for variance for a set of numbers is given by \( \sigma^2 = \frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2 \). Here, \( \mu \) is the mean of the data.

• In essence, a larger variance indicates that the data points are more spread out.
• For independent random variables, variances add up. Hence, if you have independent tasks, the total variance reflects the added variation from each task.

This concept is crucial because it influences the confidence in predictions made using those parameters. Lower variance means data tends to be closer to the mean, indicating predictability.

Independent Random Variables

Independent random variables are those whose outcomes do not influence each other. In simpler terms, the result of one variable does not change the probability of outcomes for another.
Mathematically, two random variables \(X\) and \(Y\) are independent if \(P(X \cap Y) = P(X)P(Y)\). This essentially means that joint probabilities can be calculated as the product of their probabilities.

• An important implication is that the variance of independent variables for a sum is the sum of their variances.
• Independence simplifies calculations in probability because the behavior of any one variable doesn’t affect the others.

In a practical context, understanding independence is crucial when trying to model or predict outcomes that are assumed to not interact, like the repair times being independent in a series of tasks.