91Ó°ÊÓ

Let \(X_{1}, X_{2}\), and \(X_{3}\) represent the times necessary to perform three successive repair tasks at a 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 \(\mu_{1}=\mu_{2}=\mu_{3}=60\) and \(\sigma_{1}^{2}=\sigma_{2}^{2}=\) \(\sigma_{3}^{2}=15\), calculate \(P\left(X_{1}+X_{2}+X_{3} \leq 200\right)\) What is \(P\left(150 \leq X_{1}+X_{2}+X_{3} \leq 200\right) ?\) b. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate \(P(55 \leq \bar{X})\) and \(P(58 \leq \bar{X} \leq 62)\). c. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate \(P\left(-10 \leq X_{1}-.5 X_{2}-.5 X_{3} \leq 5\right)\). d. If \(\mu_{1}=40, \quad \mu_{2}=50, \quad \mu_{3}=60, \quad \sigma_{1}^{2}=10\), \(\sigma_{2}^{2}=12\), and \(\sigma_{3}^{2}=14\), calculate \(P\left(X_{1}+\right.\) \(X_{2}+X_{3} \leq 160\) ) and \(P\left(X_{1}+X_{2} \geq 2 X_{3}\right)\).

Step by step solution

01

Summation and Distribution for Part A

For part (a), the sum of the independent normal random variables is also a normal random variable. So, \(X_1 + X_2 + X_3\) has mean \(\mu_{total} = 3 \times 60 = 180\) and variance \(\sigma_{total}^2 = 3 \times 15 = 45\). Since the distribution is normal, \(X_1 + X_2 + X_3 \sim N(180, 45)\).

02

Calculating Probability for Part A(i)

To find \(P(X_1 + X_2 + X_3 \leq 200)\), standardize using the transformation \(Z = \frac{(X_1 + X_2 + X_3) - 180}{\sqrt{45}}\). Compute \(Z = \frac{200 - 180}{\sqrt{45}} \approx 2.98\). Using a standard normal table, \(P(Z \leq 2.98) \approx 0.9986\), so \(P(X_1 + X_2 + X_3 \leq 200) = 0.9986\).

03

Calculating Probability for Part A(ii)

To find \(P(150 \leq X_1 + X_2 + X_3 \leq 200)\), calculate \(P(X_1 + X_2 + X_3 \geq 150)\). Standardize it to get \(Z = \frac{150 - 180}{\sqrt{45}} = -4.47\). Thus, \(P(Z \geq -4.47) \approx 1\). Hence, \(P(150 \leq X_1 + X_2 + X_3 \leq 200) = P(Z \leq 2.98) - P(Z \leq -4.47) \approx 0.9986\).

04

Mean and Distribution for Part B

Calculate the sample mean \(\bar{X} = \frac{X_1 + X_2 + X_3}{3}\). Then, \(\mu_{\bar{X}} = 60\) and \(\sigma_{\bar{X}}^2 = \frac{15}{3} = 5\). Thus, \(\bar{X} \sim N(60, \sqrt{5})\).

05

Calculating Probability for Part B(i)

For \(P(55 \leq \bar{X})\), calculate \(Z = \frac{55 - 60}{\sqrt{5}} = -2.24\). So, \(P(\bar{X} \geq 55) = P(Z \geq -2.24) \approx 0.9878\).

06

Calculating Probability for Part B(ii)

Find \(P(58 \leq \bar{X} \leq 62)\). Calculate \(Z_1 = \frac{58 - 60}{\sqrt{5}} = -0.89\) and \(Z_2 = \frac{62 - 60}{\sqrt{5}} = 0.89\). Thus, \(P(-0.89 \leq Z \leq 0.89) \approx 0.6229\).

07

Calculation for Part C

For \(Y = X_1 - 0.5X_2 - 0.5X_3\), calculate \(\mu_Y = \mu_1 - 0.5\mu_2 - 0.5\mu_3 = 60 - 30 - 30 = 0\). The variance \(\sigma_Y^2 = 15 + 0.25 \times 15 + 0.25 \times 15 = 22.5\). Thus, \(Y \sim N(0, \sqrt{22.5})\).

08

Calculating Probability for Part C

Standardize the inequality \(-10 \leq Y \leq 5\) using \(Z = \frac{Y - 0}{\sqrt{22.5}}\), which means compute \(Z_1 = \frac{-10}{\sqrt{22.5}} = -2.11\) and \(Z_2 = \frac{5}{\sqrt{22.5}} = 1.06\). Thus, \(P(-2.11 \leq Z \leq 1.06) \approx 0.8315\).

09

Mean and Variance for Part D

For part (d), \(X_1 + X_2 + X_3\) has mean \(\mu_{total} = 40 + 50 + 60 = 150\) and variance \(\sigma_{total}^2 = 10 + 12 + 14 = 36\). So, \(X_1 + X_2 + X_3 \sim N(150, 6)\).

10

Calculating Probability for Part D(i)

Find \(P(X_1 + X_2 + X_3 \leq 160)\). Standardize using \(Z = \frac{X_1 + X_2 + X_3 - 150}{6}\), yielding \(Z = \frac{160 - 150}{6} \approx 1.67\). So, \(P(Z \leq 1.67) \approx 0.9525\).

11

Calculating Probability for Part D(ii)

For \(P(X_1 + X_2 \geq 2X_3)\), substitute \(X_1 + X_2 - 2X_3 \leq 0\). Mean \(= 40 + 50 - 2\times60 = -30\) and variance \(= 10 + 12 + 4\times14 = 68\). Standardize the probability and find \(P(Z \leq 0)\), corresponding to \(Z = \frac{0 - (-30)}{\sqrt{68}} \approx 3.64\), which gives \(P(Z \leq 0) \approx 0.9999\).

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Key Concepts

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

Normal Distribution

The normal distribution is a fundamental concept in statistics utilized for understanding how different data outcomes are expected to behave. It is a continuous probability distribution that is symmetrical around its mean, closely resembling the shape of a bell curve. This makes it a powerful tool in real-world applications where data tends to cluster around a central value with no skewness.

A normal distribution is characterized by two parameters: its mean (\( \mu \)) and its variance (\( \sigma^2 \)).

• The mean determines the peak of the distribution and defines where the center is located on the horizontal axis.
• The variance measures how spread out the data points are around the mean. Small variance indicates data points are closely packed, whereas large variance suggests they are more dispersed.

In the original exercise, the repair times for tasks are assumed to follow a normal distribution. This assumption simplifies the calculation of probabilities for various scenarios, such as the total repair time for multiple tasks or comparing averages. Each of these uses the properties of the normal distribution to compute results involving real-world phenomena.

Probability Calculation

Calculating the probability of certain outcomes is a key process in statistics and involves using known distributions to predict the likelihood of an event. In scenarios involving normal distributions, this typically involves 'standardizing' results to fit the standard normal distribution—commonly referred to as a z-transformation.

The formula for standardizing a random variable (\( X \)) to find a z-score is:\[ Z = \frac{(X - \mu)}{\sigma} \]This helps map any normal distribution to the standard normal, which has a mean of 0 and a standard deviation of 1. Using standard normalization tables or software, you can then find the probability of a z-score occurring.

• In the exercise, to find the probability that the sum of the repair times is less than or equal to a certain value, we standardize this sum and utilize standard normal probability tables to find this likelihood.
• Similarly, for calculating between two values, say \( 150 \leq X_1 + X_2 + X_3 \leq 200 \), we use the z-transformation on both boundary values and use the table to extract the respective probabilities.

Random Variables

Random variables are fundamental in probability theory and statistics, representing values that result from stochastic processes. There are two main types: discrete and continuous. The exercises utilize continuous random variables, which can take any value within a range, such as time, length, and repair intervals.

In a practical sense, normal random variables are extensively used because they model many natural and social phenomena effectively due to the Central Limit Theorem. According to this theorem, the sum of a large number of independent and identically distributed random variables will tend to follow a normal distribution, irrespective of the original distribution.

• In the original exercise, \( X_1, X_2, \) and \( X_3 \) are independent normal random variables representing different repair times. Their independence allows us to sum their means and variances, applying these cumulative parameters to find probabilities for combined events like finding \( P(X_1 + X_2 + X_3 \leq 200) \).
• This principle provides a significant advantage, simplifying the computations needed for assessing the impact of multiple random events combined, which underscores their critical role in statistical analysis.