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Chapter 5: Problem 62

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15,30 , and 20 min, respectively, and the standard deviations are 1,2 , and \(1.5 \mathrm{~min}\), respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

Short Answer

Expert verified
The probability is approximately 0.0317, or 3.17%.

Step by step solution

01

Define the Problem

We need to find the probability that the total time for three independent normally distributed machining operations is at most 1 hour.
02

Define the Total Machining Time

Let \( T \) be the total time for the three operations. Then, \( T = T_1 + T_2 + T_3 \), where \( T_1 \sim N(15, 1^2) \), \( T_2 \sim N(30, 2^2) \), and \( T_3 \sim N(20, 1.5^2) \).
03

Determine the Distribution of Total Time

Since \( T_1 \), \( T_2 \), and \( T_3 \) are independent, \( T \) is normally distributed with mean:\[\mu_T = \mu_1 + \mu_2 + \mu_3 = 15 + 30 + 20 = 65\]and variance:\[\sigma_T^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 = 1^2 + 2^2 + 1.5^2 = 1 + 4 + 2.25 = 7.25\]
04

Standardize the Total Time to find Probability

We're looking for \( P(T \leq 60) \). Standardize \( T \) to a standard normal variable \( Z \):\[Z = \frac{T - \mu_T}{\sigma_T} = \frac{60 - 65}{\sqrt{7.25}}\]
05

Calculate the Standardized Value

Substitute the values into the standard normal variable:\[Z = \frac{60 - 65}{\sqrt{7.25}} = \frac{-5}{\sqrt{7.25}} \approx -1.857\]
06

Use the Standard Normal Table

From the standard normal distribution table, find \( P(Z \leq -1.857) \). This gives us the required probability.

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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 the realm of statistics, easily recognized by its bell-shaped curve. Often referred to as the Gaussian distribution, it describes how the values of a variable are spread out. Most of the data points fall around the mean, with fewer points present as you move further from the mean.

Key features to note:
  • The mean (\( \mu \)) defines the center of the distribution. It's the average of all data points.
  • The standard deviation (\( \sigma \)) measures the spread of data around the mean. A higher standard deviation signifies data that is more spread out.
  • The total area under the curve is 1, representing the entirety of data distribution.
In the context of machining operations, the time taken for each operation is normally distributed, implying a known average time with predictable variations.
Machining Operations
Machining operations involve a series of processes such as cutting, shaping, or drilling a piece of material into the desired form. In manufacturing, these operations need to be performed within specific timeframes to ensure efficiency.

Each machining operation is independent of one another in the scenario provided. Independence here means that the time it takes to finish one operation does not affect the time taken for another. These operations typically follow a normal distribution due to the natural variability of the processes involved. By understanding the distribution of machining times, manufacturers can predict and optimize their production procedure better.
Standard Normal Distribution
The standard normal distribution is a special form of the normal distribution where the mean is 0 and the standard deviation is 1. This standard form allows for easier computation of probabilities and data comparisons.

Why is this important?
  • By converting any normal distribution into a standard normal distribution using the transformation \( Z = \frac{X - \mu}{\sigma} \), we can utilize standard normal distribution tables to find probabilities, which simplifies the process significantly.
  • This conversion is sometimes referred to as "standardizing" a variable, like how we did with the machining operation times in the exercise.
  • Once a variable is in its standard normal form, probability values can be quickly referenced from statistical tables or calculated using statistical software.
Independent Random Variables
Independent random variables are variables that do not influence each other. Their outcomes do not impact the probability distribution of other variables, which allows for straightforward calculations of their combined distributions.

Important points include:
  • When dealing with independent normal random variables, the combined distribution is also normally distributed. This is particularly useful for calculations involving multiple processes.
  • The mean of the combined distribution is the sum of the individual means, e.g., the mean machining time for all operations.
  • For variances, because they indicate spread and not centrality, you add the variances of the individual random variables to get the total variance.
In manufacturing, understanding that machining times are independent helps in assessing the overall process efficiency and scheduling production accurately.

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Most popular questions from this chapter

Two components of a minicomputer have the following joint pdf for their useful lifetimes \(X\) and \(Y\) : $$ f(x, y)=\left\\{\begin{array}{cc} x e^{-x(1+y)} & x \geq 0 \text { and } y \geq 0 \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the probability that the lifetime \(X\) of the first component exceeds 3 ? b. What are the marginal pdf's of \(X\) and \(Y\) ? Are the two lifetimes independent? Explain. c. What is the probability that the lifetime of at least one component exceeds 3 ?

You have two lightbulbs for a particular lamp. Let \(X=\) the lifetime of the first bulb and \(Y=\) the lifetime of the second bulb (both in 1000 s of hours). Suppose that \(X\) and \(Y\) are independent and that each has an exponential distribution with parameter \(\lambda=1\). a. What is the joint pdf of \(X\) and \(Y\) ? b. What is the probability that each bulb lasts at most 1000 hours (i.e., \(X \leq 1\) and \(Y \leq 1\) )? c. What is the probability that the total lifetime of the two bulbs is at most 2 ? [Hint: Draw a picture of the region \(A=\\{(x, y): x \geq 0, y \geq 0, x+y \leq 2\\}\) before integrating.] d. What is the probability that the total lifetime is between 1 and 2 ?

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable- \(X\) for the right tire and \(Y\) for the left tire, with joint pdf $$ f(x, y)=\left\\{\begin{array}{cc} K\left(x^{2}+y^{2}\right) & 20 \leq x \leq 30,20 \leq y \leq 30 \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the value of \(K\) ? b. What is the probability that both tires are underfilled? c. What is the probability that the difference in air pressure between the two tires is at most 2 psi? d. Determine the (marginal) distribution of air pressure in the right tire alone. e. Are \(X\) and \(Y\) independent rv's?

When an automobile is stopped by a roving safety patrol, each tire is checked for tire wear, and each headlight is checked to see whether it is properly aimed. Let \(X\) denote the number of headlights that need adjustment, and let \(Y\) denote the number of defective tires. a. If \(X\) and \(Y\) are independent with \(p_{X}(0)=.5, p_{\mathcal{K}}(1)=.3\), \(p_{X}(2)=.2\), and \(p_{Y}(0)=.6, p_{Y}(1)=.1, p_{Y}(2)=p_{Y}(3)=.05\), and \(p_{Y}(4)=.2\), display the joint pmf of \((X, Y)\) in a joint probability table. b. Compute \(P(X \leq 1\) and \(Y \leq 1)\) from the joint probability table, and verify that it equals the product \(P(X \leq 1)\). \(P(Y \leq 1)\) c. What is \(P(X+Y=0)\) (the probability of no violations)? d. Compute \(P(X+Y \leq 1)\).

A binary communication channel transmits a sequence of "bits" ( 0 s and 1s). Suppose that for any particular bit transmitted, there is a \(10 \%\) chance of a transmission error (a 0 becoming a 1 or a 1 becoming a 0 ). Assume that bit errors occur independently of one another. a. Consider transmitting 1000 bits. What is the approximate probability that at most 125 transmission errors occur? b. Suppose the same 1000 -bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within 50 of the number of errors in the second?
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