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Time and motion A time-and-motion study measures the time required for an assembly-line worker to perform a repetitive task. The data show that the time required to bring a part from a bin to its position on an automobile chassis varies from car to car according to a Normal distribution with mean 11 seconds and standard deviation 2 seconds. The time required to attach the part to the chassis follows a Normal distribution with mean 20 seconds and standard deviation 4 seconds. The study finds that the times required for the two steps are independent. A part that takes a long time to position, for example, does not take more or less time to attach than other parts. (a) What is the distribution of the time required for the entire operation of positioning and attaching a randomly selected part? (b) Management's goal is for the entire process to take less than 30 seconds. Find the probability that this goal will be met for a randomly selected part.

Step by step solution

01

Determine the Combined Distribution

We have two independent Normal distributions: one for positioning with mean \( \mu_1 = 11 \) seconds and standard deviation \( \sigma_1 = 2 \) seconds, and another for attaching with mean \( \mu_2 = 20 \) seconds and standard deviation \( \sigma_2 = 4 \) seconds. The time for the entire operation can be modeled by the sum of these two times. Since they are independent, the sum is also Normally distributed with mean \( \mu = \mu_1 + \mu_2 = 11 + 20 = 31 \) seconds and variance \( \sigma^2 = \sigma_1^2 + \sigma_2^2 = 2^2 + 4^2 = 4 + 16 = 20 \). Thus, the standard deviation is \( \sigma = \sqrt{20} \approx 4.47 \) seconds.

02

Express the Probability Goal

Management's goal is for the total time to be less than 30 seconds. We need to find the probability that a randomly selected part will meet this goal. This involves finding \( P(X < 30) \) where \( X \) is the Normal distribution we found in Step 1 with mean 31 and standard deviation 4.47.

03

Calculate the Z-score

The Z-score is a standardized score that lets us find the probability using the standard Normal distribution table. The Z-score is calculated by \( Z = \frac{X - \mu}{\sigma} = \frac{30 - 31}{4.47} \approx -0.223 \).

04

Find the Corresponding Probability

We use the Z-score table (or a calculator) to find the probability that corresponds to \( Z = -0.223 \). The value for this Z-score is approximately 0.4129. This means there is a 41.29% probability that a randomly selected part will take less than 30 seconds to process.

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Key Concepts

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

Independent Events

In the context of probability, independent events refer to situations where the outcome of one event does not affect the outcome of another. This is a crucial concept in the given exercise. Here, we've got two events: the time taken to bring a part to its position and the time to attach it to the automobile chassis. The study specified that these steps are independent. Understanding the independence of these events is important because it allows us to combine their respective Normal distributions directly. This means that whether a worker takes longer or shorter to position the part does not influence how long they will take to attach it. This simplifies our calculations when determining the total time for both actions together.

Z-score

The Z-score is a measure that describes a value's relation to the mean of a group of values. In terms of Normal distribution, it tells us how many standard deviations an element is from the mean. This concept is essential for standardizing different Normal distributions to a common scale, namely the standard Normal distribution, which has a mean of 0 and a standard deviation of 1.In our exercise, we calculated a Z-score to determine the probability of the total time for positioning and attaching a part being less than 30 seconds. Using the formula:\[ Z = \frac{X - \mu}{\sigma} \]where \( X \) is the target time (30 seconds), \( \mu \) is the mean (31 seconds), and \( \sigma \) is the standard deviation (approximately 4.47 seconds), we obtained \( Z \approx -0.223 \). This negative Z-score indicates that 30 seconds is below the mean of our combined distribution, aligning with the goal of being quicker than average.

Probability Calculation

Probability calculation in the context of Normal distribution involves using the Z-score and the standard Normal distribution table to determine the likelihood of a given event happening. This is a valuable tool in the time-and-motion study for predicting how often the goal time for a task will be met.To calculate the probability for our problem, we used the Z-score of \( -0.223 \). By looking at the Z-score table, which maps Z-scores to probabilities, we found that the probability of a part being positioned and attached in less than 30 seconds is approximately 0.4129, or 41.29%.This probability is essential for management to understand how feasible it is for an assembly line to meet efficiency goals. It provides a quantifiable measure of success based on the variability of task completion times.

Time-and-Motion Study

A time-and-motion study is a business efficiency technique that involves the close examination and analysis of tasks or procedures with the aim of optimizing them. In practical terms, it helps break down processes to identify areas where time can be saved. For this exercise, the study aimed to measure two times: the time to position a part and the time to attach it. Both times were found to follow a Normal distribution. Such studies are valuable for understanding variability in task execution and formulating strategies to minimize it. In applications like assembly lines, these studies provide insights into improving productivity. By knowing the distribution of task times, companies can set realistic goals, like the aim of completing both procedures within 30 seconds, and then calculate the probability of achieving these goals through statistical means.