/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 The time for an automated system... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The time for an automated system in a warehouse to locate a part is normally distributed with a mean of 45 seconds and a standard deviation of 30 seconds. Suppose that independent requests are made for 10 parts. (a) What is the probability that the average time to locate 10 parts exceeds 60 seconds? (b) What is the probability that the total time to locate 10 parts exceeds 600 seconds?

Short Answer

Expert verified
(a) 0.0571; (b) 0.0571

Step by step solution

01

Understand the Problem and Given Data

We are asked to calculate probabilities related to the time needed for an automated system to locate parts. Given that the time is normally distributed with a mean (\( \mu \)) of 45 seconds and a standard deviation (\( \sigma \)) of 30 seconds for one part. We have 10 independent requests, so we need to consider the distribution of both the average and the total time to locate these parts.
02

Compute Distribution for Average Time of 10 Parts

The mean of the average time for 10 parts is the same as the mean for one part, \( \mu_{\bar{X}} = 45 \) seconds. The standard deviation of the average time (known as the standard error) is \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{30}{\sqrt{10}} \approx 9.49 \). This distribution is normal because the original distribution is normal.
03

Calculate Probability for Average Time Exceeding 60 Seconds

To find the probability that the average time is greater than 60 seconds, standardize 60 using: \( Z = \frac{X - \mu}{\sigma} = \frac{60 - 45}{9.49} \approx 1.58 \). Use the standard normal distribution tables or a calculator to find \( P(Z > 1.58) \). This probability is approximately 0.0571.
04

Compute Distribution for Total Time of 10 Parts

For the total time, the mean is \( \mu_{T} = 10 \times 45 = 450 \) seconds, and the standard deviation is \( \sigma_{T} = \sqrt{10} \times 30 \approx 94.87 \) seconds. This distribution is also normal.
05

Calculate Probability for Total Time Exceeding 600 Seconds

To find the probability that the total time exceeds 600 seconds, standardize 600 with: \( Z = \frac{600 - 450}{94.87} \approx 1.58 \). Again, use the standard normal distribution to find \( P(Z > 1.58) \), which matches the previous \( Z \) value, giving approximately 0.0571.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution
A normal distribution is a continuous probability distribution that is symmetric about its mean, indicating most occurrences take place around the central peak. It is depicted as a bell-shaped curve, where the highest point represents the mean.
Normal distributions are critical in statistics because they describe many natural phenomena, such as test scores, heights of individuals, and yes, even the time it takes to locate items in a warehouse.
Some key points about normal distributions include:
  • The mean, median, and mode are all equal.
  • It is determined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)).
  • About 68% of the data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.
In the context of the exercise, the time required to find parts follows a normal distribution with a mean of 45 seconds and varying distributions when considering single versus multiple parts.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of data values. When we say that the time to locate a part has a standard deviation of 30 seconds, it suggests how much the time values deviate from the average time of 45 seconds.

Key points about standard deviation include:
  • A low standard deviation indicates that the data points tend to be close to the mean.
  • A high standard deviation indicates greater spread or variability around the mean.
The formula for standard deviation is the square root of the variance, calculated as:\[\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \mu)^2}\]where \( N \) is the number of observations, \( x_i \) are the observed values, and \( \mu \) is the mean of the data set. In many real-world applications, standard deviation helps us understand the reliability of an average measure and assess risks or uncertainties related to forecasts or predictions.
Standard Error
The standard error is similar to the standard deviation but applies to sample means. It measures the accuracy with which a sample mean represents the population mean. In the exercise, when analyzing the average time to locate 10 parts, we use the standard error to understand the precision of our calculated average.

Some details about standard error:
  • The standard error decreases as the sample size increases, reflecting higher accuracy of the sample mean.
  • It is calculated as the standard deviation divided by the square root of the sample size (\( n \): \[\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\]
The smaller the standard error, the closer the sample mean is to the actual population mean. In the context of our example, calculating the standard error helps assess the likelihood that the average time differs significantly from the expected mean of 45 seconds when locating multiple parts.
Z-Score
A Z-score, also known as a standard score, indicates how many standard deviations an element is from the mean. It allows for standardization of different data points, enabling comparison between data sets on different scales.

Key notes about Z-scores:
  • They are calculated using the formula: \[Z = \frac{(X - \mu)}{\sigma}\]
  • A Z-score of 0 indicates that the element is exactly at the mean.
  • Positive Z-scores are above the mean, while negative scores are below.
In probability problems, like in this exercise, Z-scores are used to determine how likely a data point is within a normal distribution. Here, Z-scores were used to find the probability of locating parts in time frames greater than specified thresholds, translating actual time data into standard terms that can easily reference a standard normal distribution table, providing the probability of specific outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The photoresist thickness in semiconductor manufacturing has a mean of 10 micrometers and a standard deviation of 1 micrometer. Assume that the thickness is normally distributed and that the thicknesses of different wafers are independent. (a) Determine the probability that the average thickness of 10 wafers is either greater than 11 or less than 9 micrometers. (b) Determine the number of wafers that need to be measured such that the probability that the average thickness exceeds 11 micrometers is \(0.01 .\) (c) If the mean thickness is 10 micrometers, what should the standard deviation of thickness equal so that the probability that the average of 10 wafers is either greater than 11 or less than 9 micrometers is \(0.001 ?\)

Suppose that \(X\) is a random variable with probability distribution $$ f_{X}(x)=1 / 4, \quad x=1,2,3,4 $$ Find the probability distribution of \(Y=2 X+1\).

A Web site uses ads to route visitors to one of four landing pages. The probabilities for each landing page are equal. Consider 20 independent visitors and let the random variables \(W\), \(X, Y,\) and \(Z\) denote the number of visitors routed to each page. Calculate the following: 5-44. A Web site uses ads to route visitors to one of four landing pages. The probabilities for each landing page are equal. Consider 20 independent visitors and let the random variables \(W\) \(X, Y,\) and \(Z\) denote the number of visitors routed to each page. Calculate the following: (a) \(P(W=5, X=5, Y=5, Z=5)\) (b) \(P(W=5, X=5, Y=5)\) (c) \(P(W=7, X=7, Y=6 \mid Z=3)\) (d) \(P(W=7, X=7, Y=3 \mid Z=3)\) (e) \(P(W \leq 2)\) (f) \(E(W)\) (g) \(P(W=5, X=5)\) (h) \(P(W=5 \mid X=5)\)

The percentage of people given an antirheumatoid medication who suffer severe, moderate, or minor side effects are \(10,20,\) and \(70 \%,\) respectively. Assume that people react independently and that 20 people are given the medication. Determine the following: (a) The probability that \(2,4,\) and 14 people will suffer severe, moderate, or minor side effects, respectively (b) The probability that no one will suffer severe side effects (c) The mean and variance of the number of people who will suffer severe side effects (d) What is the conditional probability distribution of the number of people who suffer severe side effects given that 19 suffer minor side effects? (e) What is the conditional mean of the number of people who suffer severe side effects given that 19 suffer minor side effects?

The conditional probability distribution of \(Y\) given \(X=x\) is \(f_{Y \mid x}(y)=x e^{-x y}\) for \(y > 0,\) and the marginal probability distribution of \(X\) is a continuous uniform distribution over 0 to 10 . (a) Graph \(f_{Y \mid X}(y)=x e^{-x y}\) for \(y > 0\) for several values of \(x\) Determine: (b) \(P(Y < 2 \mid X=2)\) (c) \(E(Y \mid X=2)\) (d) \(E(Y \mid X=x)\) (e) \(f_{X Y}(x, y)\) (f) \(f_{Y}(y)\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.