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Chapter 3: Problem 117

In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been properly drilled in the boxes and in ten minutes if the holes must be re-drilled. Twenty gearboxes are in stock, 2 with improperly drilled holes. Five gearboxes must be selected from the 20 that are available for installation in the next five robots. a. Find the probability that all 5 gearboxes will fit properly. b. Find the mean, variance, and standard deviation of the time it takes to install these 5 gearboxes.

Short Answer

Expert verified
a) Probability all 5 fit properly is \( \frac{51}{92} \). b) Mean: 9.5 mins, Variance: 36.45, SD: \( \approx 6.04 \).

Step by step solution

01

Understanding the Problem

We are asked to calculate probabilities and statistical measures related to selecting gearboxes, some of which are properly drilled and some are not, from a batch. The goal is to determine the probability of choosing only properly drilled boxes and to calculate the mean, variance, and standard deviation of the installation times.
02

Calculate Total Possibilities

We need to determine how many ways we can choose 5 gearboxes from 20. This is given by the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). For this problem, \( n = 20 \) and \( r = 5 \). So, we have \( \binom{20}{5} = \frac{20!}{5! \times 15!} = 15,504 \) ways.
03

Calculate Favorable Possibilities

Next, calculate how many ways we can select 5 gearboxes all with properly drilled holes. There are 18 gearboxes with properly drilled holes. We need to select 5 from these 18, so it's \( \binom{18}{5} = \frac{18!}{5! \times 13!} = 8,568 \).
04

Calculate Probability All Fit Properly

The probability that all 5 gearboxes fit properly is the number of favorable possibilities divided by the total possibilities. This gives us \[ P = \frac{\binom{18}{5}}{\binom{20}{5}} = \frac{8,568}{15,504} = \frac{51}{92} \].
05

Calculate Expected Time (Mean)

Each properly drilled gearbox takes 1 minute, and each improperly drilled takes 10 minutes. If all 5 fit properly, it takes 5 minutes. The expected number of improperly drilled gearboxes in a selection of 5 is \( E(X) = 5 \times \frac{2}{20} = 0.5 \). Hence, mean time \( = (5 - 0.5 \times 1 + 0.5 \times 10) = 9.5 \).
06

Calculate Variance

The variance of time depends on the variance of the number of improperly drilled gearboxes, X. \( \text{Var}(X) = 5 \times \frac{2}{20} \times \frac{18}{20} = 0.45 \). Time variance becomes \( \text{Var}(Time) = 0.45 \times (10 - 1)^2 = 36.45 \).
07

Calculate Standard Deviation

The standard deviation is simply the square root of the variance, so \( \text{SD}(Time) = \sqrt{36.45} \approx 6.04 \).

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arranging, and combining objects following certain rules. In problems like selecting gearboxes, we're dealing with combinations, which is the selection of items from a larger set without considering the order.
For instance, when choosing 5 gearboxes from a batch of 20, we use the combination formula:
  • \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)
Here, \( n = 20 \) is the total number of gearboxes, and \( r = 5 \) is the number we want to select. This gives us \( 15,504 \) possible ways to choose the gearboxes. Such calculations are vital for determining how likely certain events are and are foundational in probability theory.
Expected Value
Expected value gives us the average result we'd expect if we repeated an experiment many times. In our gearbox problem, we calculate the expected number of improperly drilled gearboxes when picking five.
To find this, multiply the total number of selections (5) by the probability of selecting an improperly drilled one (2 out of 20). This provides:
  • \( E(X) = 5 \times \frac{2}{20} = 0.5 \)
This means, on average, we'd expect 0.5 improperly drilled gearboxes in our selections. Subsequently, we can determine the expected time to install, considering each gearbox type's installation time. Using these concepts helps in making informed decisions about outcomes in processes like assembly line efficiency.
Variance
Variance tells us how much the outcomes of our process vary. It鈥檚 a measure of dispersion, or how spread out the values in a data set are. In selecting gearboxes, we seek to understand how the time to installation varies with different mixes of properly and improperly drilled gearboxes.
The formula for variance in this context is:
  • \( \text{Var}(X) = n \times p \times (1-p) \)
Here, \( n = 5 \) (number of gearboxes chosen), and \( p = \frac{2}{20} = 0.1 \) (probability of selecting an improperly drilled gearbox). The resulting \( 0.45 \) variance reflects the spread of time discrepancies depending on the number of improperly drilled gearboxes selected, which is useful for anticipating variability in the assembly process.
Standard Deviation
Standard deviation provides a practical measure of how much variation there is from the average. It is the square root of variance and gives insights into the typical deviation from the mean installation time.
Given our variance of time鈥檚 calculation as \( 36.45 \), we determine the standard deviation by:
  • \( \text{SD}(Time) = \sqrt{36.45} \approx 6.04 \)
A standard deviation of around 6.04 indicates that the installation times we recorded typically vary from the average by about this many minutes. Knowing this helps us gauge the reliability and predictability of the gearbox installation process, emphasizing how much fluctuation to expect around the average time.

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Most popular questions from this chapter

Of a population of consumers, \(60 \%\) are reputed to prefer a particular brand, \(A\), of toothpaste. If a group of randomly selected consumers is interviewed, what is the probability that exactly five people have to be interviewed to encounter the first consumer who prefers brand \(A\) ? At least five people?

If \(Y\) has a geometric distribution with success probability \(p,\) show that $$P(Y=\text { an odd integer })=\frac{p}{1-q^{2}}$$.

Toss a balanced die and let \(Y\) be the number of dots observed on the upper face. Find the mean and variance of \(Y\). Construct a probability histogram, and locate the interval \(\mu \pm 2 \sigma\). Verify that Tchebysheff's theorem holds.

A city commissioner claims that \(80 \%\) of the people living in the city favor garbage collection by contract to a private company over collection by city employees. To test the commissioner's claim, 25 city residents are randomly selected, yielding 22 who prefer contracting to a private company. a. If the commissioner's claim is correct, what is the probability that the sample would contain at least 22 who prefer contracting to a private company? b. If the commissioner's claim is correct, what is the probability that exactly 22 would prefer contracting to a private company? c. Based on observing 22 in a sample of size 25 who prefer contracting to a private company, what do you conclude about the commissioner's claim that \(80 \%\) of city residents prefer contracting to a private company?

It is known that \(10 \%\) of a brand of television tubes will burn out before their guarantee has expired. If 1000 tubes are sold, find the expected value and variance of \(Y\), the number of original tubes that must be replaced. Within what limits would \(Y\) be expected to fall?
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