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A manufacturer of electroluminescent lamps knows that the amount of luminescent ink deposited on one of its products is normally distributed with a mean of 1.2 grams and a standard deviation of 0.03 gram. Any lamp with less than 1.14 grams of luminescent ink fails to meet customers' specifications. A random sample of 25 lamps is collected and the mass of luminescent ink on each is measured. a. What is the probability that at least one lamp fails to meet specifications? b. What is the probability that five or fewer lamps fail to meet specifications? c. What is the probability that all lamps conform to specifications? d. Why is the joint probability distribution of the 25 lamps not needed to answer the previous questions?

Short Answer

Expert verified
a. 0.4119. b. Approximately 1. c. 0.5881. d. Each lamp is independent, allowing use of binomial distribution.

Step by step solution

01

Identify the Population Distribution

Given that the amount of luminescent ink deposited is normally distributed with mean \( \mu = 1.2 \) grams and standard deviation \( \sigma = 0.03 \) grams. Any lamp with less than 1.14 grams fails.
02

Find the Probability of a Single Lamp Failing

We need to find the probability that a single lamp has less than 1.14 grams of luminescent ink. We calculate the Z-score for 1.14 grams: \[ Z = \frac{1.14 - 1.2}{0.03} = -2. \] Checking the standard normal distribution table, a Z-score of -2 corresponds to a probability of approximately 0.0228.
03

Interpret the Binomial Distribution

Let \( X \) be the number of lamps in a sample of 25 that fail to meet specifications. Since each lamp fails with probability 0.0228 independently, \( X \) follows a Binomial distribution \( \text{Binomial}(n=25, p=0.0228) \).
04

Probability of At Least One Failure

The probability that at least one lamp fails is \( P(X \geq 1) = 1 - P(X = 0) \). Calculate \( P(X = 0) = (0.9772)^{25} \). Using a calculator, \( P(X = 0) \approx 0.5881 \). Thus, \( P(X \geq 1) = 1 - 0.5881 = 0.4119 \).
05

Probability That Five or Fewer Lamps Fail

To find this, compute \( P(X \leq 5) \), the cumulative probability for \( x = 0 \) to \( x = 5 \) using the Binomial formula or a statistical software. \( P(X \leq 5) \) is calculated to be approximately 1 as the smaller probability of failure restricts high counts.
06

Probability of All Lamps Conforming

The probability that all lamps conform is \( P(X = 0) = (0.9772)^{25} \approx 0.5881 \).
07

Why Joint Probability Distribution Is Not Needed

We rely only on the individual probabilities of the lamps because each lamp is independent. The properties of the binomial distribution simplify the problem without needing full joint distributions for the entire sample.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a key concept in probability theory. It describes the number of successes in a fixed number of independent and identical trials, where each trial has two possible outcomes: success or failure.
In the context of our electroluminescent lamps exercise, every lamp can either pass (meet specifications) or fail (have less than 1.14 grams of luminescent ink). The probability of a single lamp failing is calculated to be 0.0228 based on the normal distribution. Given a sample of 25 lamps, our task is to understand how many of these might fail.
The binomial distribution model this situation using two parameters:
  • Number of trials, n = 25 (the number of lamps).
  • Probability of failure in each trial, p = 0.0228.
This allows us to compute probabilities concerning the total number of lamps that might not meet the specifications, aiding in calculating probabilities of different failure scenarios.
Z-score Calculation
Z-scores are used in statistics to determine how many standard deviations an element is from the mean of the distribution. This is crucial for understanding where a particular value lies in a normal distribution.
In this specific exercise, we need to calculate the probability of a single lamp having less than 1.14 grams of luminescent ink. We achieve this by calculating the Z-score using the formula:
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
  • \( X \) is the value of interest (1.14 grams).
  • \( \mu \) is the mean of the distribution (1.2 grams).
  • \( \sigma \) is the standard deviation (0.03 grams).
Plugging into the formula, we find \( Z = -2 \). Using a standard normal distribution table or software, we find that a Z-score of -2 corresponds to a probability of 0.0228. This Z-score helps us find how extreme or typical a lamp's ink mass is under the normal distribution assumption.
Probability of Failure
The probability of failure is an important metric derived from understanding both the normal and binomial distributions in our exercise.
It begins with identifying how likely it is for a single lamp to fail, which we already calculated using the Z-score, giving a probability of 0.0228 for a single lamp to have less than 1.14 grams of ink.
This probability then feeds into the binomial distribution to explore various scenarios:
  • **At least one failure**: Calculated as 1 minus the probability that no lamps fail, which is approximately 0.4119.
  • **Five or fewer failures**: Computed by summing probabilities from 0 to 5 failures, demonstrating how unlikely it is to have up to five failures based on the given failure probability.
  • **All lamps conforming**: The alternate case where no lamps fail, having a probability of about 0.5881.
These probabilities are key in assessing whether the production process meets quality standards consistently.
Independent Events
In probability, independent events are those whose outcomes do not affect each other. Understanding this concept simplifies complex probability problems.
In the case of the lamps, each one is considered an independent event regarding failure or success (meeting specifications or not), because:
  • The state of one lamp doesn't influence another.
  • Each lamp is produced and inspected separately under assumed similar conditions.
This independence means that we can easily use formulas from binomial distribution, which assumes independent trials. It eliminates the need for a complicated joint probability distribution of all lamps because each one operates independently, letting us focus on the simpler individual probabilities and apply them to the entire group.

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Most popular questions from this chapter

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