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Chapter 4: Problem 14

The automatic opening device of a military cargo parachute has been designed to open when the parachute is \(200 \mathrm{~m}\) above the ground. Suppose opening altitude actually has a normal distribution with mean value \(200 \mathrm{~m}\) and standard deviation \(30 \mathrm{~m}\). Equipment damage will occur if the parachute opens at an altitude of less than \(100 \mathrm{~m}\). What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

Short Answer

Expert verified
The probability of at least one parachute causing damage is approximately 0.002.

Step by step solution

01

Understand the Problem

We are given that the opening altitude of a parachute follows a normal distribution with a mean of 200 m and a standard deviation of 30 m. We need to find the probability of a parachute opening below 100 m, which would result in equipment damage. We are tasked with finding the probability that at least one out of five parachutes results in equipment damage.
02

Calculate Probabilty of Single Damage Event

We use the properties of the normal distribution to calculate the probability of a parachute opening below 100 m. This is done by calculating the z-score: \[ z = \frac{100 - 200}{30} = -\frac{100}{30} = -\frac{10}{3} \approx -3.33 \]. Using standard normal distribution tables or a calculator, we find the probability associated with a z-score of -3.33.
03

Reading the Z-table

A z-score of -3.33 corresponds to a probability of approximately 0.0004. This means that there is a 0.04% chance of a single parachute opening causing equipment damage.
04

Calculate Probability of No Damage for All Chutes

We calculate the probability that none of the five parachutes result in equipment damage. This is given by \( (1 - 0.0004)^5 \). Calculate this to find the probability that all parachutes open above 100 m.
05

Calculate Probability of At Least One Damage

The probability of at least one parachute causing damage is the complement of all parachutes opening correctly: \( 1 - (1 - 0.0004)^5 \). Compute this value to get the final answer.
06

Solution Conclusion

Evaluating the expression \( 1 - (0.9996)^5 \), we get approximately 0.002. This represents the probability of at least one parachute opening below 100 m and causing damage.

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Key Concepts

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

Understanding Z-score Calculation
In statistics, a z-score measures how many standard deviations an element is from the mean. It is a way to standardize data points for comparison within different normal distributions. To calculate a z-score, use the formula: \[ z = \frac{X - \mu}{\sigma} \]where:
  • \( X \) is the value of the element.
  • \( \mu \) is the mean of the distribution.
  • \( \sigma \) is the standard deviation.
In our parachute problem, the parachute must open at least 100 meters above ground. Given that the normal distribution has a mean of 200 meters and a standard deviation of 30 meters, we substitute these values into the formula, calculating:\[ z = \frac{100 - 200}{30} = -3.33 \]This z-score tells us how far 100 meters is from the mean in terms of standard deviations, revealing how unusual such an event is within the distribution.
Probability Calculation Insight
Probability calculation involves determining the likelihood that a random event will occur. When dealing with normal distribution, the z-score aids in finding this probability by correlating it to standard normal distribution tables, or using statistical software or calculators. In the parachute example, after calculating the z-score of -3.33, we refer to the standard normal distribution table. For a z-score of -3.33, the table gives us a probability of approximately 0.0004. This means there is only a 0.04% chance for a parachute to open below 100 meters, leading to equipment damage under these specific circumstances.
The Concept of Independent Events
Independent events are situations where the outcome of one event does not influence or have any effect on the occurrence of another. In probability, if two events are independent, the probability of both events occurring is the product of their individual probabilities.In our exercise, each parachute is considered an independent event. The likelihood of one parachute opening too low is separate from the likelihood of the next parachute doing the same. Therefore, the overall probability of no parachute causing damage (in a set of five) is computed as \[ (1 - 0.0004)^5 \].This multiplication respects the principle of independence, showcasing how separate events are treated in probability mathematics.
Understanding Complementary Probability
Complementary probability is the concept where the probability of an event occurring is related to the probability of it not occurring. If you know the probability of any event, you can find the complementary probability by subtracting that from 1. Essentially, if you're looking at all possibilities together, they must add up to certainty, or 1.In our parachute scenario, we need to determine the likelihood of at least one parachute failing among five chances. By first finding the probability that no parachutes cause damage, \[ (1 - 0.0004)^5 \],and then computing the complement, we determine:\[ 1 - (1 - 0.0004)^5 \].This calculation provides us with the probability of the complementary event, that is, at least one parachute will cause damage. In this instance, it comes out to be approximately 0.002, or 0.2%. This reflects a small yet crucial risk, illustrating the power of complementary probability in risk assessment.

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Most popular questions from this chapter

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