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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 that at least one of the five parachutes causes equipment damage is approximately 0.0021.

Step by step solution

01

Understand the Problem

The parachute opens according to a normal distribution with a mean of \(200 \, \text{m}\) and a standard deviation of \(30 \, \text{m}\). Equipment damage occurs if the opening altitude is less than \(100 \, \text{m}\). We need to find the probability that at least one of five parachutes causes equipment damage.
02

Identify the Probability Formula

We first find the probability that a single parachute causes equipment damage. This happens when the opening altitude is less than \(100 \, \text{m}\). This probability is expressed as \(P(X < 100)\), where \(X\) is the opening altitude which follows a normal distribution.
03

Standardize the Random Variable

To find \(P(X < 100)\), we standardize using the z-score formula: \[ z = \frac{X - \mu}{\sigma} \] Here, \(X = 100\), \(\mu = 200\), and \(\sigma = 30\). Thus, the z-score is: \[ z = \frac{100 - 200}{30} = -\frac{100}{30} \approx -3.33 \]
04

Use the Normal Distribution Table

Using the standard normal distribution table, or a calculator, find the cumulative probability for \(z = -3.33\). This probability provides \(P(X < 100)\). Normally, this is approximately \(0.00043\).
05

Calculate the Probability of No Damage in 5 Drops

The probability that a single parachute does not cause damage is \(1 - P(X < 100)\). Therefore, \[ 1 - 0.00043 = 0.99957 \] The probability that none of the 5 parachutes cause damage is \[ (0.99957)^5 \approx 0.9979 \].
06

Calculate the Probability of at least One Damage Event

The probability of at least one parachute causing damage is the complement of the probability that none causes damage: \[ 1 - 0.9979 = 0.0021 \].

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Key Concepts

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

Normal Distribution
A normal distribution is a fundamental concept in probability theory. It is often referred to as a "bell curve" because of its characteristic bell-shaped graph. In this type of distribution, most data points cluster around a central mean, with fewer values appearing as you move further away from this central point. In many natural phenomena, such as human heights or test scores, data tends to follow a normal distribution.

In our parachute example, the opening altitudes follow a normal distribution. This means that the majority of parachutes will open around the mean altitude of 200 meters. The farther an opening event deviates from this mean, the less frequent it becomes, hence the risk of parachutes opening dangerously low is relatively small.

Key features of a normal distribution include:
  • Symmetry about the mean.
  • A mean that determines the peak location.
  • A standard deviation that defines the spread of the data.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In simple terms, it tells us how spread out the values are in a data set. A lower standard deviation means the values tend to be closer to the mean, while a higher standard deviation indicates that the values are more spread out.

In the parachute problem, the standard deviation is 30 meters. This means most parachute openings are spread around the mean of 200 meters, but occasionally, there will be some instances that deviate by a few standard deviations.

The standard deviation acts as a sort of yardstick for measurement in statistics, helping us understand how much individual data points deviate from the mean on average.
Z-Score
A z-score is a way to standardize scores on a normal distribution. It represents the number of standard deviations a data point is from the mean. This is useful for comparing values from different normal distributions, or to determine the probability of a certain value occurring by measuring its deviation from the mean.

The formula to calculate a z-score is: \[ z = \frac{X - \mu}{\sigma} \]Where:
  • \(X\) is the value being standardized,
  • \(\mu\) is the mean of the distribution,
  • \(\sigma\) is the standard deviation.
For a parachute to open at less than 100 meters, the z-score is calculated as:\[ z = \frac{100 - 200}{30} \approx -3.33 \]This tells us that 100 meters is 3.33 standard deviations below the mean opening altitude. Such a low z-score suggests a low probability of occurrence.
Cumulative Probability
Cumulative probability is the probability that a variable will take a value less than or equal to a particular number. It is found by calculating the area under the probability distribution curve to the left of that value. Cumulative probabilities are particularly useful when dealing with continuous distributions like the normal distribution.

To find the probability of the parachute opening below 100 meters, we use the cumulative probability for the calculated z-score of \(-3.33\). According to the normal distribution table, this z-score corresponds to a cumulative probability of about 0.00043, meaning there's roughly a 0.043% chance that the parachute opens below the dangerous height.

Cumulative probability helps us understand the likelihood of rare events, which is crucial in assessing risks such as equipment damage in our parachute scenario.

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Most popular questions from this chapter

Let \(X\) have a standard beta density with parameters \(\alpha\) and \(\beta\). a. Verify the formula for \(E(X)\) given in the section. b. Compute \(E\left[(1-X)^{m}\right]\). If \(X\) represents the proportion of a substance consisting of a particular ingredient, what is the expected proportion that does not consist of this ingredient?

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The article "Monte Carlo Simulation-Tool for Better Understanding of LRFD" ( \(J\). Structural Engr., 1993: 1586-1599) suggests that yield strength (ksi) for A.36 grade steel is normally distributed with \(\mu=43\) and \(\sigma=4.5\). a. What is the probability that yield strength is at most 40 ? Greater than 60 ? b. What yield strength value separates the strongest \(75 \%\) from the others?

The article "On Assessing the Accuracy of Offshore Wind Turbine Reliability- Based Design Loads from the Environmental Contour Method" (Intl. J. of Offshore and Polar Engr, 2005: 132-140) proposes the Weibull distribution with \(\alpha=1.817\) and \(\beta=.863\) as a model for 1 -hour significant wave height \((\mathrm{m})\) at a certain site. a. What is the probability that wave height is at most \(.5 \mathrm{~m}\) ? b. What is the probability that wave height exceeds its mean value by more than one standard deviation? c. What is the median of the wave-height distribution? d. For \(0

The lifetime \(X\) (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=3\). Compute the following: a. \(E(X)\) and \(V(X)\) b. \(P(X \leq 6)\) c. \(P(1.5 \leq X \leq 6)\) (This Weibull distribution is suggested as a model for time in service in "On the Assessment of Equipment Reliability: Trading Data Collection Costs for Precision," J. Engr. Manuf., 1991: 105-109.)

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