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Chapter 4: Problem 50

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 1 of 5 independently dropped parachutes?

Short Answer

Expert verified
The probability of at least 1 parachute causing damage is approximately 0.00215 or 0.215%.

Step by step solution

01

Understand the problem

We are given a normal distribution for the parachute opening altitude 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 (equipment damage) and then find the likelihood of this happening to at least one parachute out of five.
02

Calculate the probability of damage for one parachute

Since the opening follows a normal distribution, convert the altitude of 100 m to a z-score using the formula: \( z = \frac{X - \mu}{\sigma} \), where \(X\) is the altitude, \(\mu = 200\) m, and \(\sigma = 30\) m. The z-score is \( z = \frac{100 - 200}{30} = -\frac{100}{30} \approx -3.33 \). Now, use a standard normal distribution table or calculator to find the probability \( P(Z < -3.33) \).
03

Determine probability from normal distribution

From standard normal distribution tables or calculators, \( P(Z < -3.33) \approx 0.00043 \). This is the probability of one parachute opening below 100 m, causing equipment damage.
04

Use binomial probability for at least one damage event

Define the failure probability for one independent event as \( p = 0.00043 \). Treat each drop as a Bernoulli trial with 5 trials (n=5). Calculate the probability of no parachutes being damaged \((1 - p)^5\) and then the probability of at least one damage as \( 1 - (1 - p)^5 \).
05

Compute the final probability

Compute \( (1 - 0.00043)^5 \approx 0.99785 \), so \( 1 - 0.99785 \approx 0.00215 \). This is the probability that at least one parachute out of five will cause equipment damage.

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Key Concepts

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

Understanding z-score
In the context of a normal distribution, the z-score is essential for understanding how far away a specific data point is from the mean. This is especially useful in determining how likely it is to deviate significantly. When we calculate the z-score, we use the formula: \[ z = \frac{X - \mu}{\sigma} \]where:
  • \(X\) is the value we are interested in
  • \(\mu\) is the mean of the distribution
  • \(\sigma\) is the standard deviation
In our parachute example, the z-score helps us find the probability of a parachute opening below a certain altitude. A negative z-score like -3.33 signifies that the event is quite rare, as it is many standard deviations below the mean altitude of 200 meters.
Importance of standard deviation
Standard deviation is a measure that describes the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. For our parachute problem:
  • The standard deviation is 30 meters, depicting how spread out the parachute opening altitudes are.
  • It helps in computing the z-score, which in turn allows us to determine the probability of an extreme scenario like opening below 100 meters.
  • A smaller or larger standard deviation would significantly affect the probability of equipment damage.
Understanding this concept is crucial as it underlies the variability and uncertainty inherent in any normal distribution.
Explaining binomial probability
Binomial probability comes into play when you have a fixed number of independent experiments (trials), each of which results in a binary outcome (success or failure). These trials are often described using the parameters:
  • \(n\): the number of trials (in our case, 5 parachute drops)
  • \(p\): probability of success on each trial (probability of equipment damage, \(0.00043\))
To find the probability of at least one success, we calculate the complement: the probability of no failures, using the formula:\[ 1 - (1-p)^n \]This formula helps us understand the likelihood of at least one parachute causing damage during these 5 independent events. In our example, it reveals that the probability of at least one damaging occurrence is approximately 0.00215, indicating a very low risk.

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Most popular questions from this chapter

Let \(X=\) the time between two successive arrivals at the drive-up window of a local bank. If \(X\) has an exponential distribution with \(\lambda=1\) (which is identical to a standard gamma distribution with \(\alpha=1\) ), compute the following: a. The expected time between two successive arrivals b. The standard deviation of the time between successive arrivals c. \(P(X \leq 4)\) d. \(P(2 \leq X \leq 5)\)

Commuting to work requires getting on a bus near home and then transferring to a second bus. If the waiting time (in minutes) at each stop has a uniform distribution with \(A=0\) and \(B=5\), then it can be shown that the total waiting time \(Y\) has the pdf $$ f(y)=\left\\{\begin{array}{cl} \frac{1}{25} y & 0 \leq y<5 \\ \frac{2}{5}-\frac{1}{25} y & 5 \leq y \leq 10 \\ 0 & y<0 \text { or } y>10 \end{array}\right. $$ a. Sketch the pdf of \(Y\). b. Verify that \(\int_{-\infty}^{\infty} f(y) d y=1\). c. What is the probability that total waiting time is at most 3 min? d. What is the probability that total waiting time is at most \(8 \mathrm{~min}\) ? e. What is the probability that total waiting time is between 3 and 8 min? f. What is the probability that total waiting time is either less than 2 min or more than 6 min?

Determine \(z_{\alpha}\) for the following: a. \(\alpha=.0055\) b. \(\alpha=.09\) c. \(\alpha=.663\)

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At a website, the waiting time \(X\) (in minutes) between hits has pdf \(f(x)=4 e^{-4 x}, x \geq 0 ; f(x)=\) 0 otherwise. Find \(M_{X}(t)\) and use it to obtain \(E(X)\) and \(V(X)\).
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