Replace Distress Flare: Normal Distribution Life with 90% and 99% Success Rates

You're on the right track. But instead of using the probability as the given value, you're using the z-score as the given value. So for (i), you would use the z-score that corresponds to a 90% probability, which is approximately 1.28. Then you would solve for x in the equation z = (x-mean)/stnd dev. This x value would be the number of years the flare will work for with a 90% probability.
  • #1
Sirsh
267
10
3. A local manufacturer creates distress flares. The times the flares last are normally distributed with a mean life of 9.8 years and a standard deviation of 1.3 years.
(b) A small boat owner who regularly travels out to sea wants to be sure his distress flare works. Determine when he should replace the distress flare, given he wants a better than:
(i) 90% chance the flare will work
(ii) 99% chance the flare will work.

For this question i am not sure what to do.. I thought that if it had said works for.. 11years you'd do z = (11-9.8)/1.3 then use this value to find out the proability. but with the percentages i am unsure. Could some please help me! much apprechiated.! :_)
 
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  • #2
This is like the inverse of what you are describing. Suppose they had asked you "After 11 years, what is the probability that the flare is working?" And you would calculate your z-score (11-9.8)/1.3, look it up in a table, or use a computer, or whatever.

The question they are asking you now is, "After how many years will the probability of the flare working be below 90%?" This is a pretty similar problem; only now you're looking through your table at the probabilities and getting the z-scores instead of the other way around.
 
  • #3
So would the probability of a certain flare be 0.9 then you'd have to find the z score, then put the z scores value into the equation z = (x-mean)/stnd dev and then you'd find the x value and that'd be the amount of years it would work for?
 
  • #4
Yes!
 

1. What is the purpose of replacing a distress flare's normal distribution life with success rates of 90% and 99%?

The purpose of this replacement is to ensure that the distress flare has a higher success rate of being effective in emergency situations. By using a normal distribution life with success rates of 90% and 99%, it is more likely that the flare will work when it is needed.

2. How is the normal distribution life of a distress flare determined?

The normal distribution life of a distress flare is determined through extensive testing and analysis. This includes factors such as the materials used, the design and construction of the flare, and the environmental conditions in which it will be used. This data is then used to create a normal distribution curve, which shows the expected probability of success for the flare.

3. What does a success rate of 90% and 99% mean for a distress flare?

A success rate of 90% means that the distress flare has a 90% chance of being effective in an emergency situation. Similarly, a success rate of 99% means that the flare has a 99% chance of being successful. This higher success rate provides a greater level of confidence that the flare will work as intended when needed.

4. Are there any potential drawbacks to using a normal distribution life with success rates of 90% and 99%?

One potential drawback is that the higher success rates may come at a higher cost, as more advanced materials and design may be needed to achieve these rates. Additionally, there may be a trade-off between success rates and other factors such as weight and size, which could impact the portability of the flare.

5. How often should a distress flare be replaced if using a normal distribution life with success rates of 90% and 99%?

The replacement frequency will depend on the specific flare and its intended use. However, it is recommended to regularly check and replace the flare according to the manufacturer's instructions to ensure it is always in optimal working condition. This may include replacing the flare after a certain number of uses or after a certain amount of time has passed.

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