A company is in the business of finding addresses of long-lost friends. The company claims to have a 77% success rate. Suppose that you have the names of five friends for whom you have no addresses and decide to use the company to track them. (a) Find the mean and standard deviation of this probability distribution. What is the expected number of friends for whom addresses will be found? (Round your answers to two decimal places.) ? = friends ? = friends (b) What is the smallest number of names would you have to submit to be at least 96% sure that at least two addresses will be found? (Enter your answer as a whole number.)
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
A company is in the business of finding addresses of long-lost friends. The company claims to have a 77% success rate. Suppose that you have the names of five friends for whom you have no addresses and decide to use the company to track them.
| ? = friends |
| ? = friends |
(b) What is the smallest number of names would you have to submit to be at least 96% sure that at least two addresses will be found? (Enter your answer as a whole number.)
names
Given that the success rate of finding addresses is 77%
Therefore, p = The probability of success = 0.77
q = probability of failure = 1 - p = 1 - 0.77 = 0.23
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