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Airplane Safety: Probability Analysis

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Airplane Safety: Suppose that 62% of all adults think that airplanes would be safer places if pilots carried guns. An opinion poll plans to ask an SRS of 1009 adults about airplane safety. The proportion of the sample who think that airplanes would be safer if pilots carried guns will vary if we take many samples from this same population. The sampling distribution of the sample proportion is approximately Normal with mean 0.62 and standard deviation about 0.015. Sketch this normal curve and use it to answer the following questions:
a. What is the probability that the poll gets a sample in which more than 65% of the people think that airplanes would be safer if pilots carried guns?
b. What is the probability of getting a sample that misses the truth (62%) by 1.5% or more?

https://brainmass.com/statistics/probability/airplane-safety-probability-analysis-516642

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2. Airplane Safety: Suppose that 62% of all adults think that airplanes would be safer places if pilots carried guns. An opinion poll plans to ask an SRS of 1009 adults about airplane safety. The proportion of the sample who think that airplanes would be safer if pilots carried guns will vary if we take many samples from this same population. The sampling distribution of the sample proportion is approximately Normal with mean 0.62 and standard deviation about 0.015. Sketch this normal curve and use it to answer the following questions:
Let X be the proportion of adults think that airplanes would be safer places if pilots carried guns. Given that X is normal with mean µ = 0.62 and standard deviation  = 0.015.
Standardizing the variable X using and from standard normal tables, we have,
a. What is the probability that the poll gets a sample in which more than 65% of the people think that airplanes would be safer if pilots carried guns?
P (X > 0.65) = = P (Z > 2) = 0.02275

b. What is the probability of getting a sample that misses the truth (62%) by 1.5% or more?
That is we need P (X ≥ 0.62 + 0.015) or P (X ≤ 0.62 - 0.015)
That is, P (X ≥ 0.635) or P (X ≤ 0.605)
P (X ≥ 0.635) = = P (Z > 1) = 0.1587

P (X ≤ 0.605) = = P (Z < -1) = 0.1587

Therefore, required probability = 0.1587 + 0.1587 = 0.3174

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