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Burka [1]
3 years ago
7

Airlines have increasingly outsourced the maintenance of their planes to other companies. A concern voiced by critics is that th

e maintenance may be less carefully done so that outsourcing creates a safety hazard. In addition, flight delays are often due to maintenance problems, so one might look at government data on percent of major maintenance outsourced and percent of flight delays blamed on the airline to determine if these concerns are justified. This was done, and data from 2005 and 2006 appeared to justify the concerns of the critics. Do more recent data still support the concerns of the critics? Here are data from 2014:
Mathematics
1 answer:
irinina [24]3 years ago
5 0

According to the information that is found in a different source, the recent data does not support the concerns of the critics. These are the numbers shown in the data:

<u>Airline / Outsource Percent / Delay Percent</u>

<em>Alaska</em> / 51.0 / 10.35

<em>American</em> / 29.4 / 20.32

<em>Delta</em> / 36.7 / 14.48

<em>Frontier</em> / 46.3 / 21.42

<em>Hawaiian</em> / 78.4 / 5.06

Based on this data, we can see that the critics do not seem to be justified in their concerns. We can see that airlines have outsourced much of the maintenance of their planes, but that there does not seem to be a strong correlation between this outsourcing and the percentage of delays. For example, the airline that outsources the most is Hawaiian (78.4%), yet it has the lowest percentage of delayed flights (5.06%).

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<span>factors of 18 = (2)(3)(3) 
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3 years ago
A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true
kvv77 [185]

Answer:

99.74% probability that the sample proportion will be less than 0.1

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 276, p = 0.06

So

\mu = E(X) = np = 276*0.06 = 16.56

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{276*0.06*0.94} = 3.9454

What is the probability that the sample proportion will be less than 0.1

This is the pvalue of Z when X = 0.1*276 = 27.6. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{27.6 - 16.56}{3.9454}

Z = 2.8

Z = 2.8 has a pvalue of 0.9974

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