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Minchanka [31]
2 years ago
6

As a part of a customer loyalty program, a restaurant's computer chooses orders at random to receive a free appetizer. Each orde

r has a 1 in 18 chance of receiving a free appetizer. (Each order is limited to one free appetizer.) Let X be the number of orders the restaurant fills in a day until they give away the first appetizer. Assume that each order getting the appetizer is independent. Find the probability that the restaurant first gives away an appetizer on the 6th order of the day. You may round your answer to the nearest hundredth
Mathematics
1 answer:
4vir4ik [10]2 years ago
3 0

Answer:

0.0417 this is the answer if you on khan academy... they gave me this answer and  yes it's correct

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Answer:

a) 0.23

b) The 95% confidence interval for the population proportion is (0.1717, 0.2883).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

Point estimate

The point estimate is:

\pi = \frac{46}{200} = 0.23

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.23 - 1.96\sqrt{\frac{0.23*0.77}{200}} = 0.1717

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.23 + 1.96\sqrt{\frac{0.23*0.77}{200}} = 0.2883

The 95% confidence interval for the population proportion is (0.1717, 0.2883).

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I took this test and got question 4 correct

:))))

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