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Aliun [14]
3 years ago
9

1. Suppose half of all newborns are girls and half are boys. Hospital A, a large city hospital, records an average of 50 births

a day. Hospital B, a small, rural hospital, records an average of 10 births a day. On a particular day, which hospital is less likely to record 80% or more female births?
Mathematics
2 answers:
kicyunya [14]3 years ago
8 0

Answer:

5%

Step-by-step explanation:

Hospital A (with 50 births a day), because the more births you see, the closer the proportions will be to 0.5.

Hospital B (with 10 births a day), because with fewer births there will be less variability.

The two hospitals are equally likely to record such an event, because the probability of a boy does not depend on the number of births

Lina20 [59]3 years ago
4 0
I had the same question it’s 5%
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Answer:

The correct answer is

(0.0128, 0.0532)

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 interval 1-\alpha, we have the following confidence interval of proportions.

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In which

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For this problem, we have that:

In a random sample of 300 circuits, 10 are defective. This means that n = 300 and \pi = \frac{10}{300} = 0.033

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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:

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The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 + 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0532

The correct answer is

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4 0
3 years ago
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