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MariettaO [177]
2 years ago
14

In preparation for an earnings report, a large retailer wants to estimate p= the proportion of annual sales

Mathematics
1 answer:
mr Goodwill [35]2 years ago
5 0

Using the z-distribution, it is found that the 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

The sample size and the estimate are given by:

n = 161, \pi = \frac{37}{161} = 0.2298

Hence:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 - 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.1648

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 + 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.2948

The 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

More can be learned about the z-distribution at brainly.com/question/25890103

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find the probability of being delt 5 clubs and 3 cards with one of each remaining suit in 8 card poker
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Total cards in a deck = 52

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The required probability = =\dfrac{^{13}C_5\times^{13}C_1\times^{13}C_1\times^{13}C_1}{^{52}C_8}

=\dfrac{\dfrac{13!}{5!8!}\times13\times13\times13}{\dfrac{52!}{8!44!}}\\\\=\dfrac{24167}{6431950}\\\\\approx0.003757

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