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klasskru [66]
3 years ago
13

In a certain game of chance, your chances of winning are 0.2. if you play the game five times and outcomes are independent, the

probability that you win at most once is
Mathematics
1 answer:
ikadub [295]3 years ago
8 0
<span>Winning Probablity = 0.2, hence Losing Probability = 0.8 Probablity of winning atmost one time, that means win one and lose four times or lose all the times. So p(W1 or W0) = p (W1) + p(W0) Winning once W1 is equal to L4, winning zero times is losing 5 times. p(W1) = p(W1&L4) and this happens 5 times; p(W0) = p(L5); p (W1) + p(W0) = p(L4) + p(L5) p(L4) + p(L5) = (5 x 0.2 x 0.8^4) + (0.8^5) => 0.8^4 + 0.8^5 p(W1 or W0) = 0.4096 + 0.32768 = 0.7373</span>
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Answer:

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Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

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Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

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Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

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  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

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Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

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