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Anastaziya [24]

3 years ago

7

Consider a political discussion group consisting of 1010 Democrats, 1010 Republicans, and 55 Independents. Suppose that two gr

oup members are randomly selected, in succession, to attend a political convention. Find the probability of selecting two RepublicansRepublicans.

1 answer:

yan [13]3 years ago

3 0

Answer:

The probability is $\frac{3}{20}$

Step-by-step explanation:

Consider a political discussion group consisting of 10 Democrats, 10 Republicans, and 5 Independents.

Total members of the group = $10+10+5=25$

Selecting 2 republicans in succession means selecting 1st republican out of 10 republicans and total 25 members and then selecting 2nd republican out of 9 left republicans and 24 members.

Hence, probability becomes:

$\frac{10}{25}\times \frac{9}{24}$

= $\frac{90}{600} =\frac{3}{20}$

So, the probability is $\frac{3}{20}$

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Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

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In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

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For question B:

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Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

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Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

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$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

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