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Sladkaya [172]
2 years ago
15

In a​ lottery, 6 numbers are randomly sampled without replacement from the integers 1 to 38. Their order of selection is not imp

ortant. Find the probability of holding a ticket that has zero winning numbers out of the 6 numbers selected for the winning ticket out of the 38 possible numbers.
Mathematics
1 answer:
inna [77]2 years ago
8 0

Answer: Our required probability is 0.328.

Step-by-step explanation:

Since we have given that

there is number on the lottery is from 1 to 38.

6 numbers are selected.

We need to find the probability of holding a ticket that has zero winning numbers out of the 6 numbers selected for the winning ticket.

So, it should be other than 6 numbers.

Remaining numbers = 38-6 = 32

So, our probability becomes

\dfrac{^{32}C_6}{^{38}C_6}\\\\=\dfrac{906192}{2760681}\\\\=0.328

Hence, our required probability is 0.328.

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2 years ago
Suppose that a local TV station conducts a survey of a random sample of 120 registered voters in order to predict the winner of
forsale [732]

Answer:

a) The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b) The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

Step-by-step explanation:

Question a:

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 z-score that has a p-value of 1 - \frac{\alpha}{2}.

Sample of 120 registered voters in order to predict the winner of a local election. The Democrat candidate was favored by 62 of the respondents.

So 120 - 62 = 58 favored the Republican candidate, so:

n = 120, \pi = \frac{58}{120} = 0.4833

99% confidence level

So \alpha = 0.01, z is the value of Z that has a p-value of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 - 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.3658

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4833 + 2.575\sqrt{\frac{0.4833*0.5167}{120}} = 0.6001

The 99% CI for the true proportion of voters who prefer the Republican candidate is (0.3658, 0.6001). This means that we are 99% sure that the true population proportion of all voters who prefer the Republican candidate is (0.3658, 0.6001).

b. If a candidate needs a simple majority of the votes to win the election, can the Republican candidate be confident of victory? Justify your response with an appropriate statistical argument.

The upper bound of the confidence interval is above 0.5 = 50%, which meas that the candidate can be confidence of victory.

8 0
2 years ago
What is 125% of 332?
bagirrra123 [75]

Answer:

W = 415

Step-by-step explanation:

Is means equals and of means multiply

W = 125% * 332

Change to decimal form

W = 1.25 *332

W = 415

7 0
2 years ago
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