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tiny-mole [99]
3 years ago
13

Franklin rolls a pair of six-sided fair dice with sides numbered 1 through 6.

Mathematics
1 answer:
Zarrin [17]3 years ago
6 0

Answer:

The answer is 7/36.

Step-by-step explanation:

First, you find out how many possible outcomes there are from rolling a pair of dice. On one cube, you can roll a 1,2,3,4,5, or 6; so there are 6 outcomes. Since there are two cubes, you multiply 6 by itself to get a total of 36 possible outcomes. Next, you find the probability of the sum of the numbers rolled being an even number; the possibilities are 2,4,6,8,10, or 12, which is 6/36. The probability of rolling a multiple of 5; the one possibility is just 5, since we already accounted for rolling a 10 as an even number. So that is 1/36. The word <u>or</u> says that we add the two probabilities, so the final answer is 6/36+1/36=7/36.

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Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

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Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population is given by;

      P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion = 0.345

           n = sample of participants = 3532

           p = population proportion

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population​ proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level of

                                                         significance are -1.96 & 1.96}

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u>= [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

    = [ 0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } } , 0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } } ]

    = [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

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