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Yuliya22 [10]
3 years ago
8

Use the given confidence interval limits to find the point estimate and the margin of error E.

Mathematics
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

The point estimate = 0.507

Margin error of  a given confidence interval = 0.032

Step-by-step explanation:

The point estimate is calculated by using the sample statistics of a population.

Thus; point estimate can be expressed with the formula:

\overline x = \dfrac{\sum \limits ^n _{i=1} \ x _i}{n}

Given that : 0.475 < p < 0.539

\overline x = \dfrac{0.475+0.539}{2}

\overline x = \dfrac{1.014}{2}

\overline x = 0.507

The point estimate = 0.507

The margin of error  which shows  the percentage of points that the derived results would differ from that of the given population value can be calculated with the formula:

Margin error of  a given confidence interval = \mathtt{\dfrac{upper \ confidence \ limit - lower  \ confidence \ limit  }{2}}

Margin error of  a given confidence interval =  \dfrac{0.539-0.475}{2}

Margin error of  a given confidence interval = \dfrac{0.064}{2}

Margin error of  a given confidence interval = 0.032

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Suppose that we roll a red and a black die. Let a = "the black die shows a 2 or a 5", b = "the sum of the two dice is at least 7
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Answer: No, A and B are not independent events.

∵ it does not satisfy the rule of probability for independent events i.e.

P(A∩B)=P(A).P(B)

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Let A be the event that  the black dice shows 2 or 5

Let B be the event that the sum of two dice is atleast 7

Sample space of A={(R_1,B_2)(R_2,B_2)(R_3,B_2)(R_4,B_2)((R_5,B_2),(R_6,B_2),(R_1,B_5),(R_2,B_5),(R_3,B_5),(R_4,B_5),(R_5,B_5),(R_6,B_5)}

Sample space of B= { (R_1,B_6),(R_2,B_5),(R_2,B_6),(R_3,B_4),(R_3,B_5),(R_3,B_6),,(R_4,B_3),(R_4,B_4),(R_4,B_5),(R_4,B_6),(R_5,B_2),(R_5,B_3),(R_5,B_4),(R_5,B_5), (R_5,B_6),(R_6,B_2),(R_6,B_3),(R_6,B_4),(R_6,B_5),(R_6,B_5)}

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⇒P(A)=\frac{1}{3}

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P(B)=\frac{20}{36}

⇒ P(B) =\frac{5}{9}

Now for Sample Space of (A∩B)= {(R_5,B_2)(R_6,B_2)(R_2,B_5)(R_3,B_5)(R_4,B_5)(R_5,B_5)(R_6,B_5)}

So, P(A∩B)= \frac{7}{36}

Now we apply the formula,

P(A).P(B)=P(A∩B)

\frac{1}{3}×\frac{5}{9} ≠ \frac{7}{36}

\frac{5}{27} ≠ \frac{7}{36}

∴ The events A and B are not independent events.




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