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Likurg_2 [28]
3 years ago
10

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

sted will get version A of the test? Express your answer as a percent, and round to the nearest tenth
Mathematics
1 answer:
alexdok [17]3 years ago
4 0

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

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<h3>What are equivalent angles?</h3>

Each angle on the second, third and fourth quadrants will have an equivalent on the first quadrant.

In this problem, the given angle is as follows:

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It is on the third quadrant, as it is between pi and 1.5 pi, hence the equivalent on the first quadrant, also known as the reference angle, is given by:

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Answer:

Test statistic Z= 0.13008 < 1.96 at 0.10 level of significance

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Step-by-step explanation:

<em>Step(I)</em>:-

Given surveyed two random samples of 390 men and 360 women who were tested

first sample proportion

                    p_{1} = \frac{360}{390} = 0.9230

second sample proportion

                  p_{2} = \frac{47}{52} = 0.9038

Step(ii):-

Null hypothesis : H₀ : There is no difference  proportion of positive tests among men is different from the proportion of positive tests among women

Alternative Hypothesis:-

There is difference between proportion of positive tests among men is different from the proportion of positive tests among women

 

Z = \frac{p_{1}- p_{2} }{\sqrt{PQ(\frac{1}{n_{1} }+\frac{1}{n_{2} }  } }

where

          P = \frac{n_{1}p_{1} +n_{2}  p_{2} }{n_{1}+n_{2}  }

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Z= \frac{0.9230-0.9038}{\sqrt{0.920 X0.08(\frac{1}{390}+\frac{1}{52}  } )}

Test statistic Z =  0.13008

Level of significance = 0.10

The critical value Z₀.₁₀ = 1.645

Test statistic Z=0.13008 < 1.645 at 0.1 level of significance

Null hypothesis is accepted

There is no difference proportion of positive tests among men is different from the proportion of positive tests among women







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