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Amiraneli [1.4K]
3 years ago
11

Suppose a loaded die has the following probability model: If this die is thrown and the top face shows an odd number, what is th

e probability that the die shows a 1
Mathematics
1 answer:
PtichkaEL [24]3 years ago
6 0

Answer:

the probability is 0.6

Step-by-step explanation:

The computation is shown below:

Let us assume the X would be the number appear on the top face

Now

= P(X = 1) ÷ P(X = 0)

= 0.3 ÷ 0.3 + 0.1 + 0.1

= 0.3 ÷ 0.5

= 0.6

hence, the probability is 0.6

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Test the hypothesis using the​ P-value approach. Be sure to verify the requirements of the test. Upper H 0​: pequals0.6 versus U
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Answer:

Step-by-step explanation:

From the given information:

The null and the alternative hypothesis can be well written as:

H_o:P=0.6

H_1:P>0.6

Given that:

n = 200

x = 135

Alpha ∝ = 0.05 level of significance

Then;

⇒ n \times p\times (1-P)

= 200 × 0.6 × (1 -0.6)

= 200 × 0.6 × 0.4

= 48 ≥ 10

The sample proportion \hat P = \dfrac{x}{n}

= \dfrac{135}{200}

= 0.675

The test statistics Z = \dfrac{\hat P - P}{\sqrt{ \dfrac{P(1-P)}{n} }}

Z = \dfrac{0.675 - 0.6}{\sqrt{ \dfrac{0.6 \times 0.4}{200} }}

Z = \dfrac{0.075}{\sqrt{ \dfrac{0.24}{200} }}

Z = 2.165

The P-value = P(Z > 2.165)

= 1 - P(Z < 2.165)

From the z tables

= 1 - 0.9848

= 0.0152

Reject the null hypothesis since P-Value is lesser than alpha. ( i.e. 0.0152 < 0.05).

Thus, there is enough evidence to conclude that the value of the population proportion is greater than 0.6

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Step-by-step explanation:

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