The sum of three numbers is 100. The second number is 2 times the third. The third number is 8 less than the first. What are the numbers?

The joint probability mass function based on the given experiment is as seen in the attached file.

The whole sample space consists of 36 elements, i.e.,

Ω = {ω_ij = (i, j) : i, j = 1, ....6}

Now, with each of these 36 elements associate values of two random variables, X₁ and X₂, such that;

X₁ ≡ sum of the outcomes on the two dice.

X₂ ≡ |difference of the outcomes on the two dice|

That is;

X(ω_ij) = X₁(ω_ij) + X₂(ω_ij) = (i + j, |i - j|) i, j = 1, 2, ...., 6

Then, the bivariate rv X = (X₁, X₂) has the following joint probability mass

function (empty cells mean that the pmf is equal to zero at the relevant values of the rvs).

The joint probability mass function is seen in the attached file.

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The conclusion of the research is that we reject the null hypothesis and we conclude that the bag filling machine does not work correctly at the 426 gram setting.

We are given;

Population Mean; μ = 426

Sample mean; x' = 433

Sample size; n = 13

Standard deviation; s = 29

significance level; α = 0.05

Let us define the hypotheses;

Null Hypothesis; H₀: μ = 426 g

Alternative Hypothesis; H_a: μ < 426 g

Formula for the z-score here is;

z = (x' - μ)/(s/√n)

z = (433 - 426)/(29/√13)

z = 0.87

From online p-value from z-score calculator, we have;

p-value =  0.1922

This p-value is greater than the significance value and as such we reject the null hypothesis and we conclude that the bag filling machine does not work correctly at the 426 gram setting.

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The p-value (0.0001) is less than α = 0.05. Based on this, we should reject the null hypothesis.

A null hypothesis (H₀) can be defined the opposite of an alternate hypothesis (H₁) and it asserts that two (2) possibilities are the same.

The test statistic can be calculated by using this formula:

[tex]t=\frac{x\;-\;u}{\frac{\delta}{\sqrt{n} } }[/tex]

Where:

For this clinical trial (study), we should use a t-test and the null and alternative hypotheses would be given by:

H₀: p = 0.028

H₁: p > 0.028

For the sample proportion, we have:

Sample proportion, P = 28/888

Sample proportion, P = 0.032.

Next, we would calculate the t-test as follows:

[tex]t=\frac{0.032\;-\;0.028}{\frac{0.02 \times 0.028}{888 } }\\\\t=\frac{0.004}{\sqrt{\frac{0.00056}{888 } }}[/tex]

t = 0.004/0.00079

t-test, t = 5.06.

For the p-value, we have:

P-value = P(t > 5.06)

P-value = 1 - P(t < 5.06)

P-value = 1 - 0.9999

P-value = 0.0001.

Therefore, the p-value (0.0001) is less than α = 0.05. Based on this, we should reject the null hypothesis.

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