21% of the students had a score of 80/100 or higher on both midterms. 25% had a score of 80/100 or higher on midterm 1. what per

To begin with, consider a straightforward hidden Markov model (HMM). We observe a series of outcomes from rolling a four-sided die at an "occasionally dishonest casino". At time t, the result x_t belongs to the set {1, 2, 3, 4}. The casino can either be in state z_t belonging to {1, 2}. When z_t is equal to 1, it uses a fair die, whereas when z_t is equal to 2, the die is biased towards rolling a 1. Specifically: p (x_t = 1 | z_t = 1) = p (x_t = 2 | z_t = 1) = p (x_t = 3 | z_t = 1) = p (x_t = 4 | z_t = 1) = 0.25, p (x_t = 1 | z_t = 2) = 0.7, and p (x_t = 2 | z_t = 2) = p (x_t = 3 | z_t = 2) = p (x_t = 4 | z_t = 2) = 0.1. Assume there is an equal likelihood of starting in either state at time t = 1, which leads to p (z1 = 1) = p (z1 = 2) = 0.5. The casino generally maintains the same die for several iterations, but it occasionally switches states with these probabilities: p (z_t + 1 = 1 | z_t = 1) = 0.8 and p (z_t + 1 = 2 | z_t = 1) = 0.2; likewise, p (z_t + 1 = 2 | z_t = 2) = 0.1 and p (z_t + 1 = 1 | z_t = 2) = 0.9. To find the probability p (z1 = z2 = z3) that the same die is used across the first three rolls under the HMM generative model, consider the following. If we assume the first die is state 1, the probability can be calculated as p(z1=1)=0.5, and consequently, p(z2=1|z1=1)=0.8 signifies that the same die might still be in use. Alternatively, if we start with the die in state 2, p(z1=2)=0.5 and p(z2=2|z1=2)=0.9 also provides a probability. Adjacent transition probabilities can be expressed as follows: p(z_t+1=2|z_t=1)=1-p(z_t+1=1|z_t=1)=0.2 and p(z_t+1=1|z_t=2)=1-p(z_t+1=2|z_t=2)=0.1. The equation for p(z3=1|z1=1) can thus be derived as a combination of previous probabilities: [p(z3=1|z2=2)*p(z2=2|z1=1)] + [p(z3=1|z2=1)*p(z2=1|z1=1)]=0.1*0.2+0.8*0.8=0.66. Similarly for p(z3=2|z1=2): [p(z3=2|z2=2)*p(z2=2|z1=2)]+[p(z3=2|z2=1)*p(z2=1|z1=2)]=0.9*0.9+0.2*0.1=0.83. Consequently, the overall probability for using the same die for the initial three rolls can be computed via: {p(z1=1)*p(z3=1|z1=1)}*{p(z1=2)*p(z3=2|z1=2)} = 0.5*0.66+0.5*0.83 = 0.745; thus, the probability amounts to 0.745.

The answer is 116.15

Step-by-step explanation:

The calculation is as follows: 1.95 - 30.00 - 7.20 - 38.50 = 38.50

This results in 77.65 = 38.50

Thus, x = $116.15

Answer:

a) Null hypothesis:

Alternative hypothesis:

b)

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c)

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d)

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

represents the average of the samples

denotes the standard deviation of the samples

indicates the number of samples

is the value we are examining

defines the significance level for the test.

t represents the specific statistic of interest

indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:

Alternative hypothesis:

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

(1)

Part b

In this scenario t=-2.3,

Initially, we need to calculate the degrees of freedom

Since this is a left-tailed test, the p-value is determined by:

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8,

We need to find the degrees of freedom

For the left-tailed test, the p-value is given by:

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6,

Firstly, we find the degrees of freedom

Since we are conducting a left-tailed test, the p-value is calculated as:

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

The function

represents a parabola positioned at the vertex (5, 3).

Attached here is the graph depicting this function.

Answer: F(x) > 0 in the ranges (-0.7, 0.76) and (0.76, ∞).

I hope this is useful:)