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The mean for the total cost of the two items is 82. The standard deviation of the total cost of the two items is 14.14214. The probability of finding two random items at this auction with a total price of less than $80 is 0.44377.
<h3>What is a random variable?</h3>A random variable is a variable with an undetermined value that gives values to each of the results of a statistical experiment.
From the parameters given:
- Let us assume that X represents the random variable that connotes the price of the item during the large auction.
Given that:
- X is normally distributed with a mean of $41 and
- A standard deviation of $10
X N(μ, σ²)
X N(41, 10²)
Suppose we made an assumption that Y should denote the total cost of items:
i.e.
Y = X₁ + X₂
Here;
The variance of (Y) is:
= 14.14214
The probability of finding the two random items at the auction with a total price of less than $80 can be computed as:
P(Y < 80)
Since the data is normally distributed,
Recall that:
P(Z < -z) = P(Z > z)
Hence;
= P (Z > 0.1414213)
= 1 - P(Z ≤ 0.1414213)
From the Z tables, the value of Z at 0.1414213 is 0.55623;
= 1 - 0.55623
= 0.44377
Therefore, we can conclude that the probability of finding two random items at this auction with a total price of less than $80 is 0.44377.
Learn more about random variables in probability here:
Answer:
Explanation:
Hello!
Full text in the attachment.
a)
Little reminder:
An observational study is one where the investigator has no control or intervenes on it. He just defines the variable of interest and merely collects and documents the information. These types of studies are usually made as precursors to a more formal experimental study, to have an idea of what's to be expected from the population.
An experimental study or experiment is one where the investigator intervenes by defining the variable of interest and artificially manipulates the study factor. It is also one of its characteristics the randomization of cases or subjects in groups (two or more, depending on what is the hypothesis of study).
This is an example of a double-blind experiment, in which neither the experimental subjects (volunteers students) nor the researcher (physician). Aside from the fact that the subjects were controlled to make sure they followed the assigned treatment, no other factors were controlled (sleeping habits, eating habits, previous medical conditions)
Still, there is enough intervention to classify it as an experimental study.
b)
You have calculated the probability of randomly selecting a student that contracted the flu. There is no specification about the group the student belongs, you have to calculate the probability of the total of students that contracted the flu.
Let F represent the event that "the student contracted the flu", you can calculate the probability as:
P(F)=
c)
In this case, you have to calculate the probability that the student contracted the flu, given that he was assigned to the "Vitamin C" group.
This is a conditional probability since you already know that the student was from the "Vitamin C" group (Symbolized V), instead of calculating the probability using the size of the total of volunteers, you'll use the total of students assigned to that group:
P(F/V)=
d)
Two events are independent when the occurrence of one of them doesn't modify the probability of occurrence of the other one. In this case, we have the events F: "The student contracted the Flu" and V: "The student took vitamin C" if F and V are independent the probability of F should stay the same even whether or not he has taken vitamin C:
P(F)=P(F/V)
But as calculated in items b. and c. these probabilities are different, meaning that the events are not independent.
e)
The claim of the health expert is that "Vitamin C reduces the occurrence of the flu in the population of volunteer students, i.e. the number of students that contracted the flu given that they took vitamin C will be less than the number of students that contracted the flu given that the took the placebo.
The parameters of the study will be the population proportions of students with the flu that took vitamin C, p₁, and the population proportion of students with the flu that took the placebo, p₂.
H₀: p₁ ≥ p₂
H₁: p₁ < p₂
f)
Let's assume that α: 0.05, using the p-value approach, the decision rule is:
p-value ≤ α, reject the null hypothesis.
p-value > α, do not reject the null hypothesis.
The p-value: 0.0096 is less than α: 0.05, the decision is to reject the null hypothesis. At a 5% significance level, there is significant evidence to reject the null hypothesis, then the population proportion of students that contracted the flu given that they took t vitamin C is less than the proportion of students that contracted the flu given that they took the placebo. You can conclude that the expert is correct and taking vitamin C reduces the occurrence of the flu among the students.
