Answer:
E. P(W | H)
Step-by-step explanation:
What each of these probabilities mean:
P(H): Probability of the game being at home
P(W): Probability of the game being a win.
P(H and W): Probability of the game being at home and being a win.
P(H|W): Probability of a win being at home.
P(W|H): Probability of winning a home game.
Which of the following probabilities do you need to find in order to determine the proportion of at-home games that were wins?
This is the probability of winning a home game. So the answer is:
E. P(W | H)
Answer:
T
Step-by-step explanation:
Answer:
The variance and standard deviation of <em>X</em> are 0.48 and 0.693 respectively.
The variance and standard deviation of (20 - <em>X</em>) are 0.48 and 0.693 respectively.
Step-by-step explanation:
The variable <em>X</em> is defined as, <em>X</em> = number of defective items in the sample.
In a sample of 20 items there are 4 defective items.
The probability of selecting a defective item is:
A random sample of <em>n</em> = 3 items are selected at random.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 3 and <em>p </em>= 0.20.
The variance of a Binomial distribution is:
Compute the variance of <em>X</em> as follows:
Compute the standard deviation (σ (X)) as follows:
Thus, the variance and standard deviation of <em>X</em> are 0.48 and 0.693 respectively.
Now compute the variance of (20 - X) as follows:
Compute the standard deviation of (20 - X) as follows:
Thus, the variance and standard deviation of (20 - <em>X</em>) are 0.48 and 0.693 respectively.
At 13% significance level, there isn't enough evidence to prove the administrators to claim that the mean score for the state's eighth graders on this exam is more than 280.
<h3>How to state hypothesis conclusion?</h3>
We are given;
Sample size; n = 78
population standard deviation σ = 37
Sample Mean; x' = 280
Population mean; μ = 287
The school administrator declares that mean score is more (bigger than) 280. Thus, the hypotheses is stated as;
Null hypothesis; H₀: μ > 280
Alternative hypothesis; Hₐ: μ < 280
This is a one tail test with significance level of α = 0.13
From online tables, the critical value at α = 0.13 is z(c) = -1.13
b) Formula for the test statistic is;
z = (x- μ)/(σ/√n)
z = ((280 - 287) *√78 )/37
z = -1.67
c) From online p-value from z-score calculator, we have;
P[ z > 280 ] = 0.048
d) The value for z = -1.67 is smaller than the critical value mentioned in problem statement z(c) = - 1.13 , the z(s) is in the rejection zone. Therefore we reject H₀
e) We conclude that at 13% significance level, there isn't enough evidence to prove the administrators to claim that the mean score for the state's eighth graders on this exam is more than 280.
Read more about Hypothesis Conclusion at; brainly.com/question/15980493
#SPJ1
