D)No, it seems to be more likely that a red will be drawn instead of a blue.
Step-by-step explanation: I got it on usatestprep
Step-by-step explanation:
According to the mentioned information, the sample space of the experiment is as follows: S= {7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18} The outcomes of the event E are as follows: E = {7, 8, 9, 10, 11, 12} Therefore, the outcomes of Ec can be calculated as:
= {xlx ∈ S , x ∉ E}
Ec = {13, 14, 15, 16, 17, 18}
The outcomes of are {13, 14, 15, 16, 17, 18} .
The obtained outcomes of the set Ec are {13, 14, 15, 16, 17, 18} . These are calculated by subtracting the outcomes of the event E from the elements of sample space S.
kindly check the image below for the concluding part if the solution to the question
D)No, it seems to be more likely that a red will be drawn instead of a blue.
Step-by-step explanation: I got it on usatestprep
Step-by-step explanation:
According to the mentioned information, the sample space of the experiment is as follows: S= {7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18} The outcomes of the event E are as follows: E = {7, 8, 9, 10, 11, 12} Therefore, the outcomes of Ec can be calculated as:
= {xlx ∈ S , x ∉ E}
Ec = {13, 14, 15, 16, 17, 18}
The outcomes of are {13, 14, 15, 16, 17, 18} .
The obtained outcomes of the set Ec are {13, 14, 15, 16, 17, 18} . These are calculated by subtracting the outcomes of the event E from the elements of sample space S.
kindly check the image below for the concluding part if the solution to the question
a. The estimate of the probability that a seed will germinate is:
82% (0.82)
b. The estimate of the probability that a seed will not germinate is:
18% (0.18)
c. The sample space is 2.
1. Yes, because they cover the entire sample space. The probabilities assigned to the sample space add up to 1 (0.82 +0.18).
They should always add up to one. Probability summed cannot exceed 1.
Step-by-step explanation:
a) Data and Calculations:
Sample = 3,000 seeds
Germinated seeds = 2,470
Non-germinated seeds = 530 (3,000 - 2,470)
Probability of a seed germinating = 2,470/3,000 = 82% approx.
Therefore, the probability of a seed not germinating = 18% (100 - 82) or (530/3,000)
e. The probability of observing a sample mean of 5.11 or less, or of 5.29 or more, is 0.018 if the true mean is 5.2.
Step-by-step explanation:
We have a two-tailed one sample t-test.
The null hypothesis claims that the pH is not significantly different from 5.2.
The alternative hypothesis is that the mean pH is significantly different from 5.2.
The sample mean pH is 5.11, with a sample size of n=50.
The P-value of the test is 0.018.
This P-value corresponds to the probability of observing a sample mean of 5.11 or less, given that the population is defined by the null hypothesis (mean=5.2).
As this test is two-tailed, it also includes the probability of the other tail. That is the probability of observing a sample with mean 5.29 or more (0.09 or more from the population mean).
Then, we can say that, if the true mean is 5.2, there is a probability P=0.018 of observing a sample of size n=50 with a sample mean with a difference bigger than 0.09 from the population mean of the null hypothesis (5.11 or less or 5.29 or more).
The right answer is e.
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