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maksim [4K]
3 years ago
10

A bag contains 4 blue marbles, 2 yellow marbles, and 6 orange marbles. What is the probability of picking an orange marble and t

hen another orange marble without replacing the marble? Round the answer to two decimal places. Write the answer just as a decimal.
Mathematics
1 answer:
ikadub [295]3 years ago
5 0

We will see that the probability of picking two orange marbles without replacement is 0.23

<h3>How to get the probability?</h3>

If we assume that all the marbles have the same probability of being randomly picked, then the probability of getting an orange marble is given by the quotient between the number of orange marbles and the total number of marbles, this gives:

P = 6/12 = 1/2

And then we need to get another orange marble, without replacing the one we picked before, this time there are 5 orange marbles and 11 in total, so the probability is:

Q = 5/11

Finally, the joint probability (of these two events happening) is the product of the probabilities, so we get:

P*Q = (1/2)*(5/11) = 0.23

If you want to learn more about probability, you can read:

brainly.com/question/251701

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b) No. Is not too small to support the conjecture with confidence

Step-by-step explanation:

1) Previous concepts and data given

The sample sizes are n_A=n_B=36

We can find the distribution for \hat X_A -\hat X_B, and since is a sampling distribution and \hat X_A and \hat X_B follows normal distributions then \hat X_A -\hat X_B follows a normal distribution too.

The mean and the deviation for \hat X_A -\hat X_B is given by:

\mu_{\hat X_A}-\mu_{\hat X_B}=\mu_A -\mu_B and since \mu_A =\mu_B then \mu_A -\mu_B=0

\sigma_{\hat X_A -\hat X_B}=\sqrt{\frac{\sigma^2_A}{n_A}+\frac{\sigma^2_B}{n_B}}

Since both A and B have the same deviation and variance then:

\sigma_{\hat X_A -\hat X_B}=\sqrt{\frac{\sigma^2}{n_A}+\frac{\sigma^2}{n_B}}=\sqrt{\frac{1}{36}+\frac{1}{36}}=\sqrt{\frac{1}{18}}=0.236

2) Part a

We want to find:

P(\hat X_A -\hat X_B \geqslant 0.2)

The random variable

Z=\frac{\hat X_A -\hat X_B -\mu_{\hat X_A}-\mu_{\hat X_B}}{\sigma_{\hat X_A -\hat X_B}} follows a normal distribution with mean 0 and deviation 1 we can use this:

P(\hat X_A -\hat X_B \geqslant 0.2)=P(\frac{\hat X_A -\hat X_B -\mu_{\hat X_A}-\mu_{\hat X_B}}{\sigma_{\hat X_A -\hat X_B}}\geqslant \frac{0.2-\mu_{\hat X_A -\hat X_B}}{\sigma_{\hat X_A -\hat X_B}})

P(\hat X_A -\hat X_B \geqslant 0.2)=P(Z\geqslant \frac{0.2-0}{0.236})=1-P(Z

3) Part b

From part a we found that we have a change of 19.84% that the experiment would give a difference between the sample means which is greater or equal than 0.2

Analyzing the probability obtained we can say that is not too small to support the conjecture with confidence that the population means for the two machines are different

   

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