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Vinvika [58]
3 years ago
11

Linda has a bag of marbles. She chooses a marble from the bag. writes down the color, and places the marbles back in the bag. Sh

e repeats this process 130 times.
Linda calculates the relative frequency of each color marble.

Which statement about Linda's experiment is true?

A. The outcomes do not appear to be equally likely, so a uniform probability model is not a good model to represent probabilities in Linda's experiment.

B. The outcomes appear to be equally likely, so a uniform probability model IS a good model to represent probabilities in Linda's experiment.

C. The outcomes do not appear to be equally likely, so a uniform probability model is a good model to represent probabilities in Linda's experiment.

D. The outcomes appear to be equally likely, so a uniform probability model is not a good model to represent probabilities in Linda's experiment.

Mathematics
2 answers:
MaRussiya [10]3 years ago
4 0
B
just took the test
hope this helps
uysha [10]3 years ago
3 0
<h2>Answer:</h2>

Option: B is the correct answer.

B.  The outcomes appear to be equally likely, so a uniform probability model IS a good model to represent probabilities in Linda's experiment.

<h2>Step-by-step explanation:</h2>

<u>Equally likely outcomes--</u>

The equally likely outcomes are the one that have equal chance or probability or likelihood of occurring.

Also, from the given relative frequency from the table we could observe that the outcomes have approximately the same relative frequency or  the chances of their occurrence is almost similar.

Hence, the outcomes are equally likely.

<em>Also a uniform probability model is one in which the outcomes have equal probability.</em>

Hence, the statement that is true about Linda's experiment is:

Option: B

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Which of the following statements are not true in developing a confidence interval for the population mean?
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Answer:

b) The width of the confidence interval becomes narrower when the sample mean increases.

Step-by-step explanation:

The confidence interval can be calculated as:

\bar{x} \pm \text{Test Statistic}\displaystyle\frac{\sigma}{\sqrt{n}}

a) The width of the confidence interval becomes wider as the confidence level increases.

The above statement is true as the confidence level increases the width increases as the absolute value of test statistic increases.

b) The width of the confidence interval becomes narrower when the sample mean increases.

The above statement is false. As the sample mean increases the width of the confidence interval increases.

c) The width of the confidence interval becomes narrower when the sample size n increases.

The above statement is true as the sample size increases the standard error decreases and the confidence interval become narrower.

8 0
3 years ago
An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi
jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

(c) 2.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, \mu=\frac{n}{2} and standard deviation, \sigma=\frac{\sqrt{n}}{2}.

Here the number of tosses is, <em>n</em> = 2500.

Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.

The mean and standard deviation are:

\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

        = μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

        = 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

                                   = 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

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