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earnstyle [38]
2 years ago
14

How can relative frequencies be used to help us estimate probabilities occurring in sampling distributions? relative frequencies

are the probabilities occurring in sampling distributions. relative frequencies show the likelihood of a certain parameter falling within the class bounds. relative frequencies can be thought of as a measure or estimate of the likelihood of a certain statistic falling within the class bounds. relative frequencies cannot be used to help us estimate probabilities occurring in sampling distributions?
Mathematics
2 answers:
morpeh [17]2 years ago
7 0
The <u>correct answer</u> is:

<span>Relative frequencies are the probabilities occurring in sampling distributions.

Explanation:

Relative frequencies are the fraction of times an event occurs within a sample. 

This is the same definition as experimental probability; thus relative frequencies are the probabilities occurring in sampling distributions.</span>
blagie [28]2 years ago
6 0
I believe the answer is is relative frequencies can be thought of as a measure or estimate of the likelihood of a certain statistic falling within the class bounds.  Relative frequency is the ratio of occurrence of an event and the total number of outcomes. It is used when it is not easy to find the probability of events by simply looking at their situations.
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Revenue > Cost
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0.50x > 10+0.20x
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0.30x > 10
x > 10/0.30
x > 33.333 which is approximate

Round up to the nearest whole number to get x = 34. You need to sell at least 34 cups of lemonade to have the revenue R(x) be larger than the cost C(x)

Answer: 34
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A 40-meter-long blade of a wind turbine makes one complete revolution in 10 seconds. To the nearest tenth, the linear velocity o
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In professional basketball games during 2009-2010, when Kobe Bryant of the Los Angeles Lakers shot a pair of free throws, 8 time
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Answer:

Is plausible that the successive throws are independent

Step-by-step explanation:

1) Table with info given

The observed values are given by the following table

__________________________________________________

First shot          Made          Second shot missed           Total

__________________________________________________

Made                  152                       33                                185

Missed                37                         8                                  45

__________________________________________________

Total                    189                       41                                230

2) Calculations and test

We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:

Null hypothesis: Independence between two successive free throws

Alternative hypothesis: No Independence between two successive free throws

_____________________________________________________

First shot                    Made                                    Second shot missed

_____________________________________________________

Made                  189(185)/230=152.0217                41(185)/230=32.9783

Missed                189(45)/230=36.9783                  41(45)/230=8.0217

_____________________________________________________

On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.

3) Calculate the chi square statistic

The statistic for this case is given by:

\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}

Where O represent the observed values and E the expected values. Replacing the values that we got we have this

\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881

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df=(r-1)(c-1)=(2-1)(2-1)=1 on this case r means the number of rows and c the number of columns.

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Solve

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we need 18600 tiles of 5 mm^2</span>
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