Find the​ range, variance, and standard deviation for the given sample data. Include appropriate units in the results.Listed below are the measured radiation absorption rates​ (in W/kg) corresponding to various cell phone models. If one of each model is measured for radiation and the results are used to find the measures of​ variation, are the results typical of the population of cell phones that are in​ use?
0.950.95
0.730.73
0.630.63
0.910.91
1.321.32
1.481.48
0.630.63
1.231.23
0.910.91
1.411.41
0.670.67
The range of the sample data is
nothing

​(Round to three decimal places as​ needed.)
Sample standard
deviationequals=nothing

kg.kg.
left parenthesis Upper W divided by kg right parenthesis squared .W/kg2.
W.W.
Upper W divided by kg.W/kg.
​(Round to three decimal places as​ needed.)
Sample
varianceequals=nothing

kg.kg.
left parenthesis Upper W divided by kg right parenthesis squared .W/kg2.
Upper W divided by kg.W/kg.
W.W.
​(Round to three decimal places as​ needed.)
If one of each model is measured for radiation and the results are used to find the measures of​ variation, are the results typical of the population of cell phones that are in​ use?
A. ​No, because it is necessary to have at least 5 of each cell phone in order to get a meaningful result. Only including one of each cell phone model is not representative of each cell phone model.
B. ​Yes, because the results from any sample of cell phones will be typical of the population.
C. ​Yes, because each model is being represented in the sample. Any sample that considers all possible cell phone models will produce results typical of the population of cell phones.
D. No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.

Answers

Answer 1
Answer:

Answer:

Range = 1.48-0.63=0.850 W/kg

s= 0.320 W/Kg

s^2 = 0.320^2= 0.103 W^2 /kg^2

D. No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.

Step-by-step explanation:

For this case we have the following data values:

0.95,0.73,0.63,0.91,1.32,1.48,0.63,1.23,0.91,1.41,0.67

The first step on this case is order the datase on increasing way and we got:

0.63 0.63 0.67 0.73 0.91 0.91 0.95 1.23 1.32 1.41 1.48

The range is defined as Range = Max-Min

And if we replace we got:

Range = 1.48-0.63=0.850 W/kg

The sample standard deviation is given by this formula:

s= \sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}

And if we replace we got s= 0.320 W/Kg

And the sample variance is just the standard deviation squared so we got:

s^2 = 0.320^2= 0.103 W^2 /kg^2

And for the last question about : If one of each model is measured for radiation and the results are used to find the measures of​ variation, are the results typical of the population of cell phones that are in​ use?

We can conclude this:

D. No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.


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Step-by-step explanation:

A 3 ft by 5 ft wooden board has three equally sized circles cut out of it. A penny will be randomly tossed onto the board. What is the probability the penny will fall through any of the holes?

Answers

Answer:

Pr = 15.71% ≈ 16%

Step-by-step explanation:

Let's begin by listing out the data given unto us, we have:

Length of board = 3 ft, Width of board = 5 ft,

Area of board = Length * Width = 3 * 5 = 15 ft²,

Diameter of each circle (d) = 1 ft; r = d ÷ 2

⇒ r = 1 ÷ 2 = 0.5 ft

Area of each circle = πr² = π * 0.5² = 0.785 ft²

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Converting to percentage, we multiply by 100

Pr = 0.1571 * 100

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Answers

Answer:

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Hope this helps!

We know that if the probability of an event happening is 100%, then the event is a certainty. Can it be concluded that if there is a 50% chance of contracting a communicable disease through contact with an infected person, there would be a 100% chance of contracting the disease if 2 contacts were made with the infected person

Answers

Answer:

The correct answer to the following question will be "No". The further explanation is given below.

Step-by-step explanation:

Probability (Keeping the disease out of 1 contact)

= 0.5

Probability (not keeping the disease out of 1 contact)

= 1-0.5

= 0.5

Now,

Probability (not keeping the disease out of 2 contact)

= Keeping the disease out of 1 contact × not keeping the disease out of 1 contact

On putting the estimated values, we get

= 0.5* 0.5

= 0.25

So that,

Probability (Keeping the disease out of 2 contact)

= 1-0.25

= 0.75 \ i.e., 75 \ percent

∴  Not 100%

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Answers

Answer:

D is the answer to your question

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Answers

Answer:

350

Step-by-step explanation: