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vodomira [7]
2 years ago
14

a survey asked 50 students if they play an instrument and if they are in band. 25 students play an instrument 20 are in a band a

nd 30 are not in a band. which table shows these data correctly entered in a two way frequency table?

Mathematics
2 answers:
gayaneshka [121]2 years ago
7 0
I would go with d .... If not it would be c...


Really hope I help but I would choose d
anzhelika [568]2 years ago
4 0

Answer:

Option 4th is correct.

Step-by-step explanation:

We have been the information 25 students play an instrument 20 are in a band 30 are not in a band.

We need to create a two way table:

                                      Band            Not in Band       Total

Play instrument                20                   5                     25

Do not play instrument     0                     25                 25

Total                                  20                    30                50

Option 4th is correct.

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Find the length of the leg of a right triangle with leg length b= 21.5 inches and the hypotenuse c= 31.9 inches. Use a calculato
sweet [91]

Answer:

23.6 (approximate value)

Step-by-step explanation:

  • a*a+b*b=c*c Pythagorean thereom
  • (21.5)^2 + b*b = (31.9)^2
  • b*b= 1017.61 - 462.25
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8 0
2 years ago
in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru
melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is  

n = 200

the proportion is

p = 154/200 = 0.77

<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>

<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant  difference. </em>

The approximation would be to a Normal curve with this parameters:

<em>Mean </em>

p = 0.77

<em>Standard deviation </em>

\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298

The 90% confidence interval for the proportion would be then  

\bf  [0.77-z^*0.0298, 0.77+z^*0.0298]

where \bf z^* is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval \bf [-z^*,z^*] is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

<em>NORMSINV(0.95) and we get a value of 1.645 </em>

The 90% confidence interval for the proportion is then

\bf  [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

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