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Shtirlitz [24]
3 years ago
15

the numbers below represent 10 trials of a simulation 2632, 1365,9367,2056, 0026, 6564, 1434, 8045, 4781, 8681 the numbers 0-7 r

epresent students who watched television last night and 8 and 9 represent students who did not. based on the simulated data, what was the probability that exactly 2 out groups out of 4 randomly selected watched television last night show with work
Mathematics
1 answer:
algol133 years ago
6 0

Answer:

  1/10

Step-by-step explanation:

Apparently, a "trial" is a representation of a sample of 4 students. If the digits in that sample are 8 or 9, the students did not watch television. We can revise each of the trials by changing digits 0-7 to W, and digits 8-9 to N. (W = watched TV; N = did not watch TV.)

Now, the 10 trials are ...

  WWWW, WWWW, WWWW, WWWW,

  WWWW, WWWW, WWWW, NWWW, WWNW, NWNW

Of the 10 trials, 1 trial had the result that exactly 2 of the 4 watched TV.

The probability that exactly 2 of 4 watched TV is about 1/10 = 10%.

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Step-by-step explanation:

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A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimat
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Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A.  ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

For this problem, we consider that we want it to be within 4%.

Item a:

  • The sample size is <u>n for which M = 0.04.</u>
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M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2

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Rounding up:

A sample of 601 is needed.

Item b:

The estimate is \pi = 0.96, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}

\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2

n = 92.2

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to \pi = 0.5, the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151  

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