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solniwko [45]
3 years ago
15

The value of a probability must be in the interval [0, 1] True False

Mathematics
2 answers:
ollegr [7]3 years ago
8 0
The answer to this question would be true
Zanzabum3 years ago
4 0

Answer:

True.

Step-by-step explanation:

[0, 1] means all numbers in between 0 and 1, 0 and 1 included.

A probability is always a number in that interval. For example: 0.384, 0.222, 0.938, etc!

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(a) A newspaper conducted a statewide survey concerning the 1998 race for state senator. The newspaper took a SRS of 1200 regist
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Answer:

n=9604

Step-by-step explanation:

1) Notation and definitions

X=620 number of people that would vote for the Republican candidate

n=1200 random sample taken

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p true population proportion of people that would vote for the Republican candidate

Me= 0.01 represent th margin of error.

Confidence =0.95 or 95%

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1- p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

The problem says that we need to use \hat p =0.05, and replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.96})^2}=9604  

And rounded we have that n=9604

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