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N76 [4]
3 years ago
8

A study shows that 84​% of the population of all calculus students consider calculus an exciting subject. Suppose 29 students ar

e randomly and independently selected from the population. If the true percentage is really 84​%, find the probability of observing 28 or more of the students who consider calculus to be an exciting subject in the sample of 29. Round to six decimal places.
Mathematics
1 answer:
ratelena [41]3 years ago
5 0

Answer:

P = 0.041553

Step-by-step explanation:

Probability that a student considers calculus exciting. = 0.84

Probability that a student finds calculus not exciting = 0.16

This is a binomial experiment.

p = 0.84 and n = 29

The probability of observing 28 or more of the students who consider calculus to be an exciting subject in our sample of 29 is given by:

\left(\begin{array}{ccc}29\\28\end{array}\right) .0.84^{28} .0.16^{1}+\left(\begin{array}{ccc}29\\29\end{array}\right) .0.84^{29} .0.16^{0}

P = 0.041553

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The Information Technology Department at a large university wishes to estimate the proportion of students living in the dormitor
nadya68 [22]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16  

And rounded up we have that n=385

Step-by-step explanation:

Previous concepts

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".  

Solution to the problem

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 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

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 0.05 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)  

We can use as an estimator for p \hat p =0.5. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16  

And rounded up we have that n=385

3 0
3 years ago
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2. put into into like terms

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Have a nice day :)

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