<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

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 $\frac{1+\alpha}{2}$ .

60 out of 100 scores are passing scores, hence $n = 100, \pi = \frac{60}{100} = 0.6$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7$

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

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2 years ago