Supposing 60 out of 100 scores are passing scores, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
60 out of 100 scores are passing scores, hence [tex]n = 100, \pi = \frac{60}{100} = 0.6[/tex]
95% confidence level
So [tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7[/tex]
The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).
A similar problem is given at https://brainly.com/question/16807970
