Answer with explanation:
Total Sample Size = 4,500 Students
Sample Chosen(n) = 130 students
Mean IQ of the Sample,\(\bar{X} = 120\)
Standard Deviation, \(\sigma=5\)
To Calculate 99% confidence interval for the students' mean IQ score, we will calculate,
\(Z_{99 percent}=2.58\)
Formula for Confidence Interval
\(=\bar{X} \pm Z_{99 percent}\times \frac{\sigma}{\sqrt{n}}\\\\=120 \pm 2.58 \times \frac{5}{\sqrt{130}}\\\\=120 \pm \frac{12.9}{11.401}\\\\ =120 \pm 1.1314\\\\=120 \pm 1.13\)
The Value of Confidence interval will lie between
⇒ 120 - 1.13 to 120 +1.13
⇒118.87 to 121.13
Option A: 118.87−121.13
