Question

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240. Provide a 95% confidence interval for the SAT scores of all the students who applied for the merit scholarships.

Answer 1

Answer: 95% confidence interval would be (1341.2,1458.8).

Step-by-step explanation:

Since we have given that

n = 64

Average = 1400

Standard deviation = 240

We need to find the 95% confidence interval for SAT scores.

So, z = 1.96

So, Interval would be

$\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=1400\pm 1.96\times \dfrac{240}{\sqrt{64}}\\\\=1400\pm 1.96\times \dfrac{240}{8}\\\\=1400\pm 1.96\times 30\\\\=1400\pm 58.8\\\\=(1400-58.8,1400+58.8)\\\\=(1341.2,1458.8)$

Hence, 95% confidence interval would be (1341.2,1458.8).