Question

2. A random sample of 100 observations from a normally distributed population possesses a mean equal to 83.2 and a standard deviation equal to 6.4. (1) Find a 95% confidence interval for .

Answer 1

Answer: 95% confidence interval would be (81.95,84.45).

Explanation:

Since we have given that

n = 100

mean = 83.2

Standard deviation = 6.4

So, At 95% confidence interval, z = 1.96

so, interval would be

`\bar{x}\pm z(\sigma)/(√(n))\\\\=83.2\pm 1.96* (6.4)/(√(100))\\\\=83.2\pm 1.2544\\\\=(83.2-1.2544,83.2+1.2544)\\\\=(81.95,84.45)`

Hence, 95% confidence interval would be (81.95,84.45).