Answer:
53.7279 ≤ μ ≤ 56.2721
Step-by-step explanation:
If the mean of the sample is known, the population standard deviation is known and the size of the sample is bigger than 30, the (1-∝) confidence interval is calculated as:
\(x-z_{\alpha/2}*\frac{s}{\sqrt{n}}\) ≤ μ ≤ \(x+z_{\alpha/2}*\frac{s}{\sqrt{n}}\)
Where x is the mean of the sample, s is the population standard deviation, n is the size of the sample and \(z_{\alpha/2}\) is the value of the standard normal distribution in which:
P(Z>z)=∝/2
If 1-∝ is equal to 99%, ∝/2 is equal to 0.005 and \(z_{\alpha/2}\) is equal to 2.57
So, if we replace x by 55, s by 7, n by 200 and \(z_{\alpha/2}\) by 2.57, we get:
\(55-2.57*\frac{7}{\sqrt{200}}\) ≤ μ ≤ \(55+2.57*\frac{7}{\sqrt{200}}\)
55 - 1.2721 ≤ μ ≤ 55 + 1.2721
53.7279 ≤ μ ≤ 56.2721
