Answer to Question #187435 in Statistics and Probability for kyut is me

Question

Assuming the standard deviation is unchanged and that the usage is normally distributed, provide an expression for calculating a 99% confidence interval for the mean usage in the March quarter of 2006. (Exactly one option must be correct)

Answer 1

Since $σ$ is assumed known,

we use the interval $x̄ ± z \dfrac{σ}{ n},$

 where $σ = 81, n = 30$ and where z is chosen to ensure that $P(|Z|≤ z) = 0.99.$

From the normal tables, $P(|Z|≤ 2.575) = 0.99$ (because $P(Z < 2.575) = 0.995)$ and so we use $z = 2.575.$

99% confidence interval expression $= \bar{x}\pm 2.575\times \dfrac{\sigma}{n}$