Question

Express the confidence interval (0.432, 0.52 in the form of ± e.

Answer 1

For a known standard deviation, the confidence interval for sample size = n is
$(x-z \frac{ \sigma }{ \sqrt{n}},x+x \frac{ \sigma }{ \sqrt{n} } )$
where
x =  average
n = sample size
$\sigma$ = stad. deviation
z = contant that reflects confidence interval

Let a = x
Let b = $z \frac{ \sigma }{ \sqrt{n} }$

From the given information,
a - b = 0.432          (1)
a + b = 0.52           (2)

Add (1) and (2): 2a = 0.952  =>  a = 0.476
Subtract (2) from (1): -2b = -0.088  => b = 0.044

Therefore, the confidence interval may be written as
(0.476 - 0.044, 0.476 + 0.044), or as
(0.476 $\pm$ 0.044)