Answer:
(0.9193, 0.9207)
Explanation:
The interval for the sample mean can be calculated as
[tex]x-z\frac{s}{\sqrt{n}}\leq x^{\prime}\leq x+z\frac{s}{\sqrt{n}}[/tex]Where x is the mean, s is the standard deviation, n is the size of the sample, and z is the standard value for the 68%, so using the standard normal table, we get that z = 1.
Replacing x = 0.92, s = 0.0050, n = 48 and z = 1, we get:
[tex]\begin{gathered} 0.92-1\frac{0.0050}{\sqrt{48}}\leq x^{\prime}\leq0.92+1\frac{0.0050}{\sqrt{48}} \\ \\ 0.92-0.0007\leq x^{\prime}\leq0.92+0.0007 \\ 0.9193\leq x^{\prime}\leq0.9207 \end{gathered}[/tex]Therefore, the interval is
(0.9193, 0.9207)
