Step 1

Here we need to find the required sample size.

Step 2

Here it is given that the sample is taken from N(M,25).

Here we need to determine the sample size.

We know \(\displaystyle{n}={\left({\frac{{{z}_{{\frac{\alpha}{{2}}}}\times{6}}}{{{E}}}}\right)}^{{{2}}}\).

At \(\displaystyle{90}\%\) confidence level \(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={1.645}\)

Also, \(E=\frac{length}{2}=\frac{4}{2}=2, 6=5\)

\(\displaystyle\therefore{n}={\left({\frac{{{1.645}\times{5}}}{{{2}}}}\right)}^{{{2}}}={16.912}\)

\(\displaystyle\approx{17}\)

Sample size \(\displaystyle{n}={17}\)