For 95% confidence interval for a population mean reported to be 132 to 142 with standard deviation $σ = 17.85$ then the sample size used in this case, is: [Given that: $Z_{.0125} = 1.96$]

49

The correct answer is Option (2) → 49 **

95% confidence interval: $132$ to $142$

Mean estimate = midpoint = $\frac{132+142}{2}=137$

Margin of error (half–width): $E=\frac{142-132}{2}=5$

Formula for confidence interval margin: $E=Z\,\frac{\sigma}{\sqrt{n}}$

Substitute values:

$5=1.96\,\frac{17.85}{\sqrt{n}}$

$\sqrt{n}=1.96\,\frac{17.85}{5}$

$\sqrt{n}=1.96\times 3.57=6.9972$

$n=(6.9972)^{2}\approx 49$

Sample size = $49$