Question:

A study of 50 individuals found mean systolic blood pressure $\bar{x} = 124.6$ mm Hg with sample standard deviation $S = 10.3$ mm Hg. Find the no. of individuals needed to reduce the standard error of the mean to 1 mm Hg.

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Always use $SE = \frac{S}{\sqrt{n}}$. To decrease SE, the sample size $n$ must be increased.
Updated On: Sep 26, 2025
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The Correct Option is D

Solution and Explanation

Step 1: Recall the formula for Standard Error (SE).
The standard error of the mean is given by:
\[ SE = \frac{S}{\sqrt{n}} \] where $S$ is the sample standard deviation and $n$ is the number of individuals.
Step 2: Substitute the given values.
Here, $SE = 1$ and $S = 10.3$.
\[ 1 = \frac{10.3}{\sqrt{n}} \] Step 3: Solve for $n$.
\[ \sqrt{n} = \frac{10.3}{1} = 10.3 \] \[ n = (10.3)^2 = 106.09 \approx 106 \] Step 4: Conclusion.
Thus, to reduce the standard error of the mean to 1 mm Hg, at least 106 individuals are required.
Final Answer:
\[ \boxed{106} \]
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