For $$ C=95 \% $$ If $$ \sigma=7.21 $$ and $$ \mathrm{n}=35 $$ then $$ E=4.12 $$

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We begin with the formula for the margin‐of‐error when the population standard deviation is known:
  
$$
E=z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}},
$$
  
where for a 95% confidence level we have $$ z_{\frac{\alpha}{2}}=1.96 $$.

**Step 1. Write down the known values.**

$$
\sigma=7.21,\quad n=35,\quad z_{\frac{\alpha}{2}}=1.96
$$
  
   
**Step 2. Compute the standard error $$\frac{\sigma}{\sqrt{n}}$$.**

First, calculate the square root of the sample size:
  
$$
\sqrt{n}=\sqrt{35}\approx5.92.
$$
  
Then, the standard error is

$$
\frac{\sigma}{\sqrt{n}} \approx\frac{7.21}{5.92}\approx1.22.
$$
  
   
**Step 3. Compute the margin‐of‐error $$E$$.**

Multiply the standard error by $$1.96$$:

$$
E=1.96\times1.22\approx2.39.
$$
  
   
**Step 4. Compare with the provided value.**

The computed margin‐of‐error is approximately $$ 2.39 $$. This value does not match the value $$ E=4.12 $$ stated in the problem. In order for $$ E $$ to equal $$ 4.12 $$ with the same values of $$\sigma$$ and $$ n $$, the corresponding $$ z $$-value would have to be

$$
z^*=\frac{E\sqrt{n}}{\sigma}=\frac{4.12\times\sqrt{35}}{7.21}\approx\frac{4.12\times5.92}{7.21}\approx3.38.
$$

A $$ z $$-value of approximately $$ 3.38 $$ corresponds to a much higher confidence level than 95%.

**Conclusion**

Under the normal conditions for a 95% confidence interval, the margin‐of‐error should be

$$
E=1.96\frac{7.21}{\sqrt{35}}\approx2.39,
$$

not $$ 4.12 $$.
The margin-of-error $$ E $$ is approximately 2.39, not 4.12, with the given values of $$ \sigma = 7.21 $$ and $$ n = 35 $$ at a 95% confidence level.

For \( C=95 \% \) If \( \sigma=7.21 \) and \( \mathrm{n}=35 \) then \( E=4.12 \)

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