Let $X_i$ and $Y_i$ represent the IQ's of both batch of students.
$$
X_1, X_2, \ldots, X_{25} \sim \mathrm{~N}\left(\mu_1, 8^2\right) \text { and } Y_1, Y_2, \ldots, Y_{25} \sim \mathrm{~N}\left(\mu_2, 7^2\right)
$$
$\bar{X}=110$ and $\bar{Y}=115$
Consider, $H_0: \mu_1=\mu_2, H_A: \mu_1 \neq \mu_2$
$$
T=\bar{X}-\bar{Y} \sim \mathrm{~N}\left(\mu_1-\mu_2, \frac{64}{25}+\frac{49}{25}\right) \text { i.e. } \mathrm{N}\left(\mu_1-\mu_2, \frac{113}{25}\right)
$$
Test: Reject $H_0$ if $|T|>c$.
$$
\begin{aligned}
\alpha=P\left(|T|>c \mid H_0\right) & =P\left(\left|\frac{T}{\sqrt{113 / 25}}\right|>\frac{c}{\sqrt{113 / 25}}\right) \\
& =P\left(|Z|>\frac{c}{\sqrt{113 / 25}}\right)=2 F_Z\left(\frac{-c}{\sqrt{113 / 25}}\right)
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow c=-\sqrt{\frac{113}{25}} F_Z^{-1}(\alpha / 2) \\
& \Rightarrow c=-\sqrt{\frac{113}{25}} F_Z^{-1}(0.025) \\
& \Rightarrow c=-\sqrt{\frac{113}{25}} \times(-1.96)=4.167
\end{aligned}
$$
Since, $|\bar{X}-\bar{Y}|=|110-115|=5>4.167$
Therefore, we will reject $H_0$.
There is a significant difference between the IQs of the two batches, and since the second batch's mean (115) is higher:
Hence A is correct
