Let \(X\) denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by \(X_1,\ldots,X_{36}\), each independently and identically distributed with distribution \(X_i\sim\mathcal N(26,7.2)\).
You want to find
\(\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)\)
Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to
\(\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)\)
Recall that if \(X\sim\mathcal N(\mu,\sigma)\), then the sampling distribution \(\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)\) with \(n\) being the size of the sample.
Transforming to the standard normal distribution, you have
\(Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}\)
so that in this case,
\(Z=6\dfrac{\overline X-26}{7.2}\)
and the probability is equivalent to
\(\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)\)
\(=\mathbb P(Z>1.481)\approx0.0693\)
