a.
The sample is consisted of 919 peas and 720 have red flowers
If we assuma a 3/4 probavility that a pea will have a red flower, according to a binomial distribution (that is, which assumes only to possible outcomes: red and not red), the standard deviation must be given by:
![\begin{gathered} \sigma=\sqrt[]{np(1-p)} \\ \sigma=\sqrt[]{919\cdot(3)/(4)(1-(3)/(4))} \\ \sigma\approx\sqrt[]{172} \\ \sigma\approx13.1268 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hcsf54nu60aw9r2dxw4bp46fipo489tp3q.png)
Then, the Z score of 720 with a expected value of 689;25 and a standard deviation of 13.1268 is given by:

Using a normal approximation, according to a normal distribution table, we conclude that the probability of getting 720 or more peas with red flowers is given by 0.0096 = 0.96%
b.
Usually, when a