For the binomial distribution table of defective tomatoes taken over $x=0$ to $5$, the mean and standard deviation are:

1. The table is truncated at $x=5$, so the mean must be computed from it directly, not from $np$. 2. Apply $E(X)=\sum x_i p(x_i)$ across $x=0$ to $5$, giving $2.5662$. 3. For the spread use $\delta=\sqrt{E(X^2)-[E(X)]^2}$. 4. That evaluates to $1.4620$. 5. Using $E(X)=np=3$ and $\delta=npq=1.5969$ ignores the truncation and is the common error; $2.4454$ and $1.5131$ belong to the Poisson table. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 1: Probability_