Computation using the standard normal distribution:

• $6.10$ corresponds to $Z \approx 1.22$.
• $6.20$ corresponds to $Z \approx 1.24$.
• $6.30$ corresponds to $Z \approx 1.26$.
• Since $X \sim N(0, 125)$, the sample mean $\bar{X}$ follows $N\left(0, \frac{\sigma^2}{n}\right) = N\left(0, \frac{125}{5}\right) = N(0, 25)$. The standard error is $\sigma_{\bar{X}} = \sqrt{25} = 5$. We require $P\{\bar{X} \lt c\} = 0.90 \implies P\left(Z \lt \frac{c - 0}{5}\right) = 0.90$. From standard normal tables, $Z_{0.90} \approx 1.28155$. Thus, $\frac{c}{5} = 1.28155 \implies c = 5 \times 1.28155 = 6.4077 \approx 6.40$.

Variance of squared standard normal variable:

• $0$ would mean $X^2$ is a constant.
• $1$ is the expected value $E[X^2] = 1$.
• Since $X \sim N(0, 1)$, its square $Y = X^2$ follows a Chi-square distribution with $1$ degree of freedom ($Y \sim \chi^2_1$). The variance of a Chi-square variable with $k$ degrees of freedom is $2k$. For $k = 1$, $\operatorname{Var}(X^2) = 2 \times 1 = 2$. Alternatively, $\operatorname{Var}(X^2) = E[X^4] - (E[X^2])^2 = 3\sigma^4 - (1\sigma^2)^2 = 3 - 1 = 2$.
• $4$ is twice the correct variance.

Parameters and moments of Binomial distribution:

• $n p$ is the mean of the binomial distribution.
• $(n p)^2$ is the square of the mean.
• $n p(1 - p) = n p q$ is the variance ($\sigma^2$) of the binomial distribution.
• The standard deviation is the positive square root of the variance: $\sigma = \sqrt{n p (1 - p)}$.