The normal distribution \(X \sim N(30, 49)\) has a mean of 30 and a variance of 49, which means it is wider and shorter compared to \(Y\) and \(Z\). The curve for \(X\) should be centered at 30 and spread roughly from 10 to 50.

The normal distribution \(Y \sim N(30, 16)\) has the same mean as \(X\) but a smaller variance, making it taller and narrower. The curve for \(Y\) should also be centered at 30 but be higher and thinner than \(X\).

The normal distribution \(Z \sim N(50, 16)\) has a mean of 50 and the same variance as \(Y\). The curve for \(Z\) should be the same shape as \(Y\) but centered at 50.

To find the probability that a randomly chosen value of Y is negative, we need to standardize the variable. The standard normal variable Z is given by:

\(Z = \frac{Y - \mu}{\sigma}\)

where \(\sigma = \frac{1}{2} \mu\). Therefore, the standardization becomes:

\(Z = \frac{0 - \mu}{\frac{1}{2} \mu} = \frac{-\mu}{\frac{1}{2} \mu} = -2\)

We need to find \(P(Y < 0) = P(Z < -2)\).

Using standard normal distribution tables, \(P(Z < -2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228\).

To find the probability that a value of X rounds to 84, we need to calculate the probability that X is between 83.5 and 84.5.

Standardize the values using the formula for the standard normal distribution:

\(Z = \frac{X - \mu}{\sigma}\)

where \(\mu = 82\) and \(\sigma = \sqrt{126}\).

Calculate the standardized values:

\(Z_1 = \frac{84.5 - 82}{\sqrt{126}} = 0.2227\)

\(Z_2 = \frac{83.5 - 82}{\sqrt{126}} = 0.1336\)

Find the probabilities using the standard normal distribution table:

\(\Phi(0.2227) = 0.5883\)

\(\Phi(0.1336) = 0.5533\)

Subtract the probabilities to find the probability that X rounds to 84:

\(P(83.5 < X < 84.5) = \Phi(0.2227) - \Phi(0.1336) = 0.5883 - 0.5533 = 0.0350\)

(i) To find the probability that the profit is between $10,000 and $12,000, we standardize these values:

\(z_1 = \frac{12 - 6.4}{5.2} = 1.077\)

\(z_2 = \frac{10 - 6.4}{5.2} = 0.692\)

Using the standard normal distribution table, find:

\(\Phi(z_1) - \Phi(z_2) = 0.8593 - 0.7556 = 0.104\)

(ii) To find the probability of a loss, we standardize the value for a loss (profit less than $0):

\(P(\text{loss}) = P\left(z < \frac{0 - 6.4}{5.2}\right) = P(z < -1.231)\)

Using the standard normal distribution table, find:

\(P(\text{loss}) = 1 - 0.8909 = 0.109\)

The probability of exactly 1 loss in 4 days is given by the binomial distribution:

\(P(1) = (0.109)^1 (0.8909)^3 \times \binom{4}{1}\)

Calculate:

\(P(1) = 0.309 \text{ or } 0.308\)

Let the random variable \(X\) represent the weight of the bags. \(X\) is normally distributed with mean \(\mu = 1.04\) kg and standard deviation \(\sigma = 0.06\) kg.

We need to find \(P(X > 1.11)\).

First, standardize the variable using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

Substitute the values:

\(Z = \frac{1.11 - 1.04}{0.06} = 1.167\)

We need \(P(Z > 1.167)\).

Using standard normal distribution tables or a calculator, find \(P(Z > 1.167) = 1 - P(Z \leq 1.167)\).

From the tables, \(P(Z \leq 1.167) \approx 0.8784\).

Thus, \(P(Z > 1.167) = 1 - 0.8784 = 0.1216\).

Therefore, the probability that a randomly chosen bag weighs more than 1.11 kg is approximately \(0.122\).