(a) To find the number of rods expected to have a length less than 54.8 mm, we first standardize the value:

\(P(X < 54.8) = P\left(Z < \frac{54.8 - 55.6}{1.2}\right)\)

\(= P(Z < -0.6667) = 1 - 0.7477 = 0.2523\)

The expected number is then:

\(400 \times 0.2523 = 100.92\)

Rounding gives 100 or 101 rods.

(b) To find the probability that a rod's length is within half a standard deviation of the mean, we calculate:

\(P\left(-\frac{1}{2} < Z < \frac{1}{2}\right) = \Phi\left(\frac{1}{2}\right) - \Phi\left(-\frac{1}{2}\right)\)

\(= 2\Phi\left(\frac{1}{2}\right) - 1\)

\(= 2 \times 0.6915 - 1 = 0.383\)

Let the random variable \(X\) represent the height of the sunflowers, where \(X \sim N(112, 17.2^2)\).

We need to find \(P(X > 120)\).

First, standardize the variable:

\(Z = \frac{X - \mu}{\sigma} = \frac{120 - 112}{17.2} = 0.4651\)

Now, find the probability:

\(P(X > 120) = 1 - \Phi(0.4651)\)

Using standard normal distribution tables or a calculator, \(\Phi(0.4651) \approx 0.6790\).

Thus, \(P(X > 120) = 1 - 0.6790 = 0.321\)

Let the random variable representing the distance be denoted as \(X\), where \(X \sim N(35.0, 11.6^2)\).

We need to find \(P(30 < X < 40)\).

First, standardize the variable using the formula for the z-score:

\(z = \frac{x - ext{mean}}{ ext{standard deviation}}\)

For \(x = 40\):

\(z = \frac{40 - 35.0}{11.6} = 0.431\)

For \(x = 30\):

\(z = \frac{30 - 35.0}{11.6} = -0.431\)

Using the standard normal distribution table, find \(\, \Phi(0.431)\) and \(\, \Phi(-0.431)\).

\(P(30 < X < 40) = \Phi(0.431) - (1 - \Phi(0.431))\)

\(= 0.334\)

To find the proportion of cars exceeding the speed limit, we first standardize the speed limit using the z-score formula:

\(z = \frac{110 - 107.6}{13.8} = 0.174\)

Next, we find the probability that a car's speed is greater than 110 km h^-1:

\(P(X > 110) = 1 - \Phi(0.174)\)

Using standard normal distribution tables, \(\Phi(0.174) \approx 0.5691\).

Thus,

\(P(X > 110) = 1 - 0.5691 = 0.431\)

(a) To find the probability that an apple is sold to the supermarket, we need to calculate the probability that the weight of an apple is between 142 grams and 205 grams. We use the standard normal distribution:

\(P(142 < X < 205) = P\left(\frac{142 - 170}{25} < z < \frac{205 - 170}{25}\right)\)

This simplifies to:

\(P(-1.12 < z < 1.4)\)

Using the standard normal distribution table, we find:

\(\Phi(1.4) - (1 - \Phi(1.12)) = 0.9192 + 0.8686 - 1\)

Thus, the probability is:

\(0.788\)

(b) To calculate the total income, we first find the probability that an apple weighs more than 205 grams:

\(P(X > 205) = 1 - 0.9192 = 0.0808\)

The income from apples sold to the supermarket is:

\(0.788 \times 0.24 \times 20000\)

The income from apples sold to the local shop is:

\(0.0808 \times 0.30 \times 20000\)

Therefore, the total income is:

\((0.0808 \times 0.30 + 0.788 \times 0.24) \times 20000 = 4266.24\)