Let the time taken be represented by the random variable \(X\), where \(X \sim N(85, 6.8^2)\).

We need to find \(P(79 < X < 91)\).

Standardize the variable using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(\mu = 85\) and \(\sigma = 6.8\).

Calculate the z-scores:

For \(X = 79\), \(Z = \frac{79 - 85}{6.8} = -0.8824\).

For \(X = 91\), \(Z = \frac{91 - 85}{6.8} = 0.8824\).

Thus, \(P(79 < X < 91) = P(-0.8824 < Z < 0.8824)\).

Using the standard normal distribution table, find:

\(\Phi(0.8824) = 0.8111\) and \(\Phi(-0.8824) = 1 - 0.8111 = 0.1889\).

Therefore, \(P(-0.8824 < Z < 0.8824) = 0.8111 - 0.1889 = 0.622\).

We need to find the probability that a cartridge contains less than 28.9 ml of ink. Let the random variable representing the volume of ink be normally distributed as \(X \sim N(30, 1.5^2)\).

We standardize the variable to find \(P(X < 28.9)\):

\(P(X < 28.9) = P\left( Z < \frac{28.9 - 30}{1.5} \right) = P(Z < -0.733)\)

Using the standard normal distribution table, \(P(Z < -0.733) = 1 - 0.7682 = 0.2318\).

The expected number of cartridges per month that contain less than 28.9 ml is:

\(0.2318 \times 8 = 1.85\)

Rounding to the nearest whole number, we get 2 cartridges.

To find the number of giraffes weighing less than 700 kg, we first calculate the z-score for 700 kg using the formula:

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

where \(X = 700\), \(\mu = 830\), and \(\sigma = 120\).

\(z = \frac{700 - 830}{120} = -1.083\)

Next, we find the probability that a giraffe weighs less than 700 kg, \(P(X < 700)\), which is equivalent to \(P(z < -1.083)\).

Using standard normal distribution tables or a calculator, \(P(z < -1.083) = 0.1394\).

Therefore, the expected number of giraffes weighing less than 700 kg is:

\(430 \times 0.1394 = 59.9\)

Rounding to the nearest whole number, we get 59 or 60 giraffes.

Let the random variable \(X\) represent the travel time in minutes. \(X\) is normally distributed with mean \(\mu = 140\) and standard deviation \(\sigma = 12\).

We need to find \(P(X < 132)\).

First, standardize the variable using the formula:

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

Substitute the values:

\(Z = \frac{132 - 140}{12} = -0.6667\)

Now, find \(P(Z < -0.6667)\).

Using standard normal distribution tables or a calculator, \(P(Z < -0.6667) = 0.252\).

(i) To find the probability that an apple is graded as medium, we need to calculate the probability that its weight is between 90 and 140 grams. We standardize these values using the formula:

\(z = \frac{x - \mu}{\sigma}\)

For 90 grams: \(z_1 = \frac{90 - 120}{24} = -\frac{5}{4}\)

For 140 grams: \(z_2 = \frac{140 - 120}{24} = \frac{5}{6}\)

The probability that an apple is graded as medium is:

\(\Phi\left(\frac{5}{6}\right) - \Phi\left(-\frac{5}{4}\right)\)

Using standard normal distribution tables or a calculator, we find:

\(\Phi(0.8333) - (1 - \Phi(1.25)) = 0.7975 - 0.1056 = 0.6919\)

Rounding to 3 significant figures gives 0.692.

(ii) We use the binomial distribution to find the probability that at least two apples are graded as medium. Let \(X\) be the number of medium apples out of 4, where \(X \sim \text{Binomial}(4, 0.692)\).

We need \(P(X \geq 2) = 1 - P(X = 0) - P(X = 1)\).

\(P(X = 0) = (1 - 0.692)^4\)

\(P(X = 1) = 4 \times 0.692^1 \times (1 - 0.692)^3\)

Calculating these:

\(P(X = 0) = 0.00899\)

\(P(X = 1) = 0.0808757\)

Thus, \(P(X \geq 2) = 1 - 0.00899 - 0.0808757 = 0.910\).