We are given that 83.4% of adult male giraffes weigh more than 950 kg. This implies that the probability \(P(X > 950) = 0.834\).

Using the standard normal distribution table, we find that a probability of 0.834 corresponds to a \(z\)-score of approximately -0.97 (since we are looking for the left tail, we use the negative \(z\)-score).

The standardization formula is:

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

Substituting the known values:

\(-0.97 = \frac{950 - 1190}{\sigma}\)

Solving for \(\sigma\):

\(-0.97\sigma = 950 - 1190\)

\(-0.97\sigma = -240\)

\(\sigma = \frac{-240}{-0.97}\)

\(\sigma \approx 247\)

We are given that 90% of the weights lie between (830 - w) kg and (830 + w) kg. This implies that 5% of the weights are below (830 - w) kg and 5% are above (830 + w) kg.

Thus, the probability that a giraffe weighs less than (830 + w) kg is 95%, which corresponds to a z-score of 1.645 for a standard normal distribution.

Using the standardization formula:

\(\frac{(830 + w) - 830}{120} = \frac{w}{120} = 1.645\)

Solving for \(w\):

\(w = 1.645 \times 120 = 197.4\)

Since \(w\) must be an integer, we round to \(w = 197\).

Given that the probability that the time is greater than k is 0.675, we have:

\(P(X > k) = 0.675\)

Using the standard normal distribution, we find the corresponding z-value for the cumulative probability of \(1 - 0.675 = 0.325\).

The z-value for 0.325 is approximately \(z = -0.454\).

Using the standardization formula:

\(z = \frac{k - 140}{12}\)

Substitute \(z = -0.454\):

\(\frac{k - 140}{12} = -0.454\)

Solving for k:

\(k - 140 = -0.454 \times 12\)

\(k - 140 = -5.448\)

\(k = 140 - 5.448\)

\(k = 134.552\)

Rounding to two decimal places, \(k = 134.55\).

Let \(X\) be the lifetime of the light bulb. We know \(X \sim N(2000, \sigma^2)\).

We are given \(P(X > 1800) = 0.96\). This can be rewritten in terms of the standard normal variable \(Z\) as:

\(P\left(Z > \frac{1800 - 2000}{\sigma}\right) = 0.96\).

Using the standard normal distribution table, we find that \(P(Z > -1.751) = 0.96\), so:

\(\frac{200}{\sigma} = 1.751\).

Solving for \(\sigma\), we get:

\(\sigma = \frac{200}{1.751} \approx 114\).

(i) To find the number of days a student uses the machine for less than 4 hours, we calculate \(P(X < 4)\). Standardize using \(Z = \frac{X - \mu}{\sigma}\), where \(\mu = 3.24\) and \(\sigma = 0.96\):

\(Z = \frac{4 - 3.24}{0.96} = 0.7917\)

Using the standard normal distribution table, \(P(Z < 0.7917) = 0.7858\).

Expected days: \(0.7858 \times 365 = 287\) days.

(ii) We need \(P(X > k) = 0.2\), which implies \(P(X < k) = 0.8\). Standardize:

\(P\left(Z < \frac{k - 3.24}{0.96}\right) = 0.8\)

From the standard normal table, \(Z = 0.842\).

\(\frac{k - 3.24}{0.96} = 0.842\)

\(k = 0.842 \times 0.96 + 3.24 = 4.05\)

(iii) We find \(P(-1.5 < Z < 1.5)\):

\(\Phi(1.5) - \Phi(-1.5) = 2\Phi(1.5) - 1\)

\(\Phi(1.5) = 0.9332\), so \(2 \times 0.9332 - 1 = 0.866\).