Given that the running times are normally distributed with mean \(\mu = 41.2\) and standard deviation \(\sigma = 3.6\), we need to find the value of \(t\) such that 95% of the times are greater than \(t\).

This implies that 5% of the times are less than \(t\). The corresponding z-score for the 5th percentile is \(z = -1.645\).

Using the standardization formula:

\(\frac{t - 41.2}{3.6} = -1.645\)

Solving for \(t\):

\(t - 41.2 = -1.645 \times 3.6\)

\(t - 41.2 = -5.922\)

\(t = 41.2 - 5.922\)

\(t = 35.278\)

Rounding to one decimal place, \(t = 35.3\).

Given that the times are normally distributed with mean \(\mu = 32.2\) and standard deviation \(\sigma = 9.6\), we need to find the value of \(t\) such that 20% of employees take longer than \(t\) minutes.

This implies that 80% of employees take less than \(t\) minutes. We need to find the corresponding \(z\)-value for 0.8 in the standard normal distribution table.

The \(z\)-value for 0.8 is approximately 0.842.

Using the standardization formula:

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

Substitute the known values:

\(0.842 = \frac{t - 32.2}{9.6}\)

Solve for \(t\):

\(t - 32.2 = 0.842 \times 9.6\)

\(t - 32.2 = 8.0832\)

\(t = 32.2 + 8.0832\)

\(t = 40.2832\)

Rounding to one decimal place, \(t = 40.3\).

We are given that the times are normally distributed with mean \(\mu = 125\) and standard deviation \(\sigma = 24\). We need to find the value of \(t\) such that 90% of the time, Karli spends more than \(t\) minutes on social media.

This implies that 10% of the time, Karli spends less than \(t\) minutes. We need to find the z-score corresponding to the 10th percentile of the standard normal distribution, which is \(z = -1.282\).

Using the standardization formula:

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

Substitute the known values:

\(-1.282 = \frac{t - 125}{24}\)

Solving for \(t\):

\(t - 125 = -1.282 \times 24\)

\(t - 125 = -30.168\)

\(t = 125 - 30.168\)

\(t = 94.832\)

Rounding to one decimal place, \(t \approx 94.2\).

(a) Using the standard normal distribution, we have:

\(z_1 = \frac{4 - \mu}{\sigma} = -1.378\)

\(z_2 = \frac{10 - \mu}{\sigma} = 0.842\)

Solving these equations simultaneously:

\(4 - \mu = -1.378\sigma\)

\(10 - \mu = 0.842\sigma\)

By solving, we find \(\sigma = 2.70\) and \(\mu = 7.72\).

(b) The probability of a leaf length being between \(\mu - 2\sigma\) and \(\mu + 2\sigma\) is given by:

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

Using \(\Phi(2) \approx 0.9772\), we have:

\(2 \times 0.9772 - 1 = 0.9544\)

Therefore, the expected number of leaves is \(0.9544 \times 800 = 763.52\), which rounds to 763 or 764.

The probability that a bag weighs more than 1.10 kg is given by:

\(P(X > 1.1) = \frac{72}{2000} = 0.036\)

Using the standard normal distribution, this corresponds to a \(z\)-value of approximately \(z = 1.798\).

We use the standardization formula:

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

Substituting the known values:

\(1.798 = \frac{1.1 - 1.04}{\sigma}\)

\(1.798 = \frac{0.06}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{0.06}{1.798} \approx 0.0334\)