First, we standardize the lengths using the formula for the standard normal variable:

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

For rods shorter than 15.75 cm:

\(P(x < 15.75) = P\left(z < \frac{15.75 - 16}{0.2}\right) = P(z < -1.25)\)

Using symmetry and standard normal tables, \(P(z < -1.25) = 1 - P(z < 1.25) = 1 - 0.8944 = 0.1056\).

For rods longer than 16.25 cm:

\(P(x > 16.25) = P\left(z > \frac{16.25 - 16}{0.2}\right) = P(z > 1.25) = 0.1056\) (by symmetry).

The probability of a rod being usable is:

\(P(\text{usable}) = 1 - P(x < 15.75) - P(x > 16.25) = 1 - 0.1056 - 0.1056 = 0.7888\).

The expected number of usable rods in a batch of 1000 is:

\(1000 \times 0.7888 = 788.8\).

Rounding gives approximately 788 or 789 usable rods.

Let the mean and standard deviation of the normal distribution be denoted by \(\mu\). Since the mean is equal to the standard deviation, we have \(\sigma = \mu\).

We need to find \(P(X < 1.5\mu)\).

Standardizing the variable, we use the formula:

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

Substituting \(X = 1.5\mu\), \(\mu = \sigma\), we get:

\(z = \frac{1.5\mu - \mu}{\mu} = \frac{0.5\mu}{\mu} = 0.5\)

Thus, \(P(X < 1.5\mu) = P(z < 0.5)\).

Using standard normal distribution tables or a calculator, \(P(z < 0.5) = 0.692\).

First, we standardize the normal variable to find the probability that a single woman's middle finger is shorter than 8.2 cm:

\(P(X < 8.2) = P\left( Z < \frac{8.2 - 7.9}{0.44} \right) = P(Z < 0.6818)\)

Using the standard normal distribution table, \(P(Z < 0.6818) \approx 0.7524\).

Now, we use the binomial distribution to find the probability that exactly 3 out of 5 women have middle fingers shorter than 8.2 cm:

\(P(3) = \binom{5}{3} (0.7524)^3 (1 - 0.7524)^2\)

\(P(3) = 10 \times (0.7524)^3 \times (0.2476)^2\)

\(P(3) \approx 0.261\)

Let the random variable for the time taken to cook an egg be denoted as \(X\), where \(X \sim N(4.2, 0.6^2)\).

We need to find \(P(3.5 < X < 4.5)\).

Standardize the variable using the formula for the standard normal distribution: \(Z = \frac{X - \mu}{\sigma}\).

For \(X = 4.5\):

\(Z = \frac{4.5 - 4.2}{0.6} = \frac{0.3}{0.6} = 0.5\).

Thus, \(P(X < 4.5) = P(Z < 0.5) = 0.6915\).

For \(X = 3.5\):

\(Z = \frac{3.5 - 4.2}{0.6} = \frac{-0.7}{0.6} = -1.167\).

Thus, \(P(X < 3.5) = P(Z < -1.167) = 0.1216\).

Therefore, \(P(3.5 < X < 4.5) = P(X < 4.5) - P(X < 3.5) = 0.6915 - 0.1216 = 0.570\).

(i) To find the probability that Peter takes longer than 10.2 minutes, we standardize the variable:

\(P(x > 10.2) = P\left( z > \frac{10.2 - 9.5}{1.3} \right)\)

\(= P(z > 0.53846)\)

Using the standard normal distribution table, \(P(z > 0.53846) = 1 - 0.7046 = 0.295\).

(ii) To find the number of days Peter takes less than 8.8 minutes, we use symmetry:

\(P(x < 8.8) = 0.2954\)

Estimate the number of days: \(365 \times 0.2954 = 107.671\)

Rounding gives approximately 107 or 108 days.