Let the random variable representing the length of the snakes be denoted by \(X\), where \(X \sim N(54, 6.1^2)\).

We need to find \(P(50 < X < 60)\).

First, convert the lengths to standard normal variables \(Z\) using the formula:

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

For \(X = 50\):

\(Z = \frac{50 - 54}{6.1} = -0.6557\)

For \(X = 60\):

\(Z = \frac{60 - 54}{6.1} = 0.9836\)

Thus, we need to find \(P(-0.6557 < Z < 0.9836)\).

This is equivalent to \(\Phi(0.9836) - \Phi(-0.6557)\).

Using the standard normal distribution table:

\(\Phi(0.9836) = 0.8375\)

\(\Phi(-0.6557) = 1 - \Phi(0.6557) = 1 - 0.7441 = 0.2559\)

Therefore, \(P(-0.6557 < Z < 0.9836) = 0.8375 - 0.2559 = 0.5816\).

Rounding to three decimal places, the probability is \(0.582\).

To find the expected number of birds with lengths between 15.4 cm and 16.8 cm, we first standardize the lengths using the formula for the standard normal distribution:

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

For 15.4 cm: \(Z = \frac{15.4 - 16.5}{0.6} = -1.833\)

For 16.8 cm: \(Z = \frac{16.8 - 16.5}{0.6} = 0.5\)

We need to find \(P(-1.833 < Z < 0.5)\).

Using the standard normal distribution table, we find:

\(\Phi(0.5) = 0.6915\)

\(\Phi(-1.833) = 1 - \Phi(1.833) = 1 - 0.9666 = 0.0334\)

Thus, \(P(-1.833 < Z < 0.5) = 0.6915 - 0.0334 = 0.6581\).

The expected number of birds is:

\(0.6581 \times 150 = 98.715\)

Rounding to the nearest whole number, we expect 99 birds.

We need to find the probability that a student's height is within half a standard deviation of the mean, i.e., between 144 cm and 152 cm.

Using the standardization formula, we calculate:

\(P(144 < X < 152) = P\left( \frac{144 - 148}{8} < Z < \frac{152 - 148}{8} \right)\)

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

Using the standard normal distribution table, we find:

\(P\left( -\frac{1}{2} < Z < \frac{1}{2} \right) = 0.6915 - (1 - 0.6915) = 2 \times 0.6915 - 1\)

\(= 0.383\)

Therefore, the expected number of students is:

\(0.383 \times 120 = 45.96\)

Rounding to the nearest whole number, we accept 45 or 46.

(i) To find the probability that a fir tree has a height less than 45 metres, we standardize the variable using the formula:

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

where \(X = 45\), \(\mu = 40\), and \(\sigma = 8\).

\(Z = \frac{45 - 40}{8} = 0.625\)

We find \(P(Z < 0.625)\) using standard normal distribution tables or a calculator, which gives:

\(P(X < 45) = 0.734\)

(ii) To find the probability that a fir tree has a height within 5 metres of the mean, we calculate:

\(P(35 < X < 45) = P(X < 45) - P(X < 35)\)

Using the result from part (i), \(P(X < 45) = 0.734\).

Standardize for \(X = 35\):

\(Z = \frac{35 - 40}{8} = -0.625\)

\(P(Z < -0.625) = 1 - P(Z < 0.625) = 1 - 0.734 = 0.266\)

Thus, \(P(35 < X < 45) = 0.734 - 0.266 = 0.468\)

(i) To find the probability that a randomly chosen athlete has a PB between 46 and 53 seconds, we standardize the values using the formula:

\(P(46 < X < 53) = P\left( \frac{46 - 49.2}{2.8} < Z < \frac{53 - 49.2}{2.8} \right)\)

This simplifies to:

\(P(-1.143 < Z < 1.357)\)

Using the standard normal distribution table, we find:

\(\Phi(1.357) + \Phi(1.143) - 1 = 0.9126 + 0.8735 - 1\)

Thus, the probability is:

0.786

(iii) To find the probability that exactly 2 out of 3 athletes have PBs of less than 46 seconds, we first find:

\(P(X < 46) = 0.1265\)

Then, using the binomial probability formula:

\(P(2 \text{ PB} < 46) = 3(1 - 0.1265)0.1265^2\)

This calculates to:

0.0419