Given that the times follow a normal distribution \(N(\mu, \sigma^2)\) with \(\mu = 3\sigma\), we need to find \(P(t > 0.6\mu)\).

First, standardize the variable:

\(P(t > 0.6\mu) = P\left( z > \frac{0.6\mu - \mu}{\mu/3} \right)\)

Substitute \(\mu = 3\sigma\) into the equation:

\(P\left( z > \frac{0.6 \times 3\sigma - 3\sigma}{3\sigma/3} \right) = P(z > -1.2)\)

Using the standard normal distribution table, \(P(z > -1.2) = 0.885\).

We need to find \(P(Y < 2\mu)\). To do this, we standardize the variable using the formula for the standard normal distribution:

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

Substituting \(Y = 2\mu\), \(\mu\), and \(\sigma = \frac{2}{3} \mu\), we have:

\(z = \frac{2\mu - \mu}{\frac{2}{3} \mu} = \frac{\mu}{\frac{2}{3} \mu} = \frac{3}{2} = 1.5\)

Thus, \(P(Y < 2\mu) = P(z < 1.5)\).

Using standard normal distribution tables or a calculator, \(P(z < 1.5) = 0.933\).

To find the proportion of Marok’s pulse rates above 160 beats per minute, we standardize the value using the formula for the z-score:

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

where \(X = 160\), \(\mu = 148.6\), and \(\sigma = 18.5\).

\(z = \frac{160 - 148.6}{18.5} = 0.616\)

We need to find \(P(X > 160) = P(z > 0.616)\).

Using the standard normal distribution table, \(P(z > 0.616) = 1 - P(z < 0.616) = 1 - 0.7310 = 0.269\).

Thus, the proportion of pulse rates above 160 beats per minute is 0.269.

(a) To find the probability that a randomly chosen cyclist has a time less than 74 minutes, we use the standard normal distribution. The standardization formula is:

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

where \(X = 74\), \(\mu = 62.3\), and \(\sigma = 8.4\). Substituting these values, we get:

\(Z = \frac{74 - 62.3}{8.4} = 1.393\)

We find \(P(Z < 1.393)\) using the standard normal distribution table, which gives us:

\(P(X < 74) = 0.918\)

(b) To find the probability that 4 randomly chosen cyclists all have times between 50 and 74 minutes, we first find the probability for one cyclist:

\(P(50 < X < 74) = P\left(\frac{50 - 62.3}{8.4} < Z < \frac{74 - 62.3}{8.4}\right)\)

\(P(-1.464 < Z < 1.393)\)

Using the standard normal distribution table, we find:

\(\Phi(1.393) = 0.9182\)

\(\Phi(-1.464) = 0.0718\)

\(P(-1.464 < Z < 1.393) = 0.9182 - 0.0718 = 0.8464\)

The probability that all 4 cyclists have times between 50 and 74 minutes is:

\((0.8464)^4 = 0.514\)

(i) To find the probability that a book is classified as large, we first standardize the height 29 cm using the normal distribution with mean 21.7 and standard deviation 6.5:

\(P(\text{large}) = 1 - \Phi\left(\frac{29 - 21.7}{6.5}\right)\)

\(= 1 - \Phi(1.123) = 1 - 0.8692 = 0.1308\)

Now, we use the binomial distribution to find the probability that fewer than 2 books are large out of 8:

\(P(0,1) = (0.8692)^8 + \binom{8}{1}(0.1308)(0.8692)^7\)

\(= 0.718\)

(ii) We need the probability of at least 1 large book to be more than 0.98. This is equivalent to:

\(1 - (0.8692)^n > 0.98\)

\((0.8692)^n < 0.02\)

Solving for the smallest integer \(n\), we find:

\(n = 28\)