Let the random variable \(X\) represent the number of people who pass the test. \(X\) follows a binomial distribution with parameters \(n = 75\) and \(p = 0.3\).

We approximate the binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 75 \times 0.3 = 22.5\)

\(\sigma^2 = np(1-p) = 75 \times 0.3 \times 0.7 = 15.75\)

The standard deviation \(\sigma\) is:

\(\sigma = \sqrt{15.75} \approx 3.9686\)

We use a continuity correction for the normal approximation:

\(P(X > 20) \approx P\left(Z > \frac{20.5 - 22.5}{\sqrt{15.75}}\right)\)

\(= P\left(Z > \frac{-2}{3.9686}\right)\)

\(= P(Z > -0.504)\)

Using standard normal distribution tables or a calculator, we find:

\(P(Z > -0.504) = \Phi(0.504) \approx 0.693\)

Thus, the probability that more than 20 people pass is approximately \(0.6925 < p < 0.693\).

Let the random variable \(X\) represent the number of students who play exactly one musical instrument in a sample of 90 students. Given that 52% of students play exactly one instrument, \(X\) follows a binomial distribution \(B(n=90, p=0.52)\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of \(X\) are given by:

\(\mu = np = 90 \times 0.52 = 46.8\)

\(\sigma^2 = np(1-p) = 90 \times 0.52 \times 0.48 = 22.464\)

We use a continuity correction to find \(P(X < 40)\), which is approximated by \(P(X < 39.5)\) in the normal distribution.

Standardizing, we find the z-score:

\(z = \frac{39.5 - 46.8}{\sqrt{22.464}} = \frac{-7.3}{4.74} \approx -1.540\)

Using standard normal distribution tables, we find:

\(P(Z < -1.540) = 1 - 0.9382 = 0.0618\)

Thus, the probability that fewer than 40 students play exactly one musical instrument is approximately 0.0618.

Let the random variable \(X\) represent the number of households with no broadband service out of 120. \(X\) follows a binomial distribution with parameters \(n = 120\) and \(p = 0.35\).

To approximate, we use a normal distribution with mean \(\mu = np = 120 \times 0.35 = 42\) and variance \(\sigma^2 = np(1-p) = 120 \times 0.35 \times 0.65 = 27.3\).

We apply a continuity correction to approximate \(P(X > 32)\) as \(P(X > 32.5)\).

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

\(P(X > 32.5) = P\left( Z > \frac{32.5 - 42}{\sqrt{27.3}} \right) = P(Z > -1.818)\)

Using the standard normal distribution table, \(\Phi(1.818) \approx 0.966\).

Thus, \(P(X > 32) = 1 - \Phi(-1.818) = \Phi(1.818)\).

Therefore, the probability is approximately \(0.965 < p < 0.966\).

The number of days the flight arrives early can be modeled by a binomial distribution with parameters \(n = 60\) and \(p = 0.15\).

First, calculate the mean and variance of the distribution:

Mean \(= np = 60 \times 0.15 = 9\)

Variance \(= np(1-p) = 60 \times 0.15 \times 0.85 = 7.65\)

Since \(n\) is large and \(p\) is small, we can use a normal approximation. The normal distribution has mean \(9\) and variance \(7.65\).

We need to find \(P(X \geq 12)\). Using continuity correction, this is approximated by \(P(X > 11.5)\).

Standardize using the normal distribution:

\(P\left(Z > \frac{11.5 - 9}{\sqrt{7.65}}\right)\)

Calculate the standardized value:

\(Z = \frac{11.5 - 9}{\sqrt{7.65}} \approx 0.9039\)

Find the probability from the standard normal distribution table:

\(1 - \Phi(0.9039) = 1 - 0.8169 = 0.183\)

Thus, the probability that Richard’s flight arrives early at least 12 times is approximately 0.183.

Let the random variable \(X\) represent the number of adults who travel to work by car in a sample of 150. \(X\) follows a binomial distribution with parameters \(n = 150\) and \(p = 0.6\).

To approximate this binomial distribution, we use a normal distribution with the same mean and variance. The mean \(\mu\) is given by:

\(\mu = np = 150 \times 0.6 = 90\)

The variance \(\sigma^2\) is given by:

\(\sigma^2 = np(1-p) = 150 \times 0.6 \times 0.4 = 36\)

Thus, the standard deviation \(\sigma\) is:

\(\sigma = \sqrt{36} = 6\)

We want to find \(P(X < 81)\). Using the normal approximation with continuity correction, we calculate:

\(P(X < 81) \approx P\left(Z < \frac{80.5 - 90}{6}\right)\)

\(= P\left(Z < -1.5833\right)\)

Using standard normal distribution tables or a calculator, we find:

\(\Phi(-1.5833) = 1 - 0.9433 = 0.0567\)

Therefore, the probability that fewer than 81 adults travel to work by car is approximately 0.0567.

For part (c), the approximation is justified because both \(np = 90\) and \(nq = 60\) are greater than 5, satisfying the conditions for using the normal approximation to the binomial distribution.