Let the random variable \(X\) represent the number of students who do not like any sports. \(X\) follows a binomial distribution with parameters \(n = 90\) and \(p = 0.3\).

To approximate, we use a normal distribution with mean \(\mu = np = 0.3 \times 90 = 27\) and variance \(\sigma^2 = np(1-p) = 0.3 \times 90 \times 0.7 = 18.9\).

Thus, \(X \sim N(27, 18.9)\).

We want \(P(X < 32)\). Using continuity correction, this is approximated by \(P(X < 31.5)\).

Standardize: \(z = \frac{31.5 - 27}{\sqrt{18.9}} = \frac{4.5}{4.347} \approx 1.035\).

Using the standard normal distribution table, \(P(Z < 1.035) = \Phi(1.035) \approx 0.850\).

Let the random variable \(X\) represent the number of days the train arrives late out of 142 days. \(X\) follows a binomial distribution with parameters \(n = 142\) and \(p = 0.35\).

To approximate, we use a normal distribution with mean \(\mu\) and variance \(\sigma^2\).

Calculate the mean: \(\mu = np = 142 \times 0.35 = 49.7\).

Calculate the variance: \(\sigma^2 = npq = 142 \times 0.35 \times 0.65 = 32.305\).

Standard deviation: \(\sigma = \sqrt{32.305}\).

We approximate \(X\) by a normal distribution \(N(49.7, 32.305)\).

We need to find \(P(X > 40)\). Using continuity correction, this becomes \(P(X > 40.5)\).

Standardize: \(Z = \frac{40.5 - 49.7}{\sqrt{32.305}} \approx -1.619\).

Find \(P(Z > -1.619)\) using standard normal distribution tables or calculator.

\(P(Z > -1.619) = 0.947\).

The probability of obtaining a pair of tails in one throw of two fair coins is \(0.25\) (since there are four equally likely outcomes: HH, HT, TH, TT, and only TT is a pair of tails).

Let \(X\) be the number of times a pair of tails is obtained in 80 throws. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.25\).

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

\(\mu = np = 80 \times 0.25 = 20\)

\(\sigma^2 = np(1-p) = 80 \times 0.25 \times 0.75 = 15\)

We need to find \(P(X > 25)\). Using the normal approximation,

\(P(X > 25) \approx P\left(Z > \frac{25.5 - 20}{\sqrt{15}}\right)\)

\(= P(Z > 1.42)\)

Using standard normal distribution tables, \(P(Z > 1.42) = 1 - P(Z \leq 1.42) = 1 - 0.9222 = 0.0778\).

The problem can be approximated using a normal distribution because the number of days is large (100 days), and the probability of messaging is not too close to 0 or 1.

First, calculate the mean and variance of the binomial distribution:

Mean: \(n \times p = 100 \times 0.72 = 72\)

Variance: \(n \times p \times (1-p) = 100 \times 0.72 \times 0.28 = 20.16\)

We approximate the binomial distribution with a normal distribution \(N(72, 20.16)\).

To find the probability of fewer than 64 days, use continuity correction: \(P(X < 64) \approx P(X < 63.5)\).

Standardize using the formula:

\(z = \frac{63.5 - 72}{\sqrt{20.16}}\)

\(z = \frac{63.5 - 72}{4.49} \approx -1.893\)

Using standard normal distribution tables or a calculator, find \(P(z < -1.893)\).

The probability is approximately 0.0292.

The problem involves a binomial distribution with parameters:

- Number of trials, \(n = 120\)

- Probability of success, \(p = 0.7\)

We approximate the binomial distribution with a normal distribution using the Central Limit Theorem.

The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 120 \times 0.7 = 84\)

\(\sigma^2 = np(1-p) = 120 \times 0.7 \times 0.3 = 25.2\)

Using a normal approximation, we find the probability that more than 75 adults own a car. Apply continuity correction:

\(P(X > 75) \approx P\left(Z > \frac{75.5 - 84}{\sqrt{25.2}}\right)\)

Calculate the standard score (z-score):

\(Z = \frac{75.5 - 84}{\sqrt{25.2}} = \frac{-8.5}{5.02} \approx -1.693\)

Using the standard normal distribution table, find:

\(P(Z > -1.693) = 0.955\)

Thus, the probability that more than 75 of them own a car is approximately 0.955.