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.
