Question

1. a magazine reported that 30% of car drivers use a cell phone while driving. in a survey of 180 drivers, what is the probability that exactly 45 of them will use their phone while driving? /5 marks

Explanation:

Step1: Identify the distribution

This is a binomial probability problem. The binomial probability formula is $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successful trials, $p$ is the probability of success on a single trial. Here, $n = 180$, $k = 45$, $p = 0.3$.

Step2: Calculate the binomial coefficient

The binomial coefficient $\binom{n}{k} = \frac{n!}{k!(n - k)!}$. So $\binom{180}{45} = \frac{180!}{45!(180 - 45)!} = \frac{180!}{45!135!}$.

Step3: Calculate $p^k$ and $(1 - p)^{n - k}$

$p^k = 0.3^{45}$, $(1 - p)^{n - k} = 0.7^{135}$.

Step4: Multiply the three parts

$P(X = 45) = \binom{180}{45} \times 0.3^{45} \times 0.7^{135}$. However, calculating this directly is cumbersome. We can use the normal approximation to the binomial distribution. First, check the conditions: $np = 180 \times 0.3 = 54$, $n(1 - p) = 180 \times 0.7 = 126$, both are greater than 5, so normal approximation is appropriate.

The mean $\mu = np = 54$, the standard deviation $\sigma = \sqrt{np(1 - p)} = \sqrt{180 \times 0.3 \times 0.7} = \sqrt{37.8} \approx 6.15$.

For the normal approximation to the binomial, we use continuity correction. So $P(X = 45) \approx P(44.5 < X < 45.5)$ in the normal distribution.

First, calculate the z - scores: $z_1 = \frac{44.5 - 54}{6.15} \approx \frac{-9.5}{6.15} \approx -1.54$, $z_2 = \frac{45.5 - 54}{6.15} \approx \frac{-8.5}{6.15} \approx -1.38$.

Then, $P(-1.54 < Z < -1.38) = P(Z < -1.38) - P(Z < -1.54)$. Looking up in the standard normal table, $P(Z < -1.38) = 0.0838$, $P(Z < -1.54) = 0.0618$. So $P(-1.54 < Z < -1.38) = 0.0838 - 0.0618 = 0.022$.

(If we calculate the binomial probability directly using a calculator or software, we can also get the result. For example, using the binomial probability formula in a calculator: $\binom{180}{45} \times 0.3^{45} \times 0.7^{135} \approx 0.021$ (approximate value). The slight difference is due to the normal approximation.)

Answer:

The probability is approximately 0.022 (using normal approximation) or approximately 0.021 (using direct binomial calculation).