George throws a ball at a target 15 times - Edexcel - A-Level Maths Statistics - Question 1 - 2022 - Paper 1

To find P(X > 5), we use the complement rule:

$P(X > 5) = 1 - P(X \leq 5)$

This requires finding P(X \leq 5), for which we sum the probabilities from 0 to 5:

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

Calculating each term using the binomial formula:

1. P(X = 0) = {15 race 0} (0.48)^0 (0.52)^{15}
2. P(X = 1) = {15 race 1} (0.48)^1 (0.52)^{14}
3. P(X = 2) = {15 race 2} (0.48)^2 (0.52)^{13}
4. $P(X = 3)$ (which we calculated earlier)
5. P(X = 4) = {15 race 4} (0.48)^4 (0.52)^{11}
6. P(X = 5) = {15 race 5} (0.48)^5 (0.52)^{10}

After finding all probabilities, we find: $P(X \leq 5) \approx 0.0790$ Thus, $P(X > 5) \approx 1 - 0.0790 = 0.9200$

For George's new scenario with n = 250 and p = 0.48, we can use the normal approximation:

1. Calculate the expected mean (\mu) and variance (\sigma^2): $\mu = np = 250 * 0.48 = 120$ $\sigma^2 = np(1-p) = 250 * 0.48 * 0.52 = 30$ $\sigma \approx \sqrt{30} \approx 5.477$

2. Now, we want P(X > 110). For the normal approximation, we need to standardize this value: $Z = \frac{X - \mu}{\sigma}$ For X = 110: $Z = \frac{110 - 120}{5.477} \approx -1.826$

3. Using the Z-table, we find: $P(Z > -1.826) \approx 0.9686$

4. Therefore, the probability that he will hit the target more than 110 times is approximately: $P(X > 110) \approx 0.9686$