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 can use the complement rule:

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

We first need to calculate $P(X \leq 5)$, which can be done using the binomial distribution:

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

Calculating each of these using the binomial probability formula for $k = 0, 1, 2, 3, 4, 5$, we sum the probabilities to find:

$P(X \leq 5) \approx 0.0799$

Now, substitute into the complement formula:

$P(X > 5) = 1 - 0.0799 \approx 0.9201$

For part (b), we consider a normal approximation to the binomial distribution as George now throws the ball 250 times.

The random variable $Y$, representing the number of hits, can be approximated by the normal distribution:

$Y \sim N(np, np(1-p))$

where $n = 250$ and $p = 0.48$.

First, we calculate the mean and standard deviation:

• Mean: $\mu = np = 250 \times 0.48 = 120$
• Variance: $\sigma^2 = np(1-p) = 250 \times 0.48 \times 0.52 \approx 30.0$
• Standard Deviation: $\sigma \approx \sqrt{30.0} \approx 5.48$

Next, we standardize the target of 110 hits:

$Z = \frac{X - \mu}{\sigma} = \frac{110 - 120}{5.48} \approx -1.83$

Using the standard normal distribution table, we find:

$P(Z > -1.83) \approx 0.968$

Thus, the probability that George will hit the target more than 110 times is approximately $0.968$.