George throws a ball at a target 15 times. Each time George throws the ball, the probability of the ball hitting the target is 0.48. The random variable X represents the number of times George hits the target in 15 throws. (a) Find (i) P(X = 3) (ii) P(X > 5) George now throws the ball at the target 250 times. (b) Use a normal approximation to calculate the probability that he will hit the target more than 110 times.

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

Question 1

George throws a ball at a target 15 times. Each time George throws the ball, the probability of the ball hitting the target is 0.48. The random variable X represent... show full transcript

Worked Solution & Example Answer:George throws a ball at a target 15 times - Edexcel - A-Level Maths Statistics - Question 1 - 2022 - Paper 1

Step 1

Find P(X = 3)

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Answer

To calculate P(X = 3), we use the binomial distribution given by:

$X \sim B(n = 15, p = 0.48)$

The probability mass function for a binomial distribution is:

$P(X = k) = {n \choose k} p^k (1-p)^{n-k}$

Substituting the values:

$P(X = 3) = {15 \choose 3} (0.48)^3 (0.52)^{12}$

Calculating this gives:

Calculate ${15 \choose 3}$ , which equals 455.
Then compute $(0.48)^3 \approx 0.1101$ and $(0.52)^{12} \approx 0.0044$ .

Putting it all together:

$P(X = 3) \approx 455 \times 0.1101 \times 0.0044 \approx 0.0197$

Step 2

Find P(X > 5)

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Answer

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

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

Using the same 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 probabilities using the binomial formula:

Calculate for each value from 0 to 5.

After calculating, sum these probabilities to get P(X \leq 5). Finally:

$P(X > 5) = 1 - P(X \leq 5) \approx 0.920$

Step 3

Use a normal approximation to calculate the probability that he will hit the target more than 110 times.

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Answer

For 250 throws, we use the normal approximation:

Let Y = number of hits, then:

$Y \sim N(\mu = np, \sigma^2 = np(1-p))$

Where:

n = 250,
p = 0.48

Calculating the mean and standard deviation:

$\mu = 250 \times 0.48 = 120$ $\sigma = \sqrt{250 \times 0.48 \times 0.52} \approx 7.90$

We need to find P(Y > 110). Using the continuity correction:

$P(Y > 110) \approx P(Y > 110.5)$

Now standardize:

$Z = \frac{110.5 - 120}{7.90} \approx -1.19$

Using the standard normal table, we find:

$P(Z > -1.19) \approx 0.883$

Thus, the required probability is approximately 0.885.

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

Worked Solution & Example Answer:George throws a ball at a target 15 times - Edexcel - A-Level Maths Statistics - Question 1 - 2022 - Paper 1

Find P(X = 3)

Find P(X > 5)

Use a normal approximation to calculate the probability that he will hit the target more than 110 times.

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