Question

Carlton and Leon are expert bowlers. Seventy percent of Carlton's rolls are strikes, while 67% of Leon's rolls are strikes. Suppose that Carlton and Leon each bowl 25 games. What is the probability that Leon's proportion of strikes is greater than Carlton's for these games?

Answer 1

Answer:

The probability that Leon strikes is greater than Carlton strike is 0.40905

Step-by-step explanation:

From the question, we have;

The percentage of Carlton's rolls are strikes, $\hat p_1$ = 70%

The number of games Carlton played, n₁ = 25

The percentage of Leon's rolls that are strikes, $\hat p_2$ = 67%

The number of games Leon played, n₂ = 25

Therefore, we have;

$\hat p=\dfrac{k_1 + k_2}{n_1 + n_2}$

Where;

k₁ = 0.7 × 25 = 17.5

k₂ = 0.67 × 25 = 16.75

$\therefore \hat p=\dfrac{17.5 + 16.75}{25 + 25} = 0.685$

The test statistic is given as follows;

$Z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p} \cdot (1-\hat{p})\cdot \left (\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}} \right )}}$

$Z=\dfrac{0.7-0.67}{\sqrt{0.685 \cdot (1-0.685)\cdot \left (\dfrac{1}{25}+\dfrac{1}{25} \right )}} \approx 0.2283$

From the z-table, we have;

The p-value for Carlton strikes is greater than Leon's strike = 0.59095

∴ The p-value for Leon strikes is greater than Carlton strike = 1 - 0.59095 = 0.40905

The probability that Leon strikes is greater than Carlton strike = 0.40905