Question

carlton and leon are expert bowlers. seventy percent of carltons rolls are strikes, while 67% of leons rolls are strikes. suppose that carlton and leon each bowl 25 games. what is the probability that leons proportion of strikes is greater than carltons for these games? if needed, use the z-table to answer the question. 0.324 0.4096 0.5904 0.676

Explanation:

Step1: Define parameters

Let $\hat{p}_L$ = Leon's strike proportion, $\hat{p}_C$ = Carlton's strike proportion.
Given: $p_L=0.67$, $n_L=25$; $p_C=0.70$, $n_C=25$.
We need $P(\hat{p}_L - \hat{p}_C > 0)$

Step2: Find mean of $\hat{p}_L - \hat{p}_C$

Mean $\mu_{\hat{p}_L - \hat{p}_C} = p_L - p_C$
$\mu_{\hat{p}_L - \hat{p}_C} = 0.67 - 0.70 = -0.03$

Step3: Find standard error

Standard error $\sigma_{\hat{p}_L - \hat{p}_C} = \sqrt{\frac{p_L(1-p_L)}{n_L} + \frac{p_C(1-p_C)}{n_C}}$
$\sigma_{\hat{p}_L - \hat{p}_C} = \sqrt{\frac{0.67(0.33)}{25} + \frac{0.70(0.30)}{25}}$
$\sigma_{\hat{p}_L - \hat{p}_C} = \sqrt{\frac{0.2211}{25} + \frac{0.21}{25}} = \sqrt{\frac{0.4311}{25}} = \sqrt{0.017244} \approx 0.1313$

Step4: Calculate z-score

$z = \frac{0 - \mu_{\hat{p}_L - \hat{p}_C}}{\sigma_{\hat{p}_L - \hat{p}_C}}$
$z = \frac{0 - (-0.03)}{0.1313} \approx \frac{0.03}{0.1313} \approx 0.228$

Step5: Find probability from z-table

$P(\hat{p}_L - \hat{p}_C > 0) = P(z > 0.23)$ (rounded z-score)
$P(z > 0.23) = 1 - P(z \leq 0.23) = 1 - 0.5910 = 0.4090$ (matches ~0.4096)

Answer:

0.4096