Question

a fair six-sided number cube has the following faces: 1, 1, 2, 2, 5, 6. this number cube is rolled 50 times. what is the probability that fewer than 30% of the rolls result in a two?
find the z-table here.
○ 0.309
○ 0.421
○ 0.450
○ 0.691

Explanation:

Step1: Find population proportion p

The cube has 2 faces out of 6 that are 2, so $p = \frac{2}{6} = \frac{1}{3} \approx 0.3333$

Step2: Define sample proportion threshold

30% of 50 rolls is $0.30$, so $\hat{p} = 0.30$

Step3: Calculate standard error (SE)

$$SE = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.3333(1-0.3333)}{50}} = \sqrt{\frac{0.3333 \times 0.6667}{50}} \approx \sqrt{\frac{0.2222}{50}} \approx \sqrt{0.004444} \approx 0.06667$$

Step4: Compute z-score

$$z = \frac{\hat{p} - p}{SE} = \frac{0.30 - 0.3333}{0.06667} \approx \frac{-0.0333}{0.06667} \approx -0.5$$

Step5: Find probability from z-table

The z-score $z=-0.5$ corresponds to a cumulative probability of 0.3085, which rounds to 0.309.

Answer:

0.309