Answer :
Using the Central Limit Theorem, we approximate the proportion of odd numbers to be within 3 standard deviations of the mean, yielding approximately 99.7% probability.
To solve this, we can use the Central Limit Theorem (CLT), which states that the distribution of the sample mean of a large enough sample from any distribution will be approximately normally distributed. Since we're dealing with a binomial distribution (rolling a die is a Bernoulli trial), the proportion of odd numbers will follow a binomial distribution. According to the CLT, the sample proportion will be approximately normally distributed if the sample size is large enough.
Given:
- Number of trials, [tex]\( n = 100 \)[/tex]
- Probability of success (rolling an odd number), [tex]\( p = \frac{1}{2} \)[/tex]
- Mean, [tex]\( \mu = np = 100 \times \frac{1}{2} = 50 \)[/tex]
- Standard deviation, [tex]\( \sigma = \sqrt{np(1-p)} = \sqrt{100 \times \frac{1}{2} \times \frac{1}{2}} = \sqrt{25} = 5 \)[/tex]
Now, we want to find the probability that the proportion of odd numbers falls between 0.35 and 0.65. We need to find the z-scores corresponding to these proportions and then find the area under the standard normal curve between these z-scores.
First, let's find the z-scores:
- [tex]\( z_1 = \frac{0.35 - 0.5}{\frac{5}{\sqrt{100}}} = -3 \)[/tex]
- [tex]\( z_2 = \frac{0.65 - 0.5}{\frac{5}{\sqrt{100}}} = 3 \)[/tex]
Now, we find the area under the standard normal curve between these z-scores:
- [tex]\( P(-3 < Z < 3) \)[/tex] represents the area between [tex]\( z_1 \) and \( z_2 \)[/tex] .
Using the 68-95-99.7 rule, the area between [tex]\( z_1 \)[/tex] and [tex]\( z_2 \)[/tex] (within 3 standard deviations from the mean) is approximately 99.7%.
So, the chance that the proportion of odd numbers will be in the range from 0.35 to 0.65 is approximately 99.7%.
