. Consider the following: Suppose 58% of the population are more than 6 feet tall
- If a random sample of size 833 is selected, what is the probability that the proportion of the sampled persons more than 6 feet tall will be greater than 63%
To find the probability that the proportion of the sampled persons more than 6 feet tall will be greater than 63%, we need to use the concept of sampling distribution and apply the Central Limit Theorem.
First, let's determine the mean and standard deviation of the sampling distribution. The mean, μ, of the sampling distribution of proportions is equal to the population proportion, p, which is 0.58 in this case. The standard deviation, σ, of the sampling distribution is given by the formula:
σ = √(p * (1-p) / n),
where p is the population proportion, 1-p is the proportion of the population not meeting the criteria, and n is the sample size.
In this case, p = 0.58 and n = 833, so we can calculate the standard deviation:
σ = √(0.58 * (1-0.58) / 833) = √(0.58 * 0.42 / 833) ≈ 0.0169.
Next, we need to standardize the proportion value of 63% using the formula:
z = (x - μ) / σ,
where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case, we want to find the probability for a proportion greater than 0.63. So, plugging the values into the formula:
z = (0.63 - 0.58) / 0.0169 ≈ 2.96.
Now, we need to find the probability of obtaining a z-score greater than 2.96. To do this, we can use a standard normal distribution table or a statistical calculator. Looking up the z-score in a standard normal distribution table, we find that the probability corresponding to a z-score of 2.96 is approximately 0.9985.
Therefore, the probability that the proportion of the sampled persons more than 6 feet tall will be greater than 63% is approximately 0.9985 or 99.85%.