A recent poll of 500 randomly chosen voters concludes that 60% of the voters prefer this canidate. Determine the probability that the candidate will get less than 56% of the votes. How do I do this type of question? i have a few to figure out.
To determine the probability that the candidate will get less than 56% of the votes, we can use the normal distribution. Since the sample size is large (500) and we have a proportion (60%) that can be considered "success", we can approximate this using the normal distribution.
To calculate this probability, we need to find the area under the normal curve to the left of 56%.
Step 1: Find the mean (μ) and standard deviation (σ).
The mean (μ) of a sample proportion can be calculated using the formula:
μ = p
where p is the proportion (60% or 0.6). So, μ = 0.6
The standard deviation (σ) of a sample proportion can be calculated using the formula:
σ = √( (p * (1-p)) / n )
where n is the sample size. In this case, n = 500. So, σ = √( (0.6 * (1-0.6)) / 500 ) = 0.0245 (approximately)
Step 2: Standardize the value of interest.
To standardize the value of interest (56%), we use the formula:
z = (x - μ) / σ
where x is the value of interest. In this case, x = 0.56, μ = 0.6, and σ = 0.0245. So, z = (0.56 - 0.6) / 0.0245 = -1.6327 (approximately)
Step 3: Find the cumulative probability.
Using a standard normal distribution table or a statistical calculator, find the cumulative probability associated with the standardized value. In this case, you need to find the cumulative probability to the left of z = -1.6327.
Step 4: Calculate the probability.
Since the probability to the left of a value is the same as the probability less than that value, the probability that the candidate will get less than 56% of the votes is equal to the cumulative probability calculated in step 3.
By following these steps, you should be able to find the probability that the candidate will get less than 56% of the votes. You can repeat this process for other questions you have by changing the given proportion and value of interest.