Could you please explain how to do this question. Suppose that a certain candidate is running in a city election. A recent poll of 500 randomly chosen voters concludes that 60% of the voters prefer this candidate. Determine the probability that the candidate will get less than 56% of the votes.
To determine the probability that the candidate will get less than 56% of the votes, we can use the normal distribution and the concept of the standard normal variable.
Step 1: Identify the given information
- Sample size (n): 500 voters
- Proportion of voters who prefer the candidate (p): 60%
- Calculate q (1 - p): 1 - 0.6 = 0.4
Step 2: Calculate the standard deviation
- The standard deviation (σ) of a binomial distribution is calculated as the square root of (p * q / n).
- σ = √((0.6 * 0.4) / 500) = 0.02449 (rounded to 5 decimal places)
Step 3: Use the standard normal distribution
- To determine the probability that the candidate will get less than 56% of the votes, we need to find the z-score corresponding to 56%.
Step 4: Calculate the z-score
- The z-score of a value (x) is calculated as (x - μ) / σ, where μ is the mean value. As we are comparing proportions, the mean value is equal to the proportion (p).
- z = (0.56 - 0.6) / 0.02449
= -1.632 (rounded to 3 decimal places)
Step 5: Find the probability using the z-score
- The probability corresponding to the z-score can be found using a standard normal distribution table or a calculator.
- In this case, we are interested in finding the probability to the left of the z-score (-1.632).
Step 6: Look up the probability
- Using a standard normal distribution table or a calculator, we find that the probability corresponding to the z-score (-1.632) is approximately 0.052 (rounded to 3 decimal places).
Therefore, the probability that the candidate will get less than 56% of the votes is approximately 0.052 or 5.2%.
To determine the probability that the candidate will get less than 56% of the votes, we need to use the concept of the sampling distribution and the normal distribution.
Step 1: Identify the sample size and the proportion of voters who prefer the candidate.
In this case, the sample size is 500 (as given in the question) and the proportion of voters who prefer the candidate is 60% (also given in the question).
Step 2: Calculate the mean and the standard deviation of the sampling distribution.
The mean of the sampling distribution is equal to the proportion of voters who prefer the candidate, which is 60% or 0.60 in decimal form. The standard deviation of the sampling distribution can be calculated using the formula:
Standard deviation (σ) = √[p(1 - p) / n]
where p is the proportion of voters who prefer the candidate (0.60) and n is the sample size (500).
Substituting the values into the formula:
σ = √[0.60(1 - 0.60) / 500]
Step 3: Convert the problem into a standard normal distribution.
To do this, we need to calculate the z-score, which indicates the number of standard deviations an observation is from the mean. The formula for calculating the z-score is:
z = (x - μ) / σ
where x is the desired value (56%), μ is the mean of the sampling distribution (0.60), and σ is the standard deviation of the sampling distribution calculated in Step 2.
Substituting the values into the formula:
z = (0.56 - 0.60) / σ
Step 4: Find the probability using the standard normal distribution table.
Using the z-score calculated in Step 3, we can find the corresponding probability from the standard normal distribution table. This probability represents the area under the curve to the left of the z-score. However, since we're interested in the probability of getting less than 56% of the votes (to the left of the desired value), we need to find the cumulative probability of the z-score.
Step 5: Interpret the result.
The probability obtained from the standard normal distribution table represents the likelihood of the candidate receiving less than 56% of the votes, given the data from the poll.
Note: If you have access to statistical software or a calculator that provides normal distribution functions, you can also use those to calculate the probability directly, without having to refer to the standard normal distribution table.