Use a normal approximation to find the probability of the indicated number of voters. In this case assume that 119 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.
The probability that fewer than 32 of 19 eligible voters voted is=
To find the probability using a normal approximation, we'll first calculate the mean and standard deviation.
Mean (μ) = np = 119 * 0.22 = 26.18
Standard deviation (σ) = √(np(1-p)) = √(119 * 0.22 * 0.78) = √(21.5884) ≈ 4.65
Now, we will find the z-score for 32 voters.
z = (x - μ) / σ = (32 - 26.18) / 4.65 ≈ 1.25
Using a z-table, we find the probability for a z-score of 1.25 is 0.8944. However, we want the probability that fewer than 32 voted, so we'll find the complement.
P(x < 32) = 1 - P(x >= 32) = 1 - 0.8944 = 0.1056
Thus, the probability that fewer than 32 of the 119 eligible voters aged 18-24 voted is approximately 0.1056 or 10.56%.
To find the probability that fewer than 32 of 119 eligible voters aged 18-24 voted, we can use a normal approximation. This approximation relies on the Central Limit Theorem, which states that for a large enough sample size, the sampling distribution of the sample mean can be approximated by a normal distribution.
First, we need to calculate the mean and standard deviation of the binomial distribution. The mean (μ) is given by μ = n * p, where n is the sample size (119) and p is the probability of success (0.22). Therefore, μ = 119 * 0.22 = 26.18.
The standard deviation (σ) of a binomial distribution is given by the formula σ = sqrt(n * p * (1 - p)). So, σ = sqrt(119 * 0.22 * (1 - 0.22)) = 5.07.
Now, we can calculate the z-score, which tells us how many standard deviations a particular value is away from the mean. The z-score formula is z = (x - μ) / σ, where x is the number of voters we are interested in (32). Therefore, z = (32 - 26.18) / 5.07 = 1.14.
Next, we need to find the cumulative probability of the z-score using a standard normal distribution table or a calculator. The cumulative probability is the probability that a value is less than or equal to the given z-score. In this case, we want the probability that fewer than 32 voters voted, so we need to find P(z < 1.14).
Looking up the cumulative probability for z = 1.14 in a standard normal distribution table, we can find that it is approximately 0.8729. This means that the probability of having fewer than 32 voters is approximately 0.8729 or 87.29%.
To find the probability that fewer than 32 of 119 eligible voters aged 18-24 voted, we can use a normal approximation.
First, we need to calculate the mean (μ) and standard deviation (σ) of the binomial distribution, where n is the number of trials (119) and p is the probability of success (0.22).
μ = n * p = 119 * 0.22 = 26.18
σ = sqrt(n * p * (1-p)) = sqrt(119 * 0.22 * 0.78) = 4.77
Next, we need to standardize the value of 32 using the formula z = (x - μ) / σ, where x is the number of voters we are interested in (32).
z = (32 - 26.18) / 4.77 = 1.22
Now, we can use a standard normal distribution table or a calculator to find the probability corresponding to the z-value of 1.22.
Using the table or calculator, we find that the probability of z < 1.22 is approximately 0.8897.
Therefore, the probability that fewer than 32 of 119 eligible voters aged 18-24 voted is approximately 0.8897 or 88.97%.