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Suppose that \(65 \%\) of all registered voters in a certain area favor a seven- day waiting period before purchase of a handgun. Among 225 randomly selected registered voters, what is the approximate probability that a. At least 150 favor such a waiting period? b. More than 150 favor such a waiting period? c. Fewer than 125 favor such a waiting period?

Short Answer

Expert verified
The approximate probabilities are as follows: a. At least 150 favor such a waiting period: \(0.2921\) or \(29.21\%\) b. More than 150 favor such a waiting period: \(0.2676\) or \(26.76\%\) c. Fewer than 125 favor such a waiting period: \(0.0008\) or \(0.08\%\)

Step by step solution

01

Calculate the mean and standard deviation of the sample

First, we need to calculate the mean and standard deviation of the sample. Mean (μ) = n * p Standard Deviation (σ) = √(n * p * (1-p)) Where n = sample size, p = probability of success Mean: μ = 225 * 0.65 = 146.25 Standard Deviation: σ = √(225 * 0.65 * (1-0.65)) = √(146.25 * 0.35) ≈ 6.864
02

Calculate "z" for each scenario

The next step is to calculate the z-score for each scenario. a. At least 150 favor such a waiting period: z = (x - μ) / σ z = (150 - 146.25) / 6.864 ≈ 0.546 b. More than 150 favor such a waiting period: z = (150.5 - 146.25) / 6.864 ≈ 0.619 (We use 150.5 because we are looking for more than 150.) c. Fewer than 125 favor such a waiting period: z = (124.5 - 146.25) / 6.864 ≈ -3.16 (We use 124.5 because we are looking for fewer than 125.)
03

Find the probabilities using a z-table

Now, we will use a z-table to find the probabilities for each scenario. a. At least 150 favor such a waiting period: The table value corresponding to z = 0.546 is 0.7079. Since we need the probability of at least 150, meaning the right tail of the distribution, we will subtract it from 1. P(x ≥ 150) = 1 - 0.7079 = 0.2921 b. More than 150 favor such a waiting period: The table value corresponding to z = 0.619 is 0.7324. Since we need the probability of more than 150, meaning the right tail of the distribution, we will subtract it from 1. P(x > 150) = 1 - 0.7324 = 0.2676 c. Fewer than 125 favor such a waiting period: The table value corresponding to z = -3.16 is 0.0008. Since we need the probability of fewer than 125, meaning the left tail of the distribution, we will use the value as it is. P(x < 125) = 0.0008
04

Results

So, we find the following approximate probabilities: a. At least 150 favor such a waiting period: 0.2921 or 29.21% b. More than 150 favor such a waiting period: 0.2676 or 26.76% c. Fewer than 125 favor such a waiting period: 0.0008 or 0.08%

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean and Standard Deviation Calculation
Understanding the calculation of the sample mean and standard deviation is crucial in probability and statistics, particularly when working with sampling distributions. The sample mean, denoted as \( \mu \), is a measure of the central tendency of the data. It represents the average outcome we might expect. In our exercise, where 65% of voters favor a policy, and we sample 225 voters, the mean is calculated by multiplying the sample size \( n \) by the probability of a voter favoring the policy \( p \):

\[ \mu = n \times p \]

For our case, this yields:

\[ \mu = 225 \times 0.65 = 146.25 \]

The standard deviation, symbolized as \( \sigma \), quantifies the amount of variation or dispersion in the sample. The formula to find the standard deviation for a binomial distribution is:

\[ \sigma = \sqrt{n \times p \times (1-p)} \]

Applying it to our scenario:

\[ \sigma = \sqrt{225 \times 0.65 \times (1 - 0.65)} \approx 6.864 \]

This calculation sets up the next steps in our problem, allowing us to assess the likelihood of different outcomes within our sample.
Z-score Calculation
A z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. It's a dimensionless quantity that helps us determine the likelihood of a specific outcome.

To calculate the z-score, we use the formula:

\[ z = \frac{(x - \mu)}{\sigma} \]

In the context of our voter sample, we computed the z-scores for different thresholds of voters favoring the waiting period to purchase a handgun. For 150 voters, the calculation was:

\[ z = \frac{(150 - 146.25)}{6.864} \approx 0.546 \]

This z-score tells us that 150 is approximately 0.546 standard deviations above the mean. We also computed the z-score for more than 150 and fewer than 125, giving us z-scores that can be used with a z-table to find associated probabilities.
Z-table Probability Lookup
The z-table is a reference table that provides the probability of a z-score occurring by chance under a standard normal distribution. The table is a fundamental tool for finding the proportion of values to the left of a specified z-score.

To find the probability related to each z-score:
  • For at least 150 favoring the policy, we looked up the z-score of 0.546 in the z-table, which gave us a value of 0.7079. The probability we need is the complement since we want 'at least', so we subtract it from 1 to get the upper tail.
  • For more than 150, we do the same with a z-score of 0.619, resulting in a table value of 0.7324, and again subtract it from 1 for the upper tail probability.
  • For fewer than 125, the z-score of -3.16 corresponds to 0.0008 in the z-table, which directly gives us the probability since it's less than the mean.
These probabilities help us understand the likelihood of observing at least or more than a certain number of voters in favor, or fewer than another, within our sample.

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