Question

Random samples of size \(n=75\) were selected from a binomial population with \(p=.4\). Use the normal distribution to approximate the following probabilities: a. \(P(\hat{p} \leq .43)\) b. \(P(.35 \leq \hat{p} \leq .43)\)

Short answer

Answer: The approximate probabilities are: a. \(P(\hat{p}\leq0.43) \approx 0.7019\) b. \(P(0.35\leq\hat{p}\leq0.43) \approx 0.5125\)

Step-by-step solution

1. Calculate the mean and standard deviation

To approximate the binomial distribution with a normal distribution, we first need to compute the mean (\(\mu\)) and standard deviation (\(\sigma\)) of the binomial distribution. The mean and standard deviation of a sample proportion are calculated as follows: $$ \mu = p \\ \sigma = \sqrt{\frac{p(1-p)}{n}} $$ Using the given values of \(p=0.4\) and \(n=75\), we can calculate the mean and standard deviation. $$ \mu = 0.4 \\ \sigma = \sqrt{\frac{0.4(1-0.4)}{75}} = 0.057 $$

2. Convert to Z-scores

Now, we need to convert the values of the sample proportions to their respective Z-scores. A Z-score represents the number of standard deviations a value is away from the mean. To find the Z-score of a sample proportion, we can use the following formula: $$ Z = \frac{\hat{p} - \mu}{\sigma} $$ a. For part (a), we want to find \(P(\hat{p}≤ 0.43)\). Therefore, we need to find the \(Z\)-score for \(\hat{p}=0.43\): $$ Z = \frac{0.43 - 0.4}{0.057} = 0.53 $$ b. For part (b), we want to find \(P(0.35≤\hat{p}≤0.43)\). Therefore, we need to find the \(Z\)-scores for \(\hat{p}=0.35\) and \(\hat{p}=0.43\): $$ Z_1 = \frac{0.35 - 0.4}{0.057} = -0.88 \\ Z_2 = \frac{0.43 - 0.4}{0.057} = 0.53 $$

3. Find probabilities using Z-table

We will use the Z-table (standard normal distribution table) to find the probabilities associated with each Z-score calculated in step 2. a. For part (a), we want to find \(P(\hat{p}≤ 0.43)\). Looking up the Z-score of \(0.53\) on the Z-table, we find that the probability is \(0.7019\). So, $$ P(\hat{p}≤ 0.43) = 0.7019 $$ b. For part (b), we want to find \(P(0.35≤\hat{p}≤0.43)\). To do this, we need to find the probability for each Z-score and subtract: $$ P(-0.88≤Z≤0.53) = P(Z≤0.53) - P(Z≤-0.88) = 0.7019 - 0.1894 = 0.5125 $$ So, $$ P(0.35≤\hat{p}≤0.43) = 0.5125 $$ In conclusion, the approximated probabilities using the normal distribution are: a. \(P(\hat{p}\leq0.43) \approx 0.7019\) b. \(P(0.35\leq\hat{p}\leq0.43) \approx 0.5125\)

Key concepts

Binomial Distribution

A binomial distribution is a probability distribution that summarizes the likelihood of a particular number of successes in a fixed number of independent trials, each with the same probability of success. It has two key parameters: the probability of success in a single trial, denoted as \( p \), and the number of trials, denoted as \( n \). For example, if you're flipping a coin where heads occur with probability \( p = 0.4 \) and you flip it 75 times, you are looking at a binomial distribution of \( n = 75 \) and \( p = 0.4 \).

Binomial distributions are discrete, meaning they deal with outcomes that have distinct values. When dealing with large sample sizes, computations can be cumbersome. Thus, it's often useful to approximate a binomial distribution with a normal distribution for practical reasons.

Sample Proportion

The sample proportion \( \hat{p} \) is a critical concept when discussing binomial distributions and their approximation by a normal distribution. It represents the fraction of successes observed in a sample. If you conduct a survey or experiment, it is the number of times an event occurs divided by the total number of trials, or sample size \( n \).

Mathematically, it is represented as: \[ \hat{p} = \frac{x}{n} \] where \( x \) is the number of successful outcomes.

In our example, where \( n = 75 \) and you're interested in predicting certain probabilities, you're essentially working with this sample proportion to make approximations using normal distribution.

Z-score Calculation

A Z-score, also known as a standard score, enables you to understand where a particular value lies in relation to the mean of a distribution, measured in terms of standard deviations. It is a way of quantifying the distance of a sample proportion from the mean of the distribution.

To calculate a Z-score for a sample proportion \( \hat{p} \), you use:
\[ Z = \frac{\hat{p} - \mu}{\sigma} \] where \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For example, if \( \hat{p} = 0.43 \), \( \mu = 0.4 \), and \( \sigma = 0.057 \), the Z-score is calculated as \( Z = \frac{0.43 - 0.4}{0.057} = 0.53 \). This shows \( \hat{p} \) is 0.53 standard deviations above the mean.

Z-scores are crucial because they allow you to use the standard normal distribution table to find probabilities, converting binomial problems into manageable calculations.

Probability Approximation

Probability approximation is a valuable technique in statistics, especially when dealing with probabilities in binomial distributions. It allows us to estimate binomial probabilities using the normal distribution when the sample size is large enough for accuracy.

This involves transforming a binomial distribution into a normal distribution by converting values into Z-scores and using standard normal distribution tables to find the required probabilities. For instance, the probability \( P(\hat{p} \leq 0.43) \) is found by calculating the Z-score and referencing the Z-table to get 0.7019.

Why the approximation? The normal distribution is continuous, unlike the discrete binomial distribution. When \( np \) and \( n(1-p) \) are large enough (typically \( np, n(1-p) \geq 5 \)), the shape of the binomial distribution becomes similar to a normal distribution, allowing for this convenient use.