Question

Consider a binomial random variable with \(n=45\) and \(p=.05 .\) Fill in the blanks below to find some probabilities using the normal approximation. a. Can we use the normal approximation? Calculate \(n p=\) ______ and \(n q=\) ______ b. Are \(n p\) and \(n q\) both greater than \(5 ?\) Yes_____ No______ c. If the answer to part \(b\) is yes, calculate \(\mu=n p=\) _______ and \(\sigma=\sqrt{n p q}=\) _____ d. To find the probability of 10 or fewer successes, what values of \(x\) should be included? \(x=\) _______ e. To include the entire block of probability for the first value of \(x=\) ______, start at ______. f. Calculate \(z=\frac{x \pm .5-n p}{\sqrt{n p q}}=\) _______. g. Calculate \(P(x \leq 10) \approx P(z<\)_____ ) \(=\) ______.

Short answer

Question: Explain why the normal approximation cannot be used in this exercise. Answer: The normal approximation cannot be used in this exercise because the value of \(np = 2.25\) is less than 5. For the normal approximation to be valid, both \(np\) and \(nq\) should be greater than 5.

Step-by-step solution

Check if the normal approximation can be used

Calculate \(np\) and \(nq\). Since \(n=45\) and \(p=0.05\), then \(q=1-p=0.95\). $$np = 45\times 0.05 = 2.25$$ $$nq = 45\times 0.95 = 42.75$$

Determine if \(np\) and \(nq\) are both greater than 5

Based on the calculations in Step 1, \(np = 2.25\) and \(nq = 42.75\). Since \(np < 5\), we cannot use the normal approximation.

Calculate \(μ\) and \(σ\)

Since we cannot use the normal approximation, we can skip this step.

Determine the values of \(x\) for 10 or fewer successes

Although we can't use the normal approximation for this exercise because \(np<5\), let's assume we could. If we were to find the probability of 10 or fewer successes, we would include the values of \(x\) from 0 to 10.

Include the entire block of probability for the first value of \(x\)

For the first value of \(x=0\), we would start at \(x=-0.5\) to include the entire block of probability.

Calculate the \(z\)-score

For finding the probability of 10 or fewer successes, the \(z\)-score would be calculated as follows (assuming we could use the normal approximation) for \(x=10\): $$z=\frac{x\pm .5-np}{\sqrt{npq}}$$ However, since the normal approximation is not valid in this case, we cannot calculate the \(z\)-score.

Calculate \(P(x\leq10)\) using the \(z\)-score

As we mentioned in Step 6, we cannot use the normal approximation for this exercise, so we cannot calculate the probability using the \(z\)-score.

Key concepts

Binomial Random Variable

The binomial random variable is a fundamental concept in statistics, which represents the number of successes in a fixed number of independent trials. Each trial can result in a success with a probability of p, or a failure with a probability of q (where q = 1-p). A simple example of a binomial scenario is flipping a coin a certain number of times and counting how many times it lands on heads.

In the context of the provided exercise, where we have a binomial random variable with 45 trials (n=45) and a 5% success rate (p=0.05), we aim to evaluate the likelihood of observing a certain number of successes. To approximate binomial probabilities when n is large, we often use the normal distribution. However, this approach is only suitable when certain criteria are fulfilled, such as the resulting np and nq values being greater than 5. In this case, since np is less than 5, the normal approximation is not appropriate, which is essential knowledge for accurately conducting probability calculations.

Probability Calculations

Probability calculations involve determining the likelihood of various outcomes. When dealing with binomial random variables, we are essentially looking at the probabilities of observing a certain number of successes out of a pre-determined number of trials. Normally, these calculations can be arduous when done manually; hence, the normal approximation is a strategy used to simplify the process by treating the binomial distribution as a normal distribution.

The crucial point to remember from the exercise is the precondition for using the normal approximation — specifically, that both np and nq must be greater than 5 for the approximation to be valid. Otherwise, one has to resort to exact probability calculations using the binomial formula, which can be quite complicated as the number of trials (n) becomes large. This example reinforces the need for students to first check the criteria before applying the normal approximation to binomial probability questions.

Z-Score Calculation

The z-score calculation plays a pivotal role when applying the normal approximation to the binomial distribution. It's a measure indicating how many standard deviations an element is from the mean. To convert a binomial distribution into a normal distribution, the z-score helps by standardizing individual data points.

Ordinarily, once the normal distribution's conditions are met, the z-score is computed using the formula \( z = \frac{x \pm 0.5 - np}{\sqrt{npq}} \), where x is the number of successes, np is the mean, and \( \sqrt{npq} \) is the standard deviation of the binomial distribution. The ±0.5 is the continuity correction factor, which adjusts for the binomial distribution's discrete nature when approximating it with the continuous normal distribution.

To elucidate, imagine you wish to find the probability of obtaining 10 or fewer successes — you would calculate the z-score for x=10. However, as highlighted in the exercise, if the conditions aren't met (np < 5), the normal approximation isn't applied, and thus, the z-score is not computed. Understanding when and how to calculate a z-score is vital for students working with normal approximation and probability problems.