Question

Let \(x\) be a binomial random variable with \(n=15\) and \(p=.5\) a. Is the normal approximation appropriate? b. Find \(P(x \geq 6)\) using the normal approximation. c. Find \(P(x>6)\) using the normal approximation. d. Find the exact probabilities for parts \(b\) and \(c,\) and compare these with your approximations.

Short answer

Calculate the approximate and exact probabilities for \(P(x \geq 6)\) and \(P(x > 6)\), and compare the results. Answer: The normal approximation is not appropriate in this case, as both conditions (\(np \geq 10\) and \(n(1-p) \geq 10\)) are not satisfied. However, the approximate and exact probabilities are as follows: For \(P(x \geq 6)\): Approximate probability: \(0.78\) Exact probability: \(0.798\) For \(P(x > 6)\): Approximate probability: \(0.60\) Exact probability: \(0.629\) The probabilities aren't too far off, but they are not close enough for the normal approximation to be considered appropriate in this case.

Step-by-step solution

Check the normal approximation's appropriateness

First, we need to determine if the normal distribution is appropriate. Check if both \(np \geq 10\) and \(n(1-p) \geq 10\) are true: \(np = 15 \times 0.5 = 7.5\) \(n(1-p) = 15 \times 0.5 = 7.5\) Both conditions are not satisfied, so the normal approximation is not appropriate in this case. However, we will continue to compute the requested probabilities using the normal approximation for comparison's sake.

Calculate the mean and standard deviation

Calculate the mean and standard deviation of the binomial random variable: Mean: \(\mu = np = 15 \times 0.5 = 7.5\) Standard deviation: \(\sigma = \sqrt{np(1-p)} = \sqrt{15 \times 0.5 \times 0.5} = 1.936\)

Find \(P(x \geq 6)\) using the normal approximation

To find \(P(x \geq 6)\) using the normal approximation, we need to calculate the z-score: \(z = \frac{6 - 7.5}{1.936} = -0.775\) Now, use a z-table or calculator to find the probability corresponding to this z-score: \(P(x \geq 6) \approx P(z \geq -0.775) = 0.78\)

Find \(P(x > 6)\) using the normal approximation

To find \(P(x > 6)\), we need to calculate the z-score for \(x=7\): \(z = \frac{7 - 7.5}{1.936} = -0.258\) Now, use a z-table or calculator to find the probability corresponding to this z-score: \(P(x > 6) \approx P(z > -0.258) = 0.60\)

Find exact probabilities for parts b and c

Using the binomial probability formula: For part b: \(P(x \geq 6) = \sum_{k=6}^{15} \binom{15}{k} (0.5)^k (0.5)^{15-k} = 0.798\) For part c: \(P(x > 6) = P(x \geq 7) = \sum_{k=7}^{15} \binom{15}{k} (0.5)^k (0.5)^{15-k} = 0.629\)

Compare approximate and exact probabilities

Now, let's compare the approximate and exact probabilities: For part b: Approximate probability: \(P(x \geq 6) \approx 0.78\) Exact probability: \(P(x \geq 6) = 0.798\) For part c: Approximate probability: \(P(x > 6) \approx 0.60\) Exact probability: \(P(x > 6) = 0.629\) We can see that the probabilities aren't too far off, but they are not close enough for the normal approximation to be considered appropriate in this case.

Key concepts

Normal Approximation

Normal approximation is a technique used to simplify probability calculations for binomial distributions, particularly when dealing with large sample sizes. Even though the binomial distribution is discrete, and the normal distribution is continuous, the normal approximation allows us to use continuous methods for calculations.

However, for this method to be reliable, there are certain conditions. The criteria for employing the normal approximation involve checking if both \(np \geq 10\) and \(n(1-p) \geq 10\). This ensures that the sample size and probability are large enough for the normal distribution to closely approximate the binomial distribution.

• Example: For \(n = 15\) and \(p = 0.5\), we check \(np = 7.5\) and \(n(1-p) = 7.5\).
• Both are less than 10, indicating that normal approximation isn't suitable.

Despite this, one might still use the technique for comparison purposes to observe how close the approximate probabilities are to the true binomial probabilities.

Probability Calculation

Probability calculation in the context of binomial and normal distributions involves determining the likelihood of a specific number of successes in a series of trials.

When using the normal approximation, we usually calculate probabilities by converting the discrete binomial variable to a continuous normal variable. This allows for easier computation using the properties of the normal distribution.

• For instance, to find \(P(x \geq 6)\), the z-score helps determine the probability that the binomial variable is above this threshold.
• This involves using tables or software to find the corresponding probability of a specific z-score value.

Despite simplification, when the conditions for normal approximation aren't met, the calculated probabilities may significantly differ from exact values, as exemplified by our problem.

Z-Score

The z-score is a measure that represents the number of standard deviations a data point is from the mean of a distribution. In the context of normal approximation to the binomial distribution, the z-score is used to transform a binomial variable to a standard normal variable.

The formula to compute the z-score is:\[z = \frac{x - \mu}{\sigma}\]where \(x\) is the desired value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. This conversion allows for using the cumulative nature of normal distribution to find probabilities.

• For example, in finding \(P(x \geq 6)\), it's equivalent to finding \(P(z \geq -0.775)\).
• Z-score tables or calculators can then provide the probability for these z-values easily.

Understanding z-scores is crucial for interpreting where your value stands relative to the entire distribution, especially in statistical analyses and probabilistic modeling.

Binomial Probability Formula

The binomial probability formula is central to calculating exact probabilities for binomial distributions. It determines the likelihood of achieving exactly \(k\) successes in \(n\) independent trials, each with the probability \(p\) of success.

The formula is given by:\[P(x = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \(\binom{n}{k}\) is a combination representing the number of ways to choose \(k\) successes out of \(n\) trials.

• To use it for probabilities like \(P(x \geq 6)\), sum the probabilities of \(x\) being from 6 to 15.
• This ensures an "exact" calculation as opposed to using an approximation.

By comparing results from this formula with those obtained through normal approximation, you can assess the accuracy and appropriateness of using approximation methods depending on the scenario conditions.