Question

Consider a binomial random varible with \(n=25\) and \(p=.6 .\) 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 more than 9 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 \(P(x>9) \approx P(z>\)______) \(=1-\) _____ \(=\) ____.

Short answer

Please provide the step by step solution and fill in the blanks. Answer: Yes, we can use the normal approximation to find the probability of more than 9 successes. The probability is approximately 0.9875.

Step-by-step solution

Checking the conditions for normal approximation

To verify if we can use the normal approximation, we need to calculate n*p and n*q. Given n = 25 and p = 0.6, we have q = 1 - p = 1 - 0.6 = 0.4. Now, calculate n*p and n*q: n*p = 25 * 0.6 = 15 n*q = 25 * 0.4 = 10

Checking if n*p and n*q are greater than 5

Now, we have to check if n*p and n*q are both greater than 5: n*p = 15 > 5 n*q = 10 > 5 Since both n*p and n*q are greater than 5, we can use the normal approximation.

Calculate μ and σ

As we can use the normal approximation, now calculate μ and σ: μ = n*p = 15 σ = √(n*p*q) = √(25*0.6*0.4) = √(6) = 2.449

Find the values of x for more than 9 successes

To find the probability of more than 9 successes, we need to find the values ​​of x to be included: x > 9

Include the entire block of probability

To include the entire block of probability for the first value of x, start at: x = 9.5

Calculate z-score

Now, calculate the z-score using the formula: z = (x ± 0.5 - μ) / σ = (9.5 - 15) / 2.449 ≈ -2.24

Calculate the probability using z-score

Now, calculate P(x>9) using the z-score: P(x>9) ≈ P(z > -2.24) = 1 - P(z < -2.24) = 1 - 0.0125 = 0.9875 #Summary of the answers: a. n*p = 15 and n*q = 10 b. Yes c. μ = 15 and σ ≈ 2.449 d. x > 9 e. x = 9.5, start at 9.5 f. z ≈ -2.24 g. P(x>9) ≈ P(z > -2.24) = 1 - 0.0125 ≈ 0.9875

Key concepts

Understanding the Binomial Distribution

The binomial distribution is a fundamental concept in statistics, commonly used to model the number of successes in a sequence of independent experiments. Imagine you're flipping a coin multiple times; each flip is an independent trial with two possible outcomes: heads or tails. Similarly, in a binomial distribution, each test has two results: success or failure.
The key parameters of a binomial distribution include:

• \(n\): The number of trials
• \(p\): The probability of success in each trial

In our exercise, we're given a situation where \(n = 25\) and \(p = 0.6\). This means we perform 25 trials, and each has a 60% chance of success. Calculating the mean or expected number of successes, \(np\), tells us the average number of successful outcomes we'd expect if the trials were repeated many times.

Computing the Z-score for Normal Approximation

In statistics, the z-score is a measure used to determine how many standard deviations an element is from the mean. When approximating a binomial distribution with a normal distribution, the z-score takes on a critical role.
To find the z-score, use the formula:\[z = \frac{x \pm 0.5 - \mu}{\sigma}\]Where:

• \(x\) is the observed number of successes
• \(\pm 0.5\) is a continuity correction, added to better adjust the discrete binomial distribution to a continuous normal distribution
• \(\mu\) is the mean \(np\)
• \(\sigma\) is the standard deviation \(\sqrt{npq}\)

The z-score tells us the position or "distance" of a value relative to the expected mean, facilitating the probability calculation using a standard normal table.

Probability Calculation Using the Normal Approximation

Once the z-score is calculated, probability calculations can predict outcomes based on the normal approximation of a binomial distribution.
The formula for calculating probabilities involves finding the area under the curve of a standard normal distribution. For our exercise, we calculated \(P(x > 9)\) by finding \(P(z > -2.24)\).
To compute this, use a standard normal distribution table, which provides the probabilities associated with each z-score:

• Locate your z-score in the table to find the probability of \(z\) being less than the computed value.
• Subtract this value from 1 to get \(P(z > -2.24)\).

This tells us the probability of observing more than 9 successes in our binomially distributed scenario.

Evaluating np and nq Conditions

Before approximating a binomial distribution with a normal distribution, it is crucial to verify if \(np\) and \(nq\) conditions are satisfied. This method is only valid if both \(np\) and \(nq\) are greater than 5.
Here's why:

• If \(np\) and \(nq\) aren't sufficiently large, the binomial distribution's discreteness will cause an inaccurate approximation using a continuous normal distribution.
• In this exercise, \(np = 15\) and \(nq = 10\), satisfying this requirement.

Satisfying these conditions ensures that the probability distribution is adequately "spread" to effectively smooth out the approximation, allowing us to use the normal distribution for calculations confidently.