Problem 25 - A First Course in Probability

Question

Each item produced by a certain manufacturer is, independently, of acceptable quality with probability .95 Approximate the probability that at most 10 of the next 150 items produced are unacceptable.

Step-by-Step Solution

Answer

The probability that at most 10 of the next 150 items produced are unacceptable is approximately 0.826 or 82.6%.

Step 1: Identify the given information

The probability of an item being unacceptable, p = 0.05. The number of trials, n = 150. The number of unacceptable items we are interested in, k = 10.

Step 2: Find the mean and standard deviation of the binomial distribution

Mean (µ) = n * p = 150 * 0.05 = 7.5 Standard deviation (σ) = \(\sqrt{n * p * (1 - p)} = \sqrt{150 * 0.05 * (1 - 0.05)} = \sqrt{7.125} = 2.67\) (approximately)

Step 3: Convert the problem into a standard normal probability problem

To approximate the probability of at most 10 unacceptable items out of 150, we can use the standard normal distribution. We need to standardize the value of k by subtracting the mean and dividing by the standard deviation. z = \(\frac{k - µ}{σ} = \frac{10 - 7.5}{2.67}\) = 0.938 (approximately)

Step 4: Find the probability using the standard normal distribution table

Now we need to find the area to the left of z = 0.938 in the standard normal table, which represents the probability that at most 10 items are unacceptable. Looking at the standard normal distribution table, we find that the probability for z = 0.938 is approximately 0.826.

Step 5: Interpret the result

The probability that at most 10 of the next 150 items produced are unacceptable is approximately 0.826 or 82.6%.

Key Concepts

Binomial Distribution

The binomial distribution is a discrete probability distribution that represents the number of successes in a fixed number of independent and identically distributed Bernoulli trials. Each trial has two possible outcomes, typically referred to as "success" and "failure". In our example, each item produced can either be acceptable or unacceptable.
The key features of binomial distribution include:

• The number of trials, denoted as \(n\).
• The probability of success in each trial, denoted as \(p\).

In the given exercise, "success" is an item being unacceptable, with a probability \(p = 0.05\), and there are \(n = 150\) trials corresponding to items produced.

Standard Deviation

Standard deviation measures the amount of variation or dispersion of a set of values. It provides insight into how much individual item outcomes differ from the average outcome. Calculating the standard deviation for a binomial distribution helps us understand how spread out the unacceptable item counts might be.
To compute the standard deviation \(σ\) for a binomial distribution, we use the formula:
\[σ = \sqrt{n \times p \times (1 - p)}\] Where:

• \(n\) is the total number of trials (150 items in our case).
• \(p\) is the probability of an item being unacceptable (0.05).

This gives us a standard deviation of approximately 2.67, indicating how much we might expect the count of unacceptable items to deviate from the average.

Probability Calculation

Probability calculation involves determining the likelihood of certain outcomes, given a specific probability distribution. In the example exercise, we aim to find the probability that at most 10 items are unacceptable from the total production of 150 items. This involves changing the binomial setting to a normal distribution via standardization.
We follow these steps:

• Compute the mean \(\mu\) of the binomial distribution using \(\mu = n \times p\).
• Apply the z-score technique to convert the binomial variable into a standard normal variable: \(z = \frac{k - \mu}{σ}\).

Here, \(k = 10\), \(\mu = 7.5\), and \(σ = 2.67\) give a z-value of approximately 0.938.

Standard Normal Distribution

The standard normal distribution is a special normal distribution with a mean of 0 and a standard deviation of 1. This normalized form enables the use of z-scores to calculate probabilities for different distributions, like the binomial one in our task.
By standardizing the problem in the binomial framework to fit the standard normal distribution, we can employ z-tables to find probabilities. In this example, the z-score of approximately 0.938 indicates how many standard deviations away \(k = 10\) is from the mean \(\mu\).
We then use the standard normal distribution table, or z-table, to find that the probability for \(z = 0.938\) is approximately 0.826. This tells us there's an 82.6% likelihood that no more than 10 items will be unacceptable, a critical insight for evaluating production quality.