Problem 26 - Applied Statistics and Probability for Engineers

Question

The yield in pounds from a day's production is normally distributed with a mean of 1500 pounds and standard deviation of 100 pounds. Assume that the yields on different days are independent random variables. (a) What is the probability that the production yield exceeds 1400 pounds on each of five days next week? (b) What is the probability that the production yield exceeds 1400 pounds on at least four of the five days next week?

Step-by-Step Solution

Answer

(a) 0.289; (b) 0.686

Step 1: Understand the Given Problem

We are given that yields from a day's production are normally distributed with a mean, \( \mu = 1500 \) pounds, and a standard deviation, \( \sigma = 100 \) pounds. We are looking to answer two probability questions concerning yields exceeding 1400 pounds across five days.

Step 2: Define the Normal Distribution Condition

For a single day, the probability that the yield exceeds 1400 pounds corresponds to finding \( P(X > 1400) \) where \( X \) is normally distributed with \( \mu = 1500 \) and \( \sigma = 100 \). We convert 1400 pounds to a z-score using the formula \( z = \frac{X - \mu}{\sigma} \).

Step 3: Calculate the Z-score for 1400 Pounds

The z-score for 1400 is given by:\[z = \frac{1400 - 1500}{100} = -1\]This means we want to find the probability \( P(Z > -1) \) for a standard normal distribution.

Step 4: Find Probability Using Z-table or Normal Distribution

Using the z-table or normal distribution calculator, find the probability corresponding to \( z = -1 \). \( P(Z > -1) = 1 - P(Z < -1) = 1 - 0.1587 = 0.8413 \). This is the probability the yield exceeds 1400 pounds on any given day.

Step 5: Compute Probability for Part (a)

To exceed 1400 pounds on each of the five days, the probability is \( (0.8413)^5 \), since days are independent:\[P(\text{exceeding 1400 each day for 5 days}) = 0.8413^5 = 0.289\]

Step 6: Compute Probability for Part (b)

For at least four out of five days to have yields exceeding 1400 pounds, compute as follows:1. Exactly four days exceed 1400:\(\binom{5}{4} \times (0.8413)^4 \times (0.1587)^1 = 5 \times 0.5005 \times 0.1587 = 0.397\)2. All five days exceed 1400 (already calculated as 0.289).Add these probabilities:\( 0.397 + 0.289 = 0.686 \).

Step 7: Conclusion

We now have our answers for the probabilities involved in parts (a) and (b) of the problem based on normal distribution and the independence of days.

Key Concepts

Probability Calculation

Calculating probabilities can feel like solving a mystery, but the process is quite logical once you understand the steps. When dealing with normal distributions, most of our interest lies in finding the probability of observing a certain value or range of values.

To calculate probabilities for normal distributions, we need the mean, standard deviation, and the specific value of interest. We then use the Z-score to transform our specific value into a standardized form. For example, in the given exercise, we calculated the probability of a daily yield exceeding 1400 pounds for five consecutive days.

• First, we transformed the yield threshold into a Z-score.
• Next, we used a Z-table or software to find the probability corresponding to that Z-score.
• Finally, we combined probabilities depending on the number of days of interest to arrive at the answer.

Z-score

The Z-score is a powerful tool in statistics that allows us to compare different data points by standardizing them.

To calculate a Z-score, use the formula: \[ z = \frac{X - \mu}{\sigma} \]where:

• \(X\) is the data point you're investigating,
• \(\mu\) is the mean of the distribution,
• \(\sigma\) is the standard deviation.

A Z-score tells you how many standard deviations a value is from the mean. Positive Z-scores mean the value is above the average, while negative ones indicate it's below. In the exercise, a Z-score of -1 was calculated for a yield of 1400 pounds, indicating it is one standard deviation below the mean of 1500 pounds.

Independent Random Variables

In probability and statistics, independent random variables help simplify complex calculations. Independence means that the outcome of one variable doesn't affect the other.

In the context of the exercise, this concept was key. The yields for each day are considered independent. This simplifies probability computations for scenarios across multiple days.

• If two events, like daily yields, are independent, the probability of both occurring is simply the product of their individual probabilities.
• This property allows for straightforward calculations when determining the likelihood of certain outcomes on multiple days.

Understanding this principle makes tackling multi-day probability problems more manageable.

Standard Deviation

Standard deviation is an essential concept in understanding how data is spread around the mean. In simple terms, it quantifies the amount of variation or dispersion in a set of data values. A low standard deviation means data points are generally close to the mean, while a high standard deviation indicates more spread out data.

In normal distribution, about 68% of values fall within one standard deviation of the mean, and about 95% fall within two. This helps us set expectations about the spread of our data and identify where a particular value stands in relation to others. In our exercise, the standard deviation of 100 pounds gave context to the typical spread of daily yields around the 1500 pounds mean.

Standard deviation helps you understand the consistency, or variability, in your data—insightful for predicting outcomes and assessing risks.