Question

In an agricultural experiment, a large field of wheat was divided into many plots (each plot being \(7 \times\) \(100 \mathrm{ft}\) ) and the yield of grain was measured for each plot. These plot yields followed approximately a normal distribution with mean \(88 \mathrm{lb}\) and standard deviation \(7 \mathrm{lb}\) (as in Exercise 4.3.5). Let \(\bar{Y}\) represent the mean yield of five plots chosen at random from the field. Find \(\operatorname{Pr}\\{\bar{Y}>80\\}\).

Short answer

Calculate the Z-Score using the value of \(-8\) and the standard error, look up the corresponding probability in the standard normal distribution table, and then subtract this cumulative probability from 1 (or find the complement directly using a calculator or software) to find \(\operatorname{Pr}\{\bar{Y}>80\}\).

Step-by-step solution

Understand and Identify the Given Information

The problem states that the yield of grain per plot follows a normal distribution with a mean (\(\mu\)) of 88 pounds and a standard deviation (\(\sigma\)) of 7 pounds. We are looking at the mean yield (\(\bar{Y}\)) of 5 randomly selected plots. The task is to find the probability that \(\bar{Y}\) is greater than 80 pounds.

Calculate the Standard Error

The standard error (SE) of the mean is calculated using the formula \(SE = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation of the individual measurements and \(n\) is the number of observations. For our case, \(SE = \frac{7 \text{ lb}}{\sqrt{5}}\).

Standardize the Mean Yield

To find the probability that \(\bar{Y}\) is greater than 80 (\(\operatorname{Pr}\{\bar{Y}>80\}\)), we need to standardize \(\bar{Y}\) using the z-score formula \(Z = \frac{\bar{Y} - \mu}{SE}\). Plugging in the values, we have \(Z = \frac{80 - 88}{7/\sqrt{5}}\).

Calculate the Z-Score

Calculate the Z-Score using the values from the previous step. \(Z = \frac{80 - 88}{7/\sqrt{5}} = \frac{-8}{7/\sqrt{5}}\). This will give us the standardized value corresponding to a mean yield of 80 pounds.

Interpret the Result

The value found in the previous step represents the probability that the mean yield of five randomly chosen plots will be greater than 80 pounds. If using a standard normal distribution table, we typically find the cumulative probability corresponding to the calculated Z-Score and then subtract it from 1 to find the probability that \(\bar{Y}\) is greater than 80. If using a calculator or software, it may directly give the probability for the Z-Score.

Key concepts

Understanding Standard Error

The concept of standard error is pivotal when dealing with statistics, particularly in the context of normal distribution. It measures the variability or dispersion of sample mean estimates from the actual population mean. In simple terms, standard error indicates the precision of the sample mean as an estimate of the population mean.

To calculate the standard error (SE), use the formula: \[SE = \frac{\sigma}{\sqrt{n}}\], where \(\sigma\) is the standard deviation of the population and \(n\) is the sample size. This equation highlights that the larger the sample size, the smaller the standard error, and thus the closer the sample mean is likely to be to the population mean.

In an agricultural experiment like the one in our exercise, the standard error helps in determining how much the average yields (\bar{Y}) of selected plots might vary from the true mean yield of the entire field.

Interpreting the Z-Score

The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. A z-score can tell you how far away a particular value is from the average.

To standardize a random variable, subtract the mean of the distribution from the value and then divide by the standard error: \[Z = \frac{X - \mu}{SE}\].

Using the z-score, researchers can assess how extraordinary a measurement is; whether it is typical for a given data set, or it is atypical. In our problem, the z-score transforms the average yield of the wheat plots into a score that can be assessed against the standard normal distribution to find the probability of a particular outcome.

Probability in Normal Distribution

Probability is a measure of the likelihood that an event will occur. When dealing with normal distributions, the probability can be represented visually as the area under the curve of a graph. The total area under a normal distribution curve equals 1, or 100% probability.

In our exercise, after finding the z-score, we look for the corresponding probability that the mean yield is greater than 80 pounds. This probability is found by looking at the standard normal distribution table, which shows the cumulative probability for any given z-score. Since we want the probability that \(\bar{Y}\) exceeds a certain value, we subtract the cumulative probability from 1.

This foundational concept allows us to predict and understand the likelihood of various outcomes given a set of data that follows a normal distribution.

Random Sampling

Random sampling is a core concept in statistics that ensures each member of a population has an equal chance of being selected. The value of this approach is that it helps to eliminate bias and allows the sample to represent the population accurately.

In our scenario, we're interested in the mean yield of randomly selected plots of wheat. Randomly sampling five plots ensures that the mean yield calculated will be an unbiased estimate of the population mean yield.

Remember, the effectiveness of random sampling heavily depends on the sample size and the method used for selecting samples. As the number of samples increases, the accuracy of the estimation of population parameters typically increases as well.