Question

Let \(X\) equal the number of chips in a chocolate chip cookies. Hundred observations of \(X\) yielded the following frequencies for the possible outcome of \(X\). $$ \begin{array}{|l|l|l|l|c|c|c|c|c|c|c|c|} \hline \text { Outcome }(x) & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text { Frequency } & 0 & 4 & 8 & 15 & 16 & 19 & 15 & 13 & 7 & 2 & 1 \\\ \hline \end{array} $$ 1\. Use these data to graph the relative frequency histogram and the Poisson probability histogram. 2\. Do these data seem to be observations of a Poisson random variable with mean \(\lambda\). Find \(\lambda\).

Short answer

The relative frequency histogram and the Poisson probability histogram are plotted using the frequency data given. Then, the mean (\(\lambda\)) of the dataset is computed. If the Poisson distribution histogram and the relative frequency histogram are similar, and the variance of the dataset approximately equals the mean (\(\lambda\)), the dataset can be assumed to follow a Poisson distribution.

Step-by-step solution

Create and Plot the Relative Frequency Histogram

First, create a histogram using the provided frequency data. The relative frequency is calculated by dividing each outcome’s frequency by the total number of observations. In this case, the total number of observations is 100. From this, plot the relative frequency histogram.

Graph the Poisson Probability Histogram

Now, plot the Poisson probability histogram. The probability mass function for a Poisson distribution is given by the formula:\[ P(x; \lambda) = \frac{\lambda^x e^{-\lambda}}{x!} \]for \( x = 0, 1, 2, 3, ... \) and \( \lambda > 0 \). Here, \( x \) is the actual number of successes that result from the experiment, and \( e \) is approximately equal to 2.71828. However, to construct the histogram, first need to find the value of \( \lambda \).

Calculate the Mean Value of Lambda (\(\lambda\))

To find the value of \( \lambda \), calculate the mean of the given data. The mean of a Poisson distribution is equal to \( \lambda \), so we can find \( \lambda \) by calculating the mean value of our observed data. The mean is the sum of each outcome multiplied by its frequency, divided by the total number of observations. Compute this to find the value of \( \lambda \).

Compare Histograms and Distributive Properties

Finally compare the relative frequency histogram with the Poisson probability histogram. If they are similar, it implies that the cookies' number of chips follows a Poisson distribution. The datasets properties must have the property of a Poisson distribution, which is that the mean is approximately equal to the variance. Compute the variance of the observed data, and if it is approximately equal to the mean (\(\lambda\)), the data probably follows a Poisson distribution.

Key concepts

Relative Frequency Histogram

A relative frequency histogram is a visual representation of the distribution of data in terms of relative frequency, which is the fraction of the total number of observations that each possible outcome occupies. This type of histogram conveys how often each outcome occurs as a proportion of the total count. In essence, it scales the frequency of each category by the total number of observations, aiding in the comparison between different sized data sets.

To construct a relative frequency histogram, as the exercise illustrates, each frequency is divided by the total number of observations to obtain the relative frequency. Then, these relative frequencies are plotted as bars, with the possible outcomes on the horizontal axis and the relative frequencies on the vertical axis.

In our exercise example calculating the relative frequency for 'Outcome' with 2 chips, which occurred 8 times out of 100 observations, would be 8/100 or 0.08. This means that 8% of the cookies had exactly 2 chips. Constructing this histogram provides a clear visual comparison of how common different outcomes are within the dataset, making it quintessential in statistical analysis and probability.

Probability Mass Function

The probability mass function (PMF) for the Poisson distribution, denoted as P(x; \(\lambda\)), gives the probability of a given number of events occurring in a fixed interval of time or space, provided that these events happen with a known constant mean rate, \(\lambda\), and independently of the time since the last event. The PMF is defined using the following formula:\[ P(x; \lambda) = \frac{\lambda^x e^{-\lambda}}{x!} \]for \(x = 0, 1, 2, 3, ...\) and \(\lambda > 0\). The variable \(x\) represents the number of events (or 'successes'), and \(e\) is the base of the natural logarithm, approximately 2.71828.

To graph the Poisson PMF, we calculate the probability for each outcome using \(\lambda\) and plot these probabilities as a histogram. This graph will show the likelihood of observing different counts of an event, such as the count of chips in a cookie. A key feature of the Poisson PMF is that it only requires the mean rate \(\lambda\) to determine the probability for all values of \(x\), simplifying the process of understanding and predicting the distribution of events.

Mean Value of Lambda (\(\lambda\))

The mean value of lambda (\(\lambda\)) in the Poisson distribution is a critical parameter as it represents both the mean and the variance of the distribution. Finding the mean value of lambda is an essential step in identifying whether a set of data follows a Poisson distribution, as highlighted in the given exercise.

To calculate \(\lambda\), we sum the product of each outcome and its corresponding frequency, then divide by the total number of observations. Let's denote the outcomes as \(x_i\) and their frequencies as \(f_i\), with \(i\) representing each possible outcome. Then,\[\lambda = \frac{\sum_{i}(x_i \times f_i)}{\text{Total Observations}}\]For example, with the data provided, the mean \(\lambda\) helps us to characterize the distribution of chocolate chips in cookies, assuming that the distribution is indeed Poisson. The accuracy of this assumption can be verified by comparing the calculated value of \(\lambda\) to the sample variance; if they are approximately equal, it strongly suggests that the data are Poisson-distributed. This equality stems from a unique property of the Poisson distribution where the mean and variance are equal, which is not the case for most other distributions.