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Let \(X\) equal the duration (in min) of a telephone call that is received between midnight and noon and reported. The following times were reported 3.2, \(0.4,1.8,0.2,2.8,1.9,2.7,0.8,1.1,0.5,1.9,2,0.5,2.8,1.2,1.5,0.7,1.5,2.8,1.2\) Draw a probability histogram for the exponential distribution and a relative frequency histogram of the data on the same graph.

Short Answer

Expert verified
The exponential distribution and the relative frequency histogram of the given phone call durations data are plotted on the same graph to provide a visual comparison. The lambda parameter of the exponential distribution is calculated as the reciprocal of the mean duration of the phone calls.

Step by step solution

01

Calculating lambda parameter

The parameter lambda for the Exponential distribution is derived as the reciprocal of the mean of the data. First, calculate the mean duration of the phone calls. Sum up all the reported times and then divide by the total number of reported times. The mean (mu) is given as: \[ mu = \frac{\sum_{i=1}^{n}X_i}{n} \]. After calculating the mean, compute lambda as: \[ lambda = \frac{1}{mu} \].
02

Plotting the exponential distribution

Now that the lambda parameter is known, the Exponential distribution can be plotted. Use the probability density function (pdf) of the Exponential distribution for plotting. The pdf of an Exponential distribution is given as: \[ f(x; lambda) = lambda * e^{-lambda * x} \] for \( x >= 0 \). Plot this function for the range of \( x \) that corresponds to the durations of the telephone calls.
03

Plotting the relative frequency histogram

Next, plot a histogram of the reported times. The heights of the bars of the histogram should correspond to the relative frequencies of the data instead of the absolute frequency. The relative frequency of a value is the absolute frequency of that value divided by the total number of data points.
04

Combining the plots

Now, combine the Exponential distribution plot from Step 2 and the relative frequency histogram from Step 3 onto the same graph. Use two different colors or styles of lines/bars to distinguish between the Exponential distribution and the histogram of the data. This will provide a visual comparison of how well the Exponential distribution fits the actual data.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Histogram
A probability histogram is a graphical representation used for visualizing the probability distribution of a continuous random variable. In this context, it's particularly useful for illustrating the nature of an exponential distribution.
The exponential distribution is characterized by a single parameter, known as lambda, which describes the rate of occurrences. In a probability histogram, the heights of the bars represent the probability of the occurrence of each interval of values. This makes it easier to see the relation between different outcomes, providing a clear picture of the likelihood of various durations of phone calls between midnight and noon.
Creating a probability histogram involves calculating the probability of each range and plotting it. For exponential distributions, which model the time until an event, like a phone call, the probability histogram provides insight into what durations are most likely. Remember to label the axes clearly and choose appropriate intervals to ensure proper representation.
Relative Frequency Histogram
A relative frequency histogram is another important tool for data analysis. Instead of showing probabilities, this histogram illustrates how often each interval of values appears in a dataset compared to the total number of observations.
For the given telephone call durations, you would calculate the relative frequency by dividing the number of occurrences of a certain duration by the total number of calls. This gives a proportion that indicates how common different call durations are in the dataset.
When plotting the relative frequency histogram, bars are drawn with heights corresponding to these relative frequencies. It's a helpful way to normalize data, allowing comparison across datasets of different sizes. Combining this with the probability histogram on the same graph lets you visually assess how well the theoretical exponential distribution fits the actual data.
Lambda Parameter
The lambda parameter is a key component of the exponential distribution. It represents the rate at which events occur. A higher lambda value indicates that events happen quickly and frequently, while a lower value means they occur less frequently.
To find lambda, you need to calculate the mean () of the data, which is simply the sum of all reported times divided by the total number of reports. Once the mean is obtained, is the reciprocal: = \[ \frac{1}{\mu} \].
With lambda calculated, it can be plugged into the probability density function, \( f(x; \lambda) = \lambda \cdot e^{-\lambda \cdot x} \), to describe the exponential distribution. This formula gives you the likelihood of different durations for a phone call and is used to generate the probability histogram, helping visualize the exponential decay pattern typical of these scenarios.

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Most popular questions from this chapter

Suppose a state government wants to analyze the number of children in families for improving their immunization program. They analyze a group of 200 families and report their findings in the form of a frequency distribution shown in Table \(3.16\) 1\. Draw the bar chart for the following data and calculate the total number of children. 2\. Calculate mean, mode, and median of the data. 3\. Calculate coefficient of kurtosis and coefficient of skewness in the above data. $$ \begin{array}{l} \text { Table 3.16 Frequency table for Problem } 3.8\\\ \begin{array}{l|r|r|r|r|r|l|l|l} \hline \text { No. of children } & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { No. of families } & 18 & 30 & 72 & 43 & 25 & 8 & 3 & 1 \\ \hline \end{array} \end{array} $$

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 \(\lambda\).

A sample of 10 claims in an insurance company had mean and variance of 5,478 and 1,723 , respectively. On reconciliation, it was found that one claim of 3,250 was wrongly written as \(4,250 .\) Calculate the mean and standard deviation of the sample with correct values.

A mobile phone company examines the ages of 150 customers to start special plans for them. Consider frequency table shown in Table \(3.15\). 1\. Draw the histogram for the data. 2\. Estimate the mean age for these policyholders. 3\. Estimate the median age for these policyholders. $$ \begin{array}{l} \text { Table 3.15 Frequency table for Problem } 3.6\\\ \begin{array}{l|l|l|l|l|l|l} \hline \text { Age(years) } & 0-14 & 15-19 & 20-29 & 30-39 & 40-49 & 50-79 \\ \hline \text { Frequency } & 14 & 40 & 28 & 27 & 24 & 17 \\ \hline \end{array} \end{array} $$

An insurance company wants to analyze the claims for damage due to fire on its household content's policies. The values for a sample of 50 claims in Rupees are shown in Table \(3.17\).$$ \begin{array}{l} \text { Table 3.17 Data for Problem } 3.9\\\ \begin{array}{l|l|l|l|l|l|l|l|l|l} \hline 57000 & 115000 & 119000 & 131000 & 152000 & 167000 & 188000 & 190000 & 197000 & 201000 \\ \hline 206000 & 209000 & 213000 & 217000 & 221000 & 229000 & 247000 & 250000 & 252000 & 253000 \\ \hline 257000 & 257000 & 258000 & 259000 & 260000 & 261000 & 262000 & 263000 & 267000 & 271000 \\ \hline 277000 & 285000 & 287000 & 305000 & 307000 & 309000 & 311000 & 313000 & 317000 & 321000 \\ \hline 322000 & 327000 & 333000 & 351000 & 357000 & 371000 & 399000 & 417000 & 433000 & 499000 \\ \hline \end{array} \end{array} $$ $$ \begin{array}{l} \text { Table } 3.18 \text { Grouped frequency distribution of the data given in Table } 3.17\\\ \begin{array}{l|l|l|l|l|l|l|l|l|l|} \hline \begin{array}{l} \text { Claim size } \\ \text { (In 1000's } \\ \text { of Rupees) } \end{array} & 50-99 & 100-149 & 150-199 & 200-249 & 250-299 & 300-349 & 350-399 & 400-449 & 450-500 \\ \hline \text { Frequency } & 1 & 3 & 5 & 8 & 16 & 10 & 4 & 2 & 1 \\ \hline \end{array} \end{array} $$ 1\. What is the range of the above data? 2\. Draw a bar graph for Table \(3.18\). 3\. For the data given in Table \(3.18\) if instead of equal-sized groups, we had a single group for all value below 250 , how would this bar be represented? 4\. Calculate the mean, median, mode, and sample geometric mean. 5\. Calculate the sample standard deviation and sample variance. 6\. Calculate the coefficient of variation. Table \(3.18\) displays the grouped frequency distribution for the considered data.

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