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The following table gives the probability distribution of the number of camcorders sold on a given day at an electronics store. $$ \begin{array}{l|ccccccc} \hline \text { Camcorders sold } & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Probability } & .05 & .12 & .19 & .30 & .20 & .10 & .04 \\ \hline \end{array} $$ Calculate the mean and standard deviation of this probability distribution. Give a brief interpretation of the value of the mean.

Short Answer

Expert verified
The mean or expected value of the distribution represents the average number of camcorders sold per day. Standard deviation gives an idea of how much the actual sales figures can deviate from this average.

Step by step solution

01

Calculating the Mean

The mean, or expected value, is calculated for a discrete random variable by summing the product of each outcome and its corresponding probability. In this case, we multiply the number of camcorders by their respective probability and add them up i.e.\[ \text{Mean}=\sum_{i=0}^{6} X_i*P(X_i) \]Or,\[ \text{Mean} = 0*.05 + 1*.12 + 2*.19 + 3*.30 + 4*.20 + 5*.10 + 6*.04 \]
02

Calculating the Standard Deviation

First we calculate the variance, which is the average of the squared differences from the mean. Variance is calculated as follows\[ \text{Variance}=\sigma^2=\sum_{i=0}^{6} (X_i - \text{mean})^2*P(X_i) \]Substitute the value of mean from step 1 and then evaluate. The standard deviation is the square root of the variance.
03

Interpretation of the Mean

The value of the mean represents the expected, or average, number of camcorders sold per day. This is the value around which individual sales numbers distribute or in other words, the electronics store can expect to sell this many camcorders on an average day.

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Key Concepts

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

Mean
The mean of a probability distribution, also known as the expected value, is a central concept in statistics. It gives us a single number that represents the average outcome of a random process. In the context of our exercise, the mean tells us how many camcorders the electronics store can expect to sell on any given day.

To compute the mean, we multiply each possible number of camcorders sold (discrete outcomes) by its probability and then sum them up. This provides a weighted average where each outcome's weight is its probability. For example, in our problem, the mean is calculated using the formula:

\[ \text{Mean} = \sum_{i=0}^{6} X_i \times P(X_i) \]

Where \(X_i\) represents the number of camcorders sold, and \(P(X_i)\) is the probability of selling \(X_i\) camcorders. This approach ensures that outcomes that are more likely have a bigger impact on the mean.
Standard Deviation
The standard deviation gives us a measure of how much each possible outcome deviates from the mean. It is a critical aspect of understanding how data is distributed around the mean.

To find the standard deviation, we first need to calculate the variance. Variance tells us the average of the squared deviations from the mean, providing insight into the data's spread. The formula to calculate variance is:

\[ \sigma^2 = \sum_{i=0}^{6} (X_i - \text{mean})^2 \times P(X_i) \]

Once we have the variance, the standard deviation is simply the square root of this value:

\[ \sigma = \sqrt{\sigma^2} \]

By knowing the standard deviation, we can understand the consistency of camcorder sales. A smaller standard deviation indicates that sales figures are closely packed around the mean, while a larger one suggests higher variability.
Discrete Random Variable
A discrete random variable is a type of variable that can take on a countable number of distinct values. In our exercise, the number of camcorders sold each day is a discrete random variable because it can only take integer values such as 0, 1, 2, etc. It cannot take fractional values like 2.5.

Discrete random variables are often accompanied by a probability distribution, which specifies the probabilities of each potential value. The distribution must satisfy two conditions:
  • The probabilities of all possible outcomes must sum up to 1.
  • Each probability must be between 0 and 1, inclusive.
When working with discrete random variables, we can calculate important statistical measures like the mean, variance, and standard deviation to better understand the nature of the data.
Variance
Variance is a fundamental concept in statistics that describes how far the values of a random variable spread out from the mean. It gives us an idea of the data's variability.

In our probability distribution of camcorders sold, variance is calculated using the formula:

\[ \sigma^2 = \sum_{i=0}^{6} (X_i - \text{mean})^2 \times P(X_i) \]

Here each outcome \((X_i)\) is subtracted from the mean, the difference is squared, and then multiplied by the probability of that outcome. This results in all deviations being converted into positive values, which are weight-adjusted by their probability.

The final sum gives an average of these squared deviations, showing us how much variation exists around the mean. A small variance indicates that the values are closer to the mean, suggesting predictable and consistent sales. In contrast, a large variance points towards a broader range of sales figures.

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Most popular questions from this chapter

Many of you probably played the game "Rock, Paper, Scissors" as a child. Consider the following variation of that game. Instead of two players, suppose three players play this game, and let us call these players \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). Each player selects one of these three items- Rock, Paper, or Scissors-independent of each other. Player A will win the game if all three players select the same item, for example, rock. Player B will win the game if exactly two of the three players select the same item and the third player selects a different item. Player \(\mathrm{C}\) will win the game if every player selects a different item. If Player B wins the game, he or she will be paid \(\$ 1\). If Player \(\mathrm{C}\) wins the game, he or she will be paid \(\$ 3\). Assuming that the expected winnings should be the same for each player to make this a fair game, how much should Player A be paid if he or she wins the game?

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