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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 for this probability distribution. Give a brief interpretation of The value of the mean.

Short Answer

Expert verified
The mean of this probability distribution is 2.98 and the standard deviation is 1.41. The mean value implies that the electronics store can expect to sell approximately 3 camcorders each day on average.

Step by step solution

01

Calculate the mean

The formula for the mean (µ) of a probability distribution is: \(µ = Σ[X * P(X)]\), where X is the value of the random variable and P(X) is the probability of that value. Apply that to the given probabilities and camcorders sold: \(µ = 0*.05 + 1*.12 + 2*.19 + 3*.30 + 4*.20 + 5*.10 + 6*.04 = 2.98\)
02

Calculate the standard deviation

The formula for the standard deviation (σ) is: \(σ = √Σ[(X-µ)^2 * P(X)]\), where µ is the mean calculated in Step 1. Apply the formula: \(σ = √[(0-2.98)^2*.05 + (1-2.98)^2*.12 + (2-2.98)^2*.19 + (3-2.98)^2*.30 + (4-2.98)^2*.20 + (5-2.98)^2*.10 + (6-2.98)^2*.04]= 1.41\)
03

Interpret the mean

The value of the mean (2.98) represents the expected number of camcorders sold in a day. This value is the 'typical' or 'average' number of camcorders the electronics store can expect to sell each day. It's important to note that because this is an average, it doesn't mean the store will sell exactly 2.98 camcorders every day. Some days they might sell more; other days less. But on average, over many days, they can expect to sell approximately 3 camcorders.

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Key Concepts

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

Mean of a Probability Distribution
The mean of a probability distribution, often denoted as \( \mu \), is essentially the "center" of the distribution. It is calculated by taking a weighted average of all possible values a random variable can take, where each value is multiplied by its probability. This provides us with the expected value or "expectation" of that random variable.
The formula to compute the mean \( \mu \) of a discrete probability distribution is:
  • \( \mu = \sum [X \times P(X)] \)
Where \( X \) is a possible value of the random variable, and \( P(X) \) is the probability of \( X \) occurring. This formula tells us how much "weight" each value has when calculating the mean.
In practical terms, if we consider the camcorders example, the mean tells us about the typical number of camcorders sold each day. It's the balancing point of the distribution, indicating what number can be expected on average if the sales are observed over many days. In our example, this calculated mean was 2.98, suggesting that, on average, close to 3 camcorders are expected to be sold daily.
Standard Deviation of a Probability Distribution
The standard deviation in probability distributions measures how spread out the possible values are around the mean. It provides insight into the variability or dispersion of the data. Lower standard deviation indicates that the values tend to be closer to the mean, while a higher standard deviation signifies that the values are more spread out.
The formula for calculating the standard deviation \( \sigma \) of a probability distribution is:
  • \( \sigma = \sqrt{\sum [(X-\mu)^2 \times P(X)]} \)
Where \( X \) represents the values of the random variable, \( \mu \) is the mean, and \( P(X) \) is the probability of \( X \). This formula calculates the average squared distance of each point from the mean, taking into account the probability of each point.
In our exercise, the standard deviation was calculated to be 1.41. This tells us how much the number of camcorders sold deviates from the mean of 2.98. The value 1.41 indicates that while some days might see sales significantly higher or lower than the mean, most days would see sales numbers falling near to the average.
Interpretation of Mean in Probability
Understanding the interpretation of the mean in a probability distribution is crucial because it provides a summary of the entire dataset in a single value. The mean can be thought of as the "expected" value of the data by expressing what one anticipates as a general outcome over the long term.
In the context of our given problem with camcorder sales, the mean of 2.98 signifies the expected number of camcorders sold on any given day at the store. However, it's important to recognize that due to the random nature of sales, you are unlikely to sell exactly 2.98 camcorders on any particular day. Instead, this figure serves as an average over a long period of observations. Some days, the actual sales will be more, and on other days, less, but across many days, the average should be close to this mean value.
Thus, the mean provides both a simple snapshot of data behavior and serves as a predictive tool, allowing businesses to make informed decisions regarding inventory and sales expectations based on past variability and average performance.

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Most popular questions from this chapter

Let \(x\) be the number of heads obtained in two tosses of a coin. The following table lists the probability distribution of \(x\). $$ \begin{array}{l|lll} \hline x & 0 & 1 & 2 \\ \hline P(x) & .25 & .50 & .25 \\ \hline \end{array} $$ Calculate the mean and standard deviation of \(x\). Give a brief interpretation of the value of the mean.

Based on its analysis of the future demand for its products, the financial department at Tipper Corporation has determined that there is a 17 probability that the company will lose \(\$ 1.2\) million during the next year, a \(.21\) probability that it will lose \(\$ .7\) million, a \(.37\) probability that it will make a profit of \(\$ .9\) million, and a \(.25\) probability that it will make a profit of \(\$ 2.3\) million. a. Let \(x\) be a random variable that denotes the profit earned by this corporation during the next year. Write the probability distribution of \(x\). b. Find the mean and standard deviation of the probability distribution of part a. Give a brief interpretation of the value of the mean.

According to a survey, \(30 \%\) of adults are against using animals for research. Assume that this result holds true for the current population of all adults. Let \(x\) be the number of adults who are against using animals for research in a random sample of two adults. Obtain the probability distribution of \(x\). Draw a tree diagram for this problem.

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Briefly explain the two characteristics (conditions) of the probability distribution of a discrete random variable.

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