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The distribution of the number of chocolate chips \((x)\) in a cookie is shown in the following table. Find the mean and the variance of the number of chocolate chips in a cookie. $$\begin{array}{llll}\hline x & 0 & 1 & 2 \\\\\hline P(X=x) & .01 & .03 & .05 \\\ \hline\end{array}$$ $$\begin{array}{llll}\hline \boldsymbol{x} & 3 & 4 & 5 \\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & .11 & .13 & .24 \\ \hline\end{array}$$ $$\begin{array}{llll}\hline x & 6 & 7 & 8 \\\\\hline P(X=x) & .22 & .16 & .05 \\\ \hline\end{array}$$

Short Answer

Expert verified
The mean (expected value) of the number of chocolate chips in a cookie is 4.02 and the variance is approximately 12.1.

Step by step solution

01

Calculate Expected Value (Mean)

To find the expected value (mean) of the number of chocolate chips in a cookie, use the formula for the mean of a discrete random variable: \(E(X) = \sum_{i} x_iP(x_i)\) Plug in the values and probabilities into the formula: \(E(X) = (0)(0.01) + (1)(0.03) + (2)(0.05) + (3)(0.11) + (4)(0.13) + (5)(0.24) + (6)(0.22) + (7)(0.16) + (8)(0.05)\) Now, calculate the sum: \(E(X) = 0 + 0.03 + 0.10 + 0.33 + 0.52 + 1.20 + 1.32 + 1.12 + 0.40 = 4.02\) Thus, the mean (expected value) of the number of chocolate chips in a cookie is 4.02.
02

Calculate Expected Value of the Squared Random Variable (E(X^2))

Next, find the expected value of the squared random variable, \(E(X^2)\), using the same formula for the mean of a discrete random variable: \(E(X^2) = \sum_{i} x_i^2P(x_i)\) Plug in the values and probabilities into the formula: \(E(X^2) = (0^2)(0.01) + (1^2)(0.03) + (2^2)(0.05) + (3^2)(0.11) + (4^2)(0.13) + (5^2)(0.24) + (6^2)(0.22) + (7^2)(0.16) + (8^2)(0.05)\) Now, calculate the sum: \(E(X^2) = 0 + 0.03 + 0.20 + 0.99 + 2.08 + 6.00 + 7.92 + 7.84 + 3.20 = 28.26\) Thus, the expected value of the squared random variable is 28.26.
03

Calculate Variance

Now that we have found the mean and the expected value of the squared random variable, we can calculate the variance using the formula for variance of discrete random variables: \(Var(X) = E(X^2) - (E(X))^2\) Plug in the values we found for \(E(X)\) and \(E(X^2)\): \(Var(X) = 28.26 - (4.02)^2\) Now, calculate the difference: \(Var(X) = 28.26 - 16.1604 = 12.0996\) Thus, the variance of the number of chocolate chips in a cookie is approximately 12.1.

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Key Concepts

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

Discrete Random Variables
Imagine rolling a die. The number that comes up is an example of a discrete random variable: you can list all the possible outcomes (1 through 6), and each outcome has a certain chance of occurring. In general, discrete random variables represent counts or measurements that can only take on specific values within a range.

Suppose we're talking about the chocolate chips in a cookie. If a cookie can have anywhere from 0 to 8 chocolate chips, each specific number of chips is a possible value of our discrete random variable. These potential values are not continuous; there can’t be 2.5 chips in a cookie—only whole numbers fit the frame of discrete data.
Expected Value
The expected value is a key term in probability, signifying the average outcome we expect after many trials of an experiment. For a discrete random variable, the expected value is calculated by multiplying each possible outcome by its probability and adding up all those products. This is like predicting the long-term average—how many chocolate chips you'd expect in a cookie if you had an unlimited number of cookies to sample. But remember, even if the expected value is a decimal, like 4.02 chips, a single cookie still won’t have a fractional chip; it's an average over many, many cookies.
Variance of a Random Variable
The variance tells us how spread out the values of a random variable are. It is a measure of the variability or dispersion around the mean. To calculate the variance of a random variable, you need the expected value of the variable and the expected value of the square of the variable. The variance is then found by subtracting the square of the expected value from the expected value of the squared variable.

In the cookie example, a variance of 12.1 means that the number of chocolate chips is quite variable from one cookie to another. If the variance was closer to zero, we would expect nearly every cookie to have a chocolate chip count close to the mean.
Probability Mass Function
The probability mass function (PMF) comes into play when we want to list the probabilities of each possible value of a discrete random variable. Like a recipe that tells you how much of each ingredient to add, a PMF tells you the likelihood of each possible number of chocolate chips in a cookie. In our case, the PMF is given by P(X=x), where X represents the variable (number of chips) and x is one of the possible values X can take. For example, P(X=3) is the probability that a randomly chosen cookie contains exactly three chocolate chips.

Understanding the PMF is crucial when we want to calculate metrics like mean and variance, as it clearly shows the probability of each event (or chocolate chip count) that can occur, and these form the basis for our statistical calculations.

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Most popular questions from this chapter

The management of the Cambridge Company has projected the sales of its products (in millions of dollars) for the upcoming year, with the associated probabilities shown in the following table: $$\begin{array}{lcccccc}\hline \text { Sales } & 20 & 22 & 24 & 26 & 28 & 30 \\\\\hline \text { Probability } & .05 & .10 & .35 & .30 & .15 & .05 \\\\\hline\end{array}$$ What does the management expect the sales to be next year?

A Christmas tree light has an expected life of \(200 \mathrm{hr}\) and a standard deviation of \(2 \mathrm{hr}\). a. Find a bound on the probability that one of these Christmas tree lights will last between \(190 \mathrm{hr}\) and \(210 \mathrm{hr}\). b. Suppose 150,000 of these Christmas tree lights are used by a large city as part of its Christmas decorations. Estimate the number of lights that are likely to require replacement between \(180 \mathrm{hr}\) and \(220 \mathrm{hr}\) of use.

The probability distribution of a random variable \(X\) is given. Compute the mean, variance, and standard deviation of \(X\). $$\begin{array}{lccccc}\hline \boldsymbol{x} & 430 & 480 & 520 & 565 & 580 \\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & .1 & .2 & .4 & .2 & .1 \end{array}$$

The percent of the voting age population who cast ballots in presidential elections from 1932 through 2000 are given in the following table: $$\begin{array}{llllllllll}\hline \begin{array}{l}\text { Election } \\ \text { Year }\end{array} & 1932 & 1936 & 1940 & 1944 & 1948 & 1952 & 1956 & 1960 & 1964 \\ \hline \text { Turnout, \% } & 53 & 57 & 59 & 56 & 51 & 62 & 59 & 59 & 62 \\ \hline\end{array}$$ $$\begin{array}{llllllllll}\hline \begin{array}{l}\text { Election } \\ \text { Year }\end{array} & 1968 & 1972 & 1976 & 1980 & 1984 & 1988 & 1992 & 1996 & 2000 \\\\\hline \text { Turnout } \% & 61 & 55 & 54 & 53 & 53 & 50 & 55 & 49 & 51 \\ \hline\end{array}$$ Find the mean and the standard deviation of the given data.

The percentage of Boston homicide cases solved each year from 2000 through 2006 is summarized in the following table: $$\begin{array}{lccccccc} \hline \text { Year } & 2000 & 2001 & 2002 & 2003 & 2004 & 2005 & 2006 \\ \hline \text { Percent } & 49 & 50 & 70 & 64 & 36 & 29 & 38 \\ \hline\end{array}$$ Find the average percent of Boston homicide cases solved per year for 2000 through \(2006 .\) What is the standard deviation for these data?

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