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Chapter 5: Q20E (page 212)

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{,}}{{\rm{X}}_{\rm{4}}}{\rm{,}}{{\rm{X}}_{\rm{5}}}\), and \({{\rm{X}}_{\rm{6}}}\) denote the numbers of blue, brown, green, orange, red, and yellow M\&M candies, respectively, in a sample of size\({\rm{n}}\). Then these \({{\rm{X}}_{\rm{i}}}\) 's have a multinomial distribution. According to the M\&M Web site, the color proportions are\({{\rm{p}}_{\rm{1}}}{\rm{ = }}{\rm{.24,}}{{\rm{p}}_{\rm{2}}}{\rm{ = }}{\rm{.13}}\), \({{\rm{p}}_{\rm{3}}}{\rm{ = }}{\rm{.16,}}{{\rm{p}}_{\rm{4}}}{\rm{ = }}{\rm{.20,}}{{\rm{p}}_{\rm{5}}}{\rm{ = }}{\rm{.13}}\), and\({{\rm{p}}_{\rm{6}}}{\rm{ = }}{\rm{.14}}\).

a. If\({\rm{n = 12}}\), what is the probability that there are exactly two M\&Ms of each color?

b. For\({\rm{n = 20}}\), what is the probability that there are at most five orange candies? (Hint: Think of an orange candy as a success and any other color as a failure.)

c. In a sample of\({\rm{20M \backslash Ms}}\), what is the probability that the number of candies that are blue, green, or orange is at least \({\rm{10}}\) ?

Short Answer

Expert verified

a. The probability is \({\rm{0}}{\rm{.00213}}\).

b. The probability is \({\rm{0}}{\rm{.8042}}\).

c. The probability is \(0.87248.\)

Step by step solution

01

Definition

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

02

Calculating the probability

(a):

The probability mass function of multinomial distribution is

\(\begin{aligned}{{\rm{f}}_{{\rm{XYZ}}}}{\rm{(x,y,z) = P(X = x,Y = y,Z = z)}}\\{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{x!y!z!}}}}{\rm{p}}_{\rm{1}}^{\rm{x}}{\rm{p}}_{\rm{2}}^{\rm{y}}{\rm{p}}_{\rm{3}}^{\rm{z}}{\rm{,}}\end{aligned}\)

where, analogously, this can be written in the same manner for given \({\rm{6}}\) random variables.

Given the probabilities

\(\begin{aligned}{*{20}{l}}{{{\rm{p}}_{\rm{1}}}{\rm{ = 0}}{\rm{.24,}}}&{{{\rm{p}}_{\rm{2}}}{\rm{ = 0}}{\rm{.13,}}}&{{{\rm{p}}_{\rm{3}}}{\rm{ = 0}}{\rm{.16,}}}\\{{{\rm{p}}_{\rm{4}}}{\rm{ = 0}}{\rm{.2,}}}&{{{\rm{p}}_{\rm{5}}}{\rm{ = 0}}{\rm{.13,}}}&{{{\rm{p}}_{\rm{6}}}{\rm{ = 0}}{\rm{.13,}}}\end{aligned}\)

where they sum to\({\rm{1}}\), the requested probability can be computed as follows

\({{\rm{f}}_{\rm{X}}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}{\rm{,}}{{\rm{x}}_{\rm{4}}}{\rm{,}}{{\rm{x}}_{\rm{5}}}{\rm{,}}{{\rm{x}}_{\rm{6}}}} \right){\rm{ = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}{\rm{,}}{{\rm{X}}_{\rm{4}}}{\rm{ = }}{{\rm{x}}_{\rm{4}}}{\rm{,}}{{\rm{X}}_{\rm{5}}}{\rm{ = }}{{\rm{x}}_{\rm{5}}}{\rm{,}}{{\rm{X}}_{\rm{6}}}{\rm{ = }}{{\rm{x}}_{\rm{6}}}} \right)\)

for \({{\rm{x}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{ = \ldots = }}{{\rm{x}}_{\rm{6}}}{\rm{ = 2}}\) is

\(\begin{aligned}{{\rm{f}}_{\rm{X}}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}{\rm{,}}{{\rm{x}}_{\rm{4}}}{\rm{,}}{{\rm{x}}_{\rm{5}}}{\rm{,}}{{\rm{x}}_{\rm{6}}}} \right)\\{\rm{ = }}\frac{{{\rm{12!}}}}{{{\rm{2!2!2!2!2!2!}}}}{\rm{ \times 0}}{\rm{.2}}{{\rm{4}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.1}}{{\rm{3}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.1}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.1}}{{\rm{3}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.1}}{{\rm{3}}^{\rm{2}}}{\rm{ = 0}}{\rm{.00213}}{\rm{.}}\end{aligned}\)

03

Calculating the probability

(b):

Binomial distribution with parameters \({\rm{n}}\) and \({\rm{p(q = 1 - p)}}\) has probability mass function

\({\rm{P(X = x) = }}\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{\rm{q}}^{{\rm{n - x}}}}{\rm{,}}\;\;\;{\rm{xI \{ 0,1,2, \ldots n\} }}{\rm{.}}\)

The parameter \({\rm{p}}\) is the probability of success, therefore

\({{\rm{p}}_{\rm{4}}}{\rm{ = 0}}{\rm{.2}}\)

which means that

\({\rm{q = 1 - 0}}{\rm{.2 = 0}}{\rm{.8}}\)

The parameter \({\rm{n}}\) is 20. Hence, the requested probability is

\({\rm{P(X拢5) = }}\sum\limits_{{\rm{x = 0}}}^{\rm{5}} {\left( {\begin{aligned}{*{20}{c}}{{\rm{20}}}\\{\rm{x}}\end{aligned}} \right)} {\rm{0}}{\rm{.}}{{\rm{2}}^{\rm{x}}}{\rm{ \times 0}}{\rm{.}}{{\rm{8}}^{{\rm{20 - x}}}}{\rm{ = 0}}{\rm{.8042}}\)

04

Calculating the probability

(c):

Look at the blue, green, and orange candies as a success, and all other colors as failure. In this case the parameter \({\rm{p}}\) is

\({\rm{p = }}{{\rm{p}}_{\rm{1}}}{\rm{ + }}{{\rm{p}}_{\rm{3}}}{\rm{ + }}{{\rm{p}}_{\rm{4}}}{\rm{ = 0}}{\rm{.24 + 0}}{\rm{.16 + 0}}{\rm{.2 = 0}}{\rm{.6}}\)

which means that

\({\rm{q = 1 - 0}}{\rm{.6 = 0}}{\rm{.4}}\)

The parameter \({\rm{n}}\) is \({\rm{20}}\) . Hence, the requested probability is

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Most popular questions from this chapter

Suppose the expected tensile strength of type-A steel is \({\rm{105ksi}}\)and the standard deviation of tensile strength is \({\rm{8ksi}}\). For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are \({\rm{100ksi}}\)and \({\rm{6ksi}}\), respectively. Let \({\rm{\bar X = }}\)the sample average tensile strength of a random sample of \({\rm{40}}\) type-A specimens, and let \({\rm{\bar Y = }}\)the sample average tensile strength of a random sample of \({\rm{35}}\)type-B specimens.

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Consider a random sample of size n from a continuous distribution having median \({\rm{0}}\)so that the probability of any one observation being positive is \({\rm{.5}}\). Disregarding the signs of the observations, rank them from smallest to largest in absolute value, and let \({\rm{w = }}\)the sum of the ranks of the observations having positive signs. For example, if the observations are \({\rm{ - }}{\rm{.3, + }}{\rm{.7, + 2}}{\rm{.1 and - 2}}{\rm{.5 }}\) then the ranks of positive observations are \({\rm{ 2and 3 }}\), so \({\rm{w = 5}}\). In Chapter \({\rm{15}}\), W will be called Wilcoxon鈥檚 signed-rank statistic. \({\rm{w}}\)can be represented as follows:

where the Yi 鈥檚 are independent Bernoulli rv鈥檚, each with \({\rm{ p = }}{\rm{.5 }}\) (\({{\rm{Y}}_{\rm{1}}}{\rm{ = 1}}\)corresponds to the observation with rank \({\rm{i }}\)being positive)

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Sixty percent of all customers who purchase these cameras also buy an extended warranty. Let \({\rm{Y}}\) denote the number of purchasers during this week who buy an extended warranty.

a. What is\({\rm{P(X = 4,Y = 2)}}\)? (Hint: This probability equals\({\rm{P(Y = 2}}\mid {\rm{X = 4) \times P(X = 4)}}\); now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty.)

b. Calculate\({\rm{P(X = Y)}}\).

c. Determine the joint pmf of \({\rm{X}}\) and \({\rm{Y}}\)then the marginal pmf of\({\rm{Y}}\).

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a. Is it plausible that the waist size distribution is at least approximately normal? Explain your reasoning. If your answer is no, conjecture the shape of the population distribution.

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