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Chapter 3: Q97E (page 137)

Of all customers purchasing automatic garage-door openers, 75% purchase a chain-driven model. Let \({\bf{X}}{\rm{ }}{\bf{5}}\) the number among the next \({\bf{15}}\) purchasers who select the chain-driven model.

a. What is the pmf of \({\bf{X}}\)?

b. Compute \({\bf{P}}\left( {{\bf{X}}{\rm{ }}.{\rm{ }}{\bf{10}}} \right).\)

c. Compute \({\bf{P}}\left( {{\bf{6}}{\rm{ }}\# {\rm{ }}{\bf{X}}{\rm{ }}\# {\rm{ }}{\bf{10}}} \right).\)

d. Compute \({\bf{m}}\) and s2 .

e. If the store currently has in stock \({\bf{10}}\) chain-driven models and \({\bf{8}}\) shaft-driven models, what is the probability that the requests of these 15 customers can all be met from existing stock?

Short Answer

Expert verified

(a) \({\text{X~Bin(15,0}}{\text{.75)}}\)

(b)\({\rm{ }}P\left( {X > 10} \right) = 0.686\) .

(c) \(P(6 \le X \le 10) = 0.313\).

(d) \(\mu = 11.75;{\sigma ^2} = 2.81\).

(e) \(P(7 \le X \le 10) = 0.310\).

Step by step solution

01

Concept introduction

The fields of mathematics dealing with the laws regulating random events, including the collection, analysis, interpretation, and display of numerical data, are known as probability and statistics.

02

Determine What is the pmf of \({\bf{X}}\)?

(a)

The Binomial Random Variable \(X\)has the following definition:

\(X = {\rm{the}}\,{\rm{number}}\,{\rm{of}}\,\,{\rm{S's}}\,\,{\rm{amonth}}\,{\rm{n trials }}\)

If requirements one through four on page \(117\) are met (binomial experiment).

We can see that the random variable \(X\)has a Binomial Distribution with the pmf \(b(x;n,p)\)

Theorem:

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{,{\rm{ otherwise }}}\end{array}} \right.\)

The parameters are, as we can see,

\(n = 15\)

\(p = 0.75\)

Therefore, \(X~{\mathop{\rm Bin}\nolimits} (15,0.75)\)

03

Compute \({\bf{P}}\left( {{\bf{X}}{\rm{ }}.{\rm{ }}{\bf{10}}} \right).\)

(b)

The cumulative density function of the binomial random variable \(X\) with parameters \(n\) and \(p\)is:

\(B(x;n,p) = P(X \le x) = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\)

The following is true

\(P(X > 10) = 1 - P(X \le 10) = 1 - B(10;15,0.75) = 1 - 0.314 = 0.686\)

04

Compute \({\bf{P}}\left( {{\bf{6}}{\rm{ }}\# {\rm{ }}{\bf{X}}{\rm{ }}\# {\rm{ }}{\bf{10}}} \right).\)

(c)

We have,

The following holds:

\(P(6 \le X \le 10) = B(10;15,0.75) - B(5;15,0.75) = 0.314 - 0.001 = 0.313\)

05

Compute \({\bf{m}}\) and s2 .

(d)

The following is valid for a binomial random variable \(X\) with parameters\(n\), \(p\), and \({\rm{q = 1 - p}}\).

\(E(X) = np\)

\(V(X) = np(1 - p) = npq\)

\({\sigma _X} = \sqrt {npq} \)

The following is true

\(E(X) = {\bf{\mu }} = 15 \cdot {\bf{0}}.{\bf{75}} = 11.75\)

We also have the fact that the variance is

\(V(X) = {\sigma ^2} = 15 \cdot 0.75 \cdot 0.25 = 2.81\).

06

Determine the probability

(e)

The following must be true in order to meet the requirements:

\(X \le 10\)

\(15 - X \le 8\)

Because there must be fewer than \(8\) customers who choose shaft-driven models \({\rm{(15 - X}}\)reflects the number of customers who choose shaft-driven models).

When we combine the requirements we get

\(7 \le X \le 10\)

As a result, the probability requested is

\(P(7 \le X \le 10) = B(10;15,0.75) - B(6;15,0.75)\)

\(\begin{aligned} &= 0{\rm{.314 - 0}}{\rm{.004}}\\ &= 0{\rm{.310}}\end{aligned}\).

Hence, the required probability is \({\rm{0}}{\rm{.310}}\).

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