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Chapter 5: Q39E (page 229)

It is known that\({\rm{80\% }}\)of all brand A extremal hard drives work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that\({\rm{n = 15}}\)drives are randomly selected. Let\({\rm{X = }}\)the number of successes in the sample. The statistic\({\rm{X/n}}\)is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. (Hint: One possible value of\({\rm{X/n is }}{\rm{.2}}\), corresponding to\({\rm{X = 3}}\). What is the probability of this value (what kind of\({\rm{rv}}\)is\({\rm{X}}\))?)

Short Answer

Expert verified

\({\rm{X is binomial random variable}}{\rm{. }}\)

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

Obtaining the sampling distribution

Obviously, \({\rm{X}}\)is binomial random variable with parameters

\({\rm{n = 15 and p - 0}}{\rm{.8}}{\rm{. }}\)

All possible values that random variable \({\rm{X/n}}\) can take are

\(\frac{{\rm{x}}}{{{\rm{15}}}}{\rm{,}}\;\;\;{\rm{x脦 \{ 0,1,2, \ldots ,15\} }}{\rm{.}}\)

The probabilities can be calculated as

\(\begin{array}{l}{\rm{P}}\left( {\frac{{\rm{X}}}{{{\rm{15}}}}{\rm{ = }}\frac{{\rm{x}}}{{{\rm{15}}}}} \right)\rm &= P(X = x)\\\rm &= b(x;15,0{\rm{.8),}}\;\;\;{\rm{x脦 \{ 0,1,2, \ldots ,15\} }}\end{array}\)

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

The following table represents the pmf of random variable \({\rm{X/n}}\)

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Most popular questions from this chapter

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \({\rm{\bar X}}\) when the population distribution is lognormal with \({\rm{E(ln(X)) = 3}}\) and\({\rm{V(ln(X)) = 1}}\). Consider the four sample sizes\({\rm{n = 10,20,30}}\), and\({\rm{50}}\), and in each case use \({\rm{1000}}\) replications. For which of these sample sizes does the \({\rm{\bar X}}\) sampling distribution appear to be approximately normal?

Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv鈥檚 with expected values \({\mu _1}, {\mu _2}, and {\mu _3}\)and variances \(\sigma _1^2 , \sigma _2^2, and \sigma _3^2 \), respectively. a. If \(\mu = {\mu _2} = {\mu _3} = 60\)and\(\sigma _1^2 = \sigma _2^2 = \sigma _3^2 = 15\), calculate \(P\left( {{T_0} \le 200} \right)\)and\(P\left( {150 \le {T_0} \le 200} \right)\)? b. Using the \(\mu 's and \sigma 's\)given in part (a), calculate both \(P\left( {55 \le X} \right)\)and \(P\left( {58 \le X \le 62} \right)\).c. Using the \(\mu 's and \sigma 's\)given in part (a), calculate and interpret\(P\left( { - 10 \le {X_1} - .5{X_2} - .5{X_3} \le 5} \right)\). d. If\({\mu _1} = 40, {\mu _1} = 50, {\mu _1} = 60,\),\( \sigma _1^2 = 10, \sigma _2^2 = 12, and \sigma _3^2 = 14\) calculate \(P\left( {{X_1} + {X_2} + {X_3} \le 160} \right)\)and also \(P\left( {{X_1} + {X_2} \ge 2{X_3}} \right).\)

Suppose that when the pH of a certain chemical compound is\(5.00\), the pH measured by a randomly selected beginning chemistry student is a random variable with a mean of\(5.00\)and a standard deviation .2. A large batch of the compound is subdivided and a sample is given to each student in a morning lab and each student in an afternoon lab. Let\(X = \)the average pH as determined by the morning students and\(Y = \)the average pH as determined by the afternoon students.

a. If pH is a normal variable and there are\(25\)students in each lab, compute\(P\left( { - .1 \le X - Y \le - .1} \right)\)

b. If there are\(36\)students in each lab, but pH determinations are not assumed normal, calculate (approximately)\(P\left( { - .1 \le X - Y \le - .1} \right)\).

Question: The number of customers waiting for gift-wrap service at a department store is an rv X with possible values \({\rm{0,1,2,3,4}}\)and corresponding probabilities \({\rm{.1,}}{\rm{.2,}}{\rm{.3,}}{\rm{.25,}}{\rm{.15}}{\rm{.}}\)A randomly selected customer will have \({\rm{1,2}}\),or \({\rm{3}}\) packages for wrapping with probabilities \({\rm{.6,}}{\rm{.3,}}\)and \({\rm{.1,}}\)respectively. Let \({\rm{Y = }}\)the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer).

a. Determine \({\rm{P(X = 3,Y = 3)}}\), i.e., \({\rm{P(3,3)}}\).

b. Determine \({\rm{p(4,11)}}\).

The difference between the number of customers in line at the express checkout and the number in line at the super-express checkout is\({{\rm{X}}_{\rm{1}}}{\rm{ - }}{{\rm{X}}_{\rm{2}}}\). Calculate the expected difference.
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