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Chapter 5: Q54E (page 237)

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean \({\rm{2}}{\rm{.65 }}\)and standard deviation \({\rm{.85}}\) (suggested in 鈥淢odeling Sediment and Water Column Interactions for Hydrophobic Pollutants,鈥� Water Research, \({\rm{1984: 1164 - 1174 }}\)).

a. If a random sample of \({\rm{25}}\)specimens is selected, what is the probability that the sample average sediment density is at most \({\rm{3}}{\rm{.00 }}\)? Between \({\rm{2}}{\rm{.65 }}\)and \({\rm{3}}{\rm{.00 }}\)?

b. How large a sample size would be required to ensure that the first probability in part (a) is at least \({\rm{.99}}\)?

Short Answer

Expert verified

a. \({\rm{P(\bar X 拢 3) = 0}}{\rm{.9803;P(2}}{\rm{.65拢 \bar X拢 3) = 0}}{\rm{.4803;}}\)

b. \(n = 33.{\rm{\;}}\)

Step by step solution

01

definition of probability

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Determining the probability that the sample average sediment density is at most \({\rm{3}}{\rm{.00}}\) ? Between \({\rm{2}}{\rm{.65 and 3}}{\rm{.00}}\)

Assume \({\rm{X}}\)is a regularly distributed random variable with a mean of \({\rm{\mu = 2}}{\rm{.65}}\)and a standard deviation of \({\rm{\sigma = 0}}{\rm{.8}}\)

The sample average \({\rm{\bar X}}\)mean value is presented with

\({{\rm{\mu }}_{{\rm{\bar X}}}}{\rm{ = \mu = 2}}{\rm{.65}}\)

and the sample average \({\rm{\bar X}}\)standard deviation \({{\rm{\sigma }}_{\overline {{\rm{\bar y}}} }}\)is supplied with

\({{\rm{\sigma }}_{{\rm{\bar X}}}}{\rm{ = }}\frac{{\rm{1}}}{{\sqrt {\rm{n}} }}{\rm{ \times \sigma = }}\frac{{\rm{1}}}{{\sqrt {{\rm{25}}} }}{\rm{ \times 0}}{\rm{.85 = 0}}{\rm{.17}}\)

The likelihood of the event \({\rm{\{ \bar X拢 3\} }}\)is

\({\rm{P(\bar X拢 3) = P}}\left( {\frac{{{\rm{\bar X - \mu \bar X}}}}{{{{\rm{\sigma }}_{{\rm{\bar X}}}}}}{\rm{拢 }}\frac{{{\rm{3 - 2}}{\rm{.65}}}}{{{\rm{0}}{\rm{.17}}}}} \right){\rm{ = P(Z拢 2}}{\rm{.06)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0}}{\rm{.9803}}\)

(1): from the appendix's normal probability table. Software can also be used to calculate the likelihood.

The likelihood of the event \({\rm{\{ 2}}{\rm{.65拢 \bar X拢 3\} }}\)is

\(\begin{aligned}P(2.65拢\bar X拢3) &= P(\bar X拢3) - P(\bar X拢2.65) \\ &= P(Z拢2.06) - P(Z拢0)\\ &=0.4803 \end{aligned}\)

(2): from the appendix's normal probability table. Software can also be used to calculate the likelihood.

03

Determining the first probability in part (a) is at least \({\rm{.99}}\)

Relation can be used to calculate the number \({\rm{n}}\) (sample size).

\({\rm{P(\bar X拢 3) = 0}}{\rm{.99}}\)

The following statement is correct:

\({\rm{P(\bar X拢 3) = P}}\left( {\frac{{{\rm{\bar X - }}{{\rm{\mu }}_{{\rm{\bar X}}}}}}{{{{\rm{\sigma }}_{{\rm{\bar X}}}}}}{\rm{拢 }}\frac{{{\rm{3 - 2}}{\rm{.65}}}}{{{\rm{0}}{\rm{.85/}}\sqrt {\rm{n}} }}} \right){\rm{ = P}}\left( {{\rm{Z拢 }}\frac{{{\rm{0}}{\rm{.35}}}}{{{\rm{0}}{\rm{.85/}}\sqrt {\rm{n}} }}} \right)\)

Consequently, because

\({\rm{P}}\left( {{\rm{Z拢 }}\frac{{{\rm{0}}{\rm{.35}}}}{{{\rm{0}}{\rm{.85/}}\sqrt {\rm{n}} }}} \right){\rm{ = 0}}{\rm{.99}}\)

This indicates that, according to the normal probability table in the appendix,

\(\frac{{{\rm{0}}{\rm{.35}}}}{{{\rm{0}}{\rm{.85/}}\sqrt {\rm{n}} }}{\rm{ = 2}}{\rm{.33}}\)

This is due to the fact that the probability of the event \({\rm{\{ Z拢 2}}{\rm{.33\} }}\)is \({\rm{0}}{\rm{.99}}\)Hence,

\(\begin{array}{*{20}{c}}{\frac{{{\rm{0}}{\rm{.35}}}}{{{\rm{0}}{\rm{.85/}}\sqrt {\rm{n}} }}{\rm{ = 2}}{\rm{.33}}}\\{{\rm{n = 32}}{\rm{.02}}{\rm{.}}}\end{array}\)

Finally, the sample size for which the probability \({\rm{P(\bar X拢 3)}}\) at least \({\rm{0}}{\rm{.99}}\) is

\({\rm{n = 33}}{\rm{.}}\)

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Most popular questions from this chapter

Suppose the distribution of the time X (in hours) spent by students at a certain university on a particular project is gamma with parameters \(\alpha = 50\)and\(\beta = 2\). Because a is large, it can be shown that X has an approximately normal distribution. Use this fact to compute the approximate probability that a randomly selected student spends at most \(125\) hours on the project.

Two different professors have just submitted final exams for duplication. Let \({\rm{X}}\) denote the number of typographical errors on the first professor鈥檚 exam and \({\rm{Y}}\) denote the number of such errors on the second exam. Suppose \({\rm{X}}\) has a Poisson distribution with parameter \({{\rm{\mu }}_{\rm{1}}}\), \({\rm{Y}}\) has a Poisson distribution with parameter \({{\rm{\mu }}_{\rm{2}}}\), and \({\rm{X}}\) and \({\rm{Y}}\) are independent.

a. What is the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\)?

b. What is the probability that at most one error is made on both exams combined?

c. Obtain a general expression for the probability that the total number of errors in the two exams is m (where \({\rm{m}}\) is a nonnegative integer). (Hint: \({\rm{A = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:x + y = m}}} \right\}{\rm{ = }}\left\{ {\left( {{\rm{m,0}}} \right)\left( {{\rm{m - 1,1}}} \right){\rm{,}}.....{\rm{(1,m - 1),(0,m)}}} \right\}\)Now sum the joint pmf over \({\rm{(x,y)}} \in {\rm{A}}\)and use the binomial theorem, which says that

\({\rm{P(X + Y = m)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\sum\limits_{{\rm{k = 0}}}^{\rm{m}} {\left( {\begin{array}{*{20}{c}}{\rm{m}}\\{\rm{k}}\end{array}} \right){{\rm{a}}^{\rm{k}}}{{\rm{b}}^{{\rm{m - k}}}}{\rm{ = }}\left( {{\rm{a + b}}} \right)} ^{\rm{m}}}\)

A binary communication channel transmits a sequence of 鈥渂its鈥� (\({\bf{0s}}{\rm{ }}{\bf{and}}{\rm{ }}{\bf{1s}}\)). Suppose that for any particular bit transmitted, there is a\({\bf{10}}\% \)chance of a transmission error (a zero becoming a one or a one becoming a zero). Assume that bit errors occur independently of one another.

a. Consider transmitting\(1000\)bits. What is the approximate probability that at most\(125\)transmission errors occur?

b. Suppose the same\(1000\)-bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within\(50\)of the number of errors in the second?

There are \({\rm{40}}\) students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time needed to grade a randomly chosen first examination paper is a random variable with an expected value of \({\rm{6}}\)min and a standard deviation of \({\rm{6}}\)min.

a. If grading times are independent and the instructor begins grading at \({\rm{6:50}}\) p.m. and grades continuously, what is the (approximate) probability that he is through grading before the \({\rm{11:00}}\) p.m. TV news begins?

b. If the sports report begins at \({\rm{11:10,}}\) what is the probability that he misses part of the report if he waits until grading is done before turning on the TV?

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 Weibull with \({\rm{\alpha = 2}}\) and\({\rm{\beta = 5}}\), as in Example\({\rm{5}}{\rm{.20}}\).[A1] Consider the four sample sizes, and\({\rm{30}}\), 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?

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