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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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