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Chapter 5: Q83SE (page 244)

Let\(\mu \)denote the true pH of a chemical compound. A sequence of 鈥榥鈥� independent sample pH determinations will be made. Suppose each sample pH is a random variable with expected value m and standard deviation\(.1\). How many determinations are required if we wish the probability that the sample average is within\(.02\)of the true pH to be at least\(.95\)? What theorem justifies your probability calculation?

Short Answer

Expert verified

The theorem that justifies the probability calculation (the normal distribution Z) is Central Limit Theorem.

Step by step solution

01

Definition of Expected Value

The expected value is calculated in statistics and probability analysis by multiplying each conceivable outcome by the probability that it will occur and then adding all of those values together. Investors might choose the scenario that is most likely to provide the desired result by calculating anticipated values.

02

Calculation for finding the number of determinations.

Let n be the number of determinations required to obtain the probability of interest. The event of interest is \(\mu - 0.02 \le \bar X \le \mu + 0.02\)

Where,\(\bar X\)is the sample average and\(\mu \)the expected value. The expected value of the sample average is\(\mu \) and the standard deviation is\(0.1/\sqrt n \)(this stands from the proposition for the sample average), therefore, the following holds

\(\begin{array}{l}P(\mu - 0.02 \le \bar X \le \mu + 0.02) = P\left( {\frac{{\mu - 0.02 - \mu }}{{0.1/\sqrt n }} \le \frac{{\bar X - \mu }}{{0.1/\sqrt n }} \le \frac{{\mu + 0.02 - \mu }}{{0.1/\sqrt n }}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\;\;\; = P\left( {\frac{{ - 0.02}}{{0.1/\sqrt n }} \le Z \le \frac{{0.02}}{{0.1/\sqrt n }}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = P( - 0.2\sqrt n \le Z \le 0.2\sqrt n )\end{array}\)

03

Calculation for finding the number of determinations.

The value x for which

\(P( - x \le Z \le x) = 0.95\)

can be computed from

\(\begin{array}{c}P( - x \le Z \le x) = P(Z \le x) - P(Z \le - x)\\ = \Phi (x) - \Phi ( - x)\\ = 2 \cdot (\Phi (x) - 0.5)\end{array}\)

Where,\(\Phi (x) - 0.5\)is the area from 0 up to x, where\(x > 0\). Because standard normal curve is symmetric, the area from -x to 0 is equal to area from 0 to x. Therefore,

\(2 \cdot (\Phi (x) - 0.5) = 0.95\Phi (x) = 0.975\)

where\(\Phi (x) - 0.5\)is the area from\(0\)up to x, where\(x > 0\).Because standard normal curve is symmetric, the area from -x to\(0\)is equal to area from\(0\)to x. Therefore,

\(2 \cdot (\Phi (x) - 0.5){\rm{ }} = 0.95\Phi (x) = 0.975\)

from the normal probability table in the appendix, value x for which

\(\Phi (x) = 0.975\)

is \(x = 1.96\).

04

Calculation for finding the number of determinations.

From

\(P( - 0.2\sqrt n \le Z \le 0.2\sqrt n ) = 0.95\)

and

\(P( - 1.96 \le Z \le 1.96) = 0.95\)

the number of determinations required can be found as

\(\begin{array}{c}0.2\sqrt n = 1.96\\n = 94\end{array}\)

The theorem that justifies the probability calculation (the normal distribution Z) is Central Limit Theorem.

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Most popular questions from this chapter

The lifetime of a certain type of battery is normally distributed with mean value \({\rm{10}}\)hours and standard deviation \({\rm{1}}\)hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only \({\rm{5\% }}\)of all packages?

Consider a system consisting of three components as pictured. The system will continue to function as long as the first component functions and either component \({\rm{2}}\) or component \({\rm{3}}\)functions. Let \({{\rm{X}}_{{\rm{1,}}}}{{\rm{X}}_{\rm{2}}}\), and \({{\rm{X}}_{\rm{3}}}\) denote the lifetimes of components \({\rm{1}}\), \({\rm{2}}\), and \({\rm{3}}\), respectively. Suppose the \({{\rm{X}}_{\rm{i}}}\) 鈥檚 are independent of one another and each \({{\rm{X}}_{\rm{i}}}\) has an exponential distribution with parameter \({\rm{\lambda }}\).

a. Let \({\rm{Y}}\) denote the system lifetime. Obtain the cumulative distribution function of \({\rm{Y}}\)and differentiate to obtain the pdf. (Hint: \({{\rm{F}}_{\left( {\rm{Y}} \right)}}{\rm{P}}\left\{ {{\rm{Y}} \le {\rm{y}}} \right\}\); express the event \(\left\{ {{\rm{Y}} \le {\rm{y}}} \right\}\)in terms of unions and/or intersections of the three events \(\left\{ {{{\rm{X}}_{\rm{i}}} \le {\rm{y}}} \right\}\), \(\left\{ {{{\rm{X}}_{\rm{2}}} \le {\rm{y}}} \right\}\), and \(\left\{ {{{\rm{X}}_3} \le {\rm{y}}} \right\}\).)

b. Compute the expected system lifetime

If two loads are applied to a cantilever beam as shown in the accompanying drawing, the bending moment at \({\rm{0}}\) due to the loads is \({{\rm{a}}_{\rm{1}}}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)

a. Suppose that \({{\rm{X}}_{\rm{1}}}{\rm{ and }}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)are independent rv鈥檚 with means \({\rm{2 and 4kip}}\), respectively, and standard deviations \({\rm{.5}}\) and \({\rm{1}}{\rm{.0kip}}\), respectively. If \({{\rm{a}}_{\rm{1}}}{\rm{ = 5ft and }}{{\rm{a}}_{\rm{2}}}{\rm{ = 10ft }}\), what is the expected bending moment and what is the standard deviation of the bending moment?

b. If \({{\rm{X}}_{\rm{1}}}{\rm{ and }}{{\rm{X}}_{\rm{2}}}{\rm{ }}\)are normally distributed, what is the probability that the bending moment will exceed 75 kip-ft?

c. Suppose the positions of the two loads are random variables. Denoting them by \({{\rm{A}}_{\rm{1}}}{\rm{ and }}{{\rm{A}}_{\rm{2}}}{\rm{ }}\), assume that these variables have means of \({\rm{5 and 10ft }}\), respectively, that each has a standard deviation of \({\rm{.5}}\), and that all are independent of one another. What is the expected moment now?

d. For the situation of part (c), what is the variance of the bending moment?

e. If the situation is as described in part (a) except that \({\rm{Corr}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.5}}\) (so that the two loads are not independent), what is the variance of the bending moment?

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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.
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