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Chapter 4: Q37E (page 167)

Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean\({\rm{104}}\)and standard deviation\({\rm{5}}\)(information in the article "Mathematical Model of Chloride Concentration in Human Blood," \({\rm{J}}\). of Med. Engr. and Tech.,\({\rm{2006: 25 - 30}}\), including a normal probability plot as described in Section\({\rm{4}}{\rm{.6}}\), supports this assumption).

a. What is the probability that chloride concentration equals\({\rm{105}}\)? Is less than\({\rm{105}}\)? Is at most\({\rm{105}}\)?

b. What is the probability that chloride concentration differs from the mean by more than 1 standard deviation? Does this probability depend on the values of\({\rm{\mu }}\)and\({\rm{\sigma }}\)?

c. How would you characterize the most extreme\({\rm{.1\% }}\)of chloride concentration values?

Short Answer

Expert verified

(a) The probabilities are \({\rm{P(X = 105) = 0}}\)and \({\rm{P(X < 105) = 0}}{\rm{.5793}}\)and \(P(X \le 105) = 0.5793\)

(b) The probabilities is \({\rm{P(|X| > \mu \pm \sigma ) = 0}}{\rm{.3174}}\)

(c) The most extreme \({\rm{0}}{\rm{.1\% }}\)of chloride concentration values are below \({\rm{87}}{\rm{.55mmol/L}}\)and above\({\rm{120}}{\rm{.45mmol/L}}\).

Step by step solution

01

Introduction

The term "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

Explanation

Given: Normal distribution

\(\begin{array}{l}{\rm{\mu = 104}}\\{\rm{\sigma = 5}}\end{array}\)

(a) The standardized score is the value \({\rm{x}}\)decreased by the mean and then divided by the standard deviation.

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{105 - 104}}}}{{\rm{5}}}\\{\rm{\gg 0}}{\rm{.20}}\end{array}\)

Determine the corresponding probability using table \({\rm{A}}{\rm{.3}}\):

\(\begin{array}{l}P(X < 105) = P(Z < 0.20) = 0.5793\\P(X \le 105) = P(Z < 0.20) = 0.5793\end{array}\)

The probability of a continuous random variable being equal to a specific value is always zero (Note: you can notice this by the probabilities \({\rm{P(X < 105)}}\)and \(P(X \le \)\({\rm{105}}\)) that are equal):

\({\rm{P(X = 105) = 0}}\)

03

Explanation

b)

\({\rm{x}}\)is \({\rm{1}}\) standard deviation from the mean:

\({\rm{x = \mu \pm \sigma }}\)

The standardized score is the value \({\rm{x}}\)decreased by the mean and then divided by the standard deviation.

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{\mu \pm \sigma - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{ \pm \sigma }}}}{{\rm{\sigma }}}\\{\rm{ = \pm 1}}{\rm{.00}}\end{array}\)

Determine the corresponding probability using table \({\rm{A}}{\rm{.3}}\):

\(\begin{array}{c}{\rm{P(|X| > \mu \pm \sigma ) = P(Z < - 1 or Z > 1)}}\\{\rm{ = 2P(Z < - 1) = 2(0}}{\rm{.1587)}}\\{\rm{ = 0}}{\rm{.3174}}\end{array}\)

Note: the probability is not dependent on the values of \({\rm{\mu }}\)and\({\rm{\sigma }}\).

04

Explanation

c)

The most extreme \({\rm{0}}{\rm{.1\% }}\)of chloride concentration values are the lowest \({\rm{0}}{\rm{.05\% }}\)of the chloride values and the highest \({\rm{0}}{\rm{.05\% }}\) (lowest\({\rm{99}}{\rm{.95\% }}\)) of the chloride values (using that the normal distribution is symmetric about the mean).

Determine the \({\rm{z}}\)-score corresponding to a probability of \({\rm{0}}{\rm{.05\% }}\) (0.0005), and \({\rm{99}}{\rm{.95\% (0}}{\rm{.9995)}}\) using table A.3:

\({\rm{z = \pm 3}}{\rm{.29}}\)Note: there are multiple z-score with probability \({\rm{0}}{\rm{.0005/0}}{\rm{.9995}}\)in table \({\rm{A}}{\rm{.3}}\), thus we used technology to narrow the score down further (you could also take the average score, which would be\({\rm{ \pm 3}}{\rm{.295}}\)).

The corresponding value is the mean increased by the product of the \({\rm{z}}\)-score and the standard deviation:

\(\begin{array}{c}{\rm{x = \mu + z\sigma = 104 - 3}}{\rm{.29(5)}}\\{\rm{ = 87}}{\rm{.55x = \mu - z\sigma }}\\{\rm{ = 104 + 3}}{\rm{.29(5)}}\\{\rm{ = 120}}{\rm{.45}}\end{array}\)

Thus, the most extreme \({\rm{0}}{\rm{.1\% }}\)of chloride concentration values are below \({\rm{87}}{\rm{.55mmol/L}}\)and above\({\rm{120}}{\rm{.45mmol/L}}\).

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Most popular questions from this chapter

Let \({\rm{X}}\) denote the voltage at the output of a microphone, and suppose that \({\rm{X}}\) has a uniform distribution on the interval from \({\rm{ - 1}}\) to \({\rm{1}}\). The voltage is processed by a 鈥渉ard limiter鈥� with cut-off values \({\rm{ - }}{\rm{.5}}\) and \({\rm{.5}}\), so the limiter output is a random variable \({\rm{Y}}\) related to \({\rm{X}}\) by \({\rm{Y = X}}\) if \({\rm{|X|}} \le {\rm{.5,Y = }}{\rm{.5}}\) if \({\rm{X > }}{\rm{.5}}\), and \({\rm{Y = - }}{\rm{.5}}\) if \({\rm{X < - }}{\rm{.5}}\). a. What is \({\rm{P(Y = }}{\rm{.5)}}\)? b. Obtain the cumulative distribution function of \({\rm{Y}}\) and graph it.

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Let X 5 the time it takes a read/write head to locate the desired record on a computer disk memory device once the head has been positioned over the correct track. If the disks rotate once every \({\bf{25}}\) milliseconds, a reasonable assumption is that X is uniformly distributed on the interval\(\left( {{\bf{0}},{\rm{ }}{\bf{25}}} \right)\). a. Compute\({\bf{P}}\left( {{\bf{10}} \le {\bf{X}} \le {\bf{20}}} \right)\). b. Compute \({\bf{P}}\left( {{\bf{X}} \le {\bf{10}}} \right)\). c. Obtain the cdf F(X). d. Compute E(X) and \({\sigma _X}\).

a. Show that if X has a normal distribution with parameters \({\rm{\mu }}\) and\({\rm{\sigma }}\), then \({\rm{Y = aX + b}}\) (a linear function of X ) also has a normal distribution. What are the parameters of the distribution of Y (i.e., E(Y) and V(Y)) ? (Hint: Write the cdf of\({\rm{Y,P(Y}} \le {\rm{y)}}\), as an integral involving the pdf of X, and then differentiate with respect to y to get the pdf of Y.)

b. If, when measured in\(^{\rm{^\circ }}{\rm{C}}\), temperature is normally distributed with mean 115 and standard deviation 2 , what can be said about the distribution of temperature measured in\(^{\rm{^\circ }}{\rm{F}}\)?

A family of pdf鈥檚 that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. The family has two parameters, \({\rm{k}}\) and \({\rm{\theta }}\), both\({\rm{ > 0}}\), and the pdf is

\({\rm{f(x;\theta ) = \{ }}\begin{array}{*{20}{c}}{\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}}&{{\rm{x}} \ge {\rm{\theta }}}\\{\rm{0}}&{{\rm{x < \theta }}}\end{array}\)

a. Sketch the graph of \({\rm{f(x;\theta )}}\).

b. Verify that the total area under the graph equals \({\rm{1}}\).

c. If the rv \({\rm{X}}\) has pdf \({\rm{f(x;\theta )}}\), for any fixed \({\rm{b > \theta }}\), obtain an expression for \({\rm{P(X}} \le {\rm{b)}}\).

d. For \({\rm{\theta < a < b}}\) obtain an expression for the probability \({\rm{P(a}} \le {\rm{X}} \le {\rm{b)}}\).
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