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Chapter 4: Q30E (page 141)

Find the following percentiles for the standard normal distribution. Interpolate where appropriate.

\(\begin{array}{*{20}{l}}{{\rm{a}}{\rm{. 91st}}}\\\begin{array}{l}{\rm{b}}{\rm{. 9th }}\\{\rm{c}}{\rm{. 75th }}\\{\rm{d}}{\rm{. 25th }}\\{\rm{e}}{\rm{. }}{{\rm{6}}^{{\rm{th}}}}\end{array}\end{array}\)

Short Answer

Expert verified

a) \({\rm{1}}{\rm{.34}}\)

b) \({\rm{ - 1}}{\rm{.34}}\)

c) \({\rm{0}}{\rm{.6745}}\)

d) \({\rm{ - 0}}{\rm{.6745}}\)

e) \({\rm{ - 1}}{\rm{.555}}\)

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 \({\rm{91st}}\)

(a) The \({{\rm{z}}_{{\rm{0}}{\rm{.09}}}}\)percentile represents the \({\rm{91st}}\) percentile. It signifies that the area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.09}}}}\)is \({\rm{0}}{\rm{.91}}\) (or that the region to the right of \({{\rm{z}}_{{\rm{0}}{\rm{.09}}}}\)$ is \({\rm{0}}{\rm{.09}}\)or that:

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.09}}}}} \right){\rm{ = 0}}{\rm{.91}}\)

For any \({\rm{z}}\), we examine Appendix Table A.3 to see if \({\rm{f(z)}}\) equals \({\rm{0}}{\rm{.91}}\)

At the point where the \({\rm{1}}{\rm{.3}}\) row and the \({\rm{0}}{\rm{.7}}\) column cross, \({\rm{0}}{\rm{.9099}}\)is the amount there. \({\rm{0}}{\rm{.91}}\) is a pretty close match. As a result, without utilizing interpolation, we may say:

\({{\rm{z}}_{{\rm{0}}{\rm{.09}}}}{\rm{ = 1}}{\rm{.37}}\)

03

Determining the \({\rm{9th}}\)

(b) The \({{\rm{z}}_{{\rm{0}}{\rm{.91}}}}\)percentile represents the \({\rm{9th}}\) percentile. It signifies that the area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.91}}}}\)is \({\rm{0}}{\rm{.09}}\) :

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.91}}}}} \right){\rm{ = 0}}{\rm{.09}}\)

For any \({\rm{z}}\), we examine Appendix Table A.3 to see if \({\rm{f(z)}}\) equals \({\rm{0}}{\rm{.09}}\)

At the point where the \({\rm{ - 1}}{\rm{.3}}\) row and the \({\rm{0}}{\rm{.4}}\) column cross, \({\rm{0}}{\rm{.9091}}\)is the amount there. \({\rm{0}}{\rm{.09}}\) is a pretty close match. As a result, without utilizing interpolation, we may say:

\({{\rm{z}}_{{\rm{0}}{\rm{.91}}}}{\rm{ = 1}}{\rm{.34}}\)

04

Determining the \({\rm{75th }}\)

(c) The \({{\rm{z}}_{{\rm{0}}{\rm{.25}}}}\) percentile for the \({\rm{75th}}\)percentile. It means that the region to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.25}}}}\)equals \({\rm{0}}{\rm{.75}}\), or in other words:

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}} \right){\rm{ = 0}}{\rm{.75}}\)

For any\({\rm{z}}\), we examine Appendix Table A.3 to see if \({\rm{f(z)}}\)equals \({\rm{0}}{\rm{.75}}\).

\({\rm{0}}{\rm{.7486 and 0}}{\rm{.7517}}\)are the two closest values from the table, with z-values \({\rm{0}}{\rm{.67 and 0}}{\rm{.68,}}\)respectively. As a result of the interpolation, we can write:

\(\begin{array}{*{20}{c}}{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}{\rm{ - 0}}{\rm{.67}}}}{{{\rm{0}}{\rm{.75 - 0}}{\rm{.7486}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.68 - 0}}{\rm{.67}}}}{{{\rm{0}}{\rm{.7517 - 0}}{\rm{.7486}}}}}\\{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}{\rm{ - 0}}{\rm{.67}}}}{{{\rm{0}}{\rm{.0014}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.01}}}}{{{\rm{0}}{\rm{.0031}}}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}{\rm{ - 0}}{\rm{.67 = 0}}{\rm{.0045}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.25}}}}{\rm{ = 0}}{\rm{.6745}}}\end{array}\)

05

Determining \({\rm{25th}}\)

(d) The \({{\rm{z}}_{{\rm{0}}{\rm{.75}}}}\) percentile for the \({\rm{25th}}\)percentile. It means that the area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.75}}}}\) equals \({\rm{0}}{\rm{.25}}\), or in other words:

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}} \right){\rm{ = 0}}{\rm{.25}}\)

For any \({\rm{z}}\), we examine Appendix Table A. 3 to see if \({\rm{f(z)}}\)equals \({\rm{0}}{\rm{.25}}\)

\({\rm{0}}{\rm{.2514 and 0}}{\rm{.2483}}\)are the two closest values from the table, with z-values \({\rm{ - 0}}{\rm{.67 and - 0}}{\rm{.68}}\)respectively. As a result of the interpolation, we can write:

\(\begin{array}{*{20}{c}}{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}{\rm{ - ( - 0}}{\rm{.67)}}}}{{{\rm{0}}{\rm{.25 - 0}}{\rm{.2514}}}}}&{{\rm{ = }}\frac{{{\rm{ - 0}}{\rm{.68 - ( - 0}}{\rm{.67)}}}}{{{\rm{0}}{\rm{.2483 - 0}}{\rm{.2514}}}}}\\{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}{\rm{ + 0}}{\rm{.67}}}}{{{\rm{ - 0}}{\rm{.0014}}}}}&{{\rm{ = }}\frac{{{\rm{0}}{\rm{.01}}}}{{{\rm{0}}{\rm{.0031}}}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}{\rm{ + 0}}{\rm{.67}}}&{{\rm{ = - 0}}{\rm{.0045}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.75}}}}}&{{\rm{ = - 0}}{\rm{.6745}}}\end{array}\)

06

Determining \({{\rm{6}}^{{\rm{th}}}}\)

(d) The \({{\rm{z}}_{{\rm{0}}{\rm{.94}}}}\) percentile for the \({\rm{25th}}\)percentile. It means that the area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.94}}}}\) equals \({\rm{0}}{\rm{.06}}\), or in other words:

\({\rm{f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}} \right){\rm{ = 0}}{\rm{.06}}\)

For any \({\rm{z}}\), we examine Appendix Table A. 3 to see if \({\rm{f(z)}}\)equals \({\rm{0}}{\rm{.06}}\)

\({\rm{0}}{\rm{.0606 and 0}}{\rm{.0594}}\)are the two closest values from the table, with z-values \({\rm{ - 1}}{\rm{.55 and - 1}}{\rm{.56}}\) respectively. As a result of the interpolation, we can write:

\(\begin{array}{*{20}{c}}{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}{\rm{ - ( - 1}}{\rm{.56)}}}}{{{\rm{0}}{\rm{.06 - 0}}{\rm{.0594}}}}{\rm{ = }}\frac{{{\rm{ - 1}}{\rm{.55 - ( - 1}}{\rm{.56)}}}}{{{\rm{0}}{\rm{.0606 - 0}}{\rm{.0594}}}}}\\{\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}{\rm{ + 1}}{\rm{.56}}}}{{{\rm{0}}{\rm{.0006}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.01}}}}{{{\rm{0}}{\rm{.0012}}}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}{\rm{ + 1}}{\rm{.56 = 0}}{\rm{.0050}}}\\{{{\rm{z}}_{{\rm{0}}{\rm{.94}}}}{\rm{ = - 1}}{\rm{.555}}}\end{array}\)

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Most popular questions from this chapter

Let X have a uniform distribution on the interval \({\rm{(A,B)}}\). a. Obtain an expression for the \({\rm{(100p)th}}\) percentile. b. Compute \({\rm{E(X),V(X)}}\) and \({{\rm{\sigma }}_{\rm{X}}}\). c. For n, a positive integer, compute \({\rm{E}}\left( {{{\rm{X}}^{\rm{n}}}} \right)\).

Based on an analysis of sample data, the article 鈥淧edestrians鈥� Crossing Behaviours and Safety at Unmarked Roadways in China鈥� (Accident Analysis and Prevention, \({\rm{2011: 1927 - 1936}}\) proposed the pdf \({\rm{f(x) = }}{\rm{.15}}{{\rm{e}}^{{\rm{ - }}{\rm{.15(x - 1)}}}}\) when \({\rm{x}} \ge {\rm{1}}\) as a model for the distribution of \({\rm{X = }}\)time (sec) spent at the median line.

a. What is the probability that waiting time is at most \({\rm{5}}\) sec? More than \({\rm{5}}\) sec?

b. What is the probability that waiting time is between \({\rm{2}}\) and \({\rm{5}}\) sec?

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 \({\rm{P(a}} \le {\rm{X}} \le {\rm{b)}}\).

The accompanying sample consisting of observations \({\rm{n = 20}}\) on dielectric breakdown voltage of a piece of epoxy resin appeared in the article "Maximum Likelihood Estimation in the \({\rm{3}}\)-Parameter Weibull Distribution (IEEE Trans. on Dielectrics and Elec. Insul., \({\rm{1996}}\): \({\rm{43 - 55)}}\). The values of \({\rm{(i - }}{\rm{.5)/n}}\) for which \({\rm{z}}\) percentiles are needed are \({\rm{(1 - }}{\rm{.5)/20 = }}{\rm{.025,(2 - }}{\rm{.5)/20 = }}\)\({\rm{.075, \ldots }}\), and\({\rm{.975}}\). Would you feel comfortable estimating population mean voltage using a method that assumed a normal population distribution?

\(\begin{array}{*{20}{c}}{{\rm{ Observation }}}&{{\rm{24}}{\rm{.46}}}&{{\rm{25}}{\rm{.61}}}&{{\rm{26}}{\rm{.25}}}&{{\rm{26}}{\rm{.42}}}&{{\rm{26}}{\rm{.66}}}\\{{\rm{ zpercentile }}}&{{\rm{ - 1}}{\rm{.96}}}&{{\rm{ - 1}}{\rm{.44}}}&{{\rm{ - 1}}{\rm{.15}}}&{{\rm{ - }}{\rm{.93}}}&{{\rm{ - }}{\rm{.76}}}\\{{\rm{ Observation }}}&{{\rm{27}}{\rm{.15}}}&{{\rm{27}}{\rm{.31}}}&{{\rm{27}}{\rm{.54}}}&{{\rm{27}}{\rm{.74}}}&{{\rm{27}}{\rm{.94}}}\\{{\rm{ zpercentile }}}&{{\rm{ - }}{\rm{.60}}}&{{\rm{ - }}{\rm{.45}}}&{{\rm{ - }}{\rm{.32}}}&{{\rm{ - }}{\rm{.19}}}&{{\rm{ - }}{\rm{.06}}}\\{{\rm{ Observation }}}&{{\rm{27}}{\rm{.98}}}&{{\rm{28}}{\rm{.04}}}&{{\rm{28}}{\rm{.28}}}&{{\rm{28}}{\rm{.49}}}&{{\rm{28}}{\rm{.50}}}\\{{\rm{ zpercentile }}}&{{\rm{.06}}}&{{\rm{.19}}}&{{\rm{.32}}}&{{\rm{.45}}}&{{\rm{.60}}}\\{{\rm{ Observation }}}&{{\rm{28}}{\rm{.87}}}&{{\rm{29}}{\rm{.11}}}&{{\rm{29}}{\rm{.13}}}&{{\rm{29}}{\rm{.50}}}&{{\rm{30}}{\rm{.88}}}\\{{\rm{ zpercentile }}}&{{\rm{.76}}}&{{\rm{.93}}}&{{\rm{1}}{\rm{.15}}}&{{\rm{1}}{\rm{.44}}}&{{\rm{1}}{\rm{.96}}}\end{array}\)

Let X represent the number of individuals who respond to a particular online coupon offer. Suppose that X has approximately a Weibull distribution with \(\alpha = 10\)and \(\beta = 20\). Calculate the best possible approximation to the probability that X is between \(15 and 20\), inclusive.
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