91Ó°ÊÓ

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.

We'll use the following statement to solve these probabilities:

Allow \({\rm{X}}\) to be a continuous \({\rm{rv's}}\) with \({\rm{F(x)}}\). Then, for any \({\rm{a}}\) number,

\(\begin{array}{*{20}{c}}{{\rm{P(X£ a) = F(a)}}}\\{{\rm{P(X > a) = 1 - F(a)}}}\end{array}\)

and \({\rm{a < b}}\) for any two numbers a and b,

\({\rm{P(a£ X£ b) = F(b) - F(a)}}\)

\({\rm{Z}}\)is a standard normal random variable in the given situation. Its \({\rm{cdf}}\) is denoted by \({\rm{f(z)}}\) , the values of which are given in Appendix table-A.3 for various values of \({\rm{z}}\).

\({\rm{P(a£ Z£ b) = f(b) - f(a)}}\)

(a) \({\rm{P(0£ Z£ 2}}{\rm{.17)}}\) : This is the area beneath the standard normal curve above the interval with the left endpoint of 0 and the right endpoint of \({\rm{2}}{\rm{.17}}\) .

In the diagram below, this is the shaded area. Thus

\({\rm{P(0£ Z£ 2}}{\rm{.17) = f(2}}{\rm{.17) - f(0)}}\)

To find \({\rm{f(2}}{\rm{.17)}}\), look in Appendix Table A.3 at the intersection of the \({\rm{2}}{\rm{.1}}\) row and the. \({\rm{0}}{\rm{.7}}\) column. \({\rm{0}}{\rm{.9850}}\) is the figure there, therefore

\({\rm{f(2}}{\rm{.17) = 0}}{\rm{.9850}}\)

To find \({\rm{f(2}}{\rm{.17)}}\) , look in Appendix Table A.3 at the intersection of the \({\rm{0}}{\rm{.0}}\) row and the \({\rm{.00}}\) column. There's a \({\rm{0}}{\rm{.5}}\) figure there, thus

\(\begin{array}{*{20}{c}}{{\rm{f(0) = 0}}{\rm{.5}}}\\{{\rm{P(0£ Z£ 2}}{\rm{.17) = f(2}}{\rm{.17) - f(0)}}}\\{{\rm{ = 0}}{\rm{.9850 - 0}}{\rm{.5}}}\\{{\rm{P(0£ Z£ 2}}{\rm{.17) = 0}}{\rm{.4850}}}\end{array}\)

(b) \({\rm{P(0£ Z£ 1)}}\): It's the area under the standard normal curve above the interval with \({\rm{0}}\) as the left endpoint and\({\rm{1}}\) as the right endpoint. In the diagram below, this is the shaded area. Thus

\({\rm{P(0£ Z£ 1) = f(1) - f(0)}}\)

To find \({\rm{f(1)}}\), look in Appendix Table A.3 at the intersection of the \({\rm{1}}{\rm{.0}}\)row and the \({\rm{.00}}\)column. \({\rm{0}}{\rm{.8143}}\) is the figure there, therefore

\({\rm{f(1) = 0}}{\rm{.8413}}\)

To find \({\rm{f(0)}}\), look in Appendix Table A.3 at the intersection of the \({\rm{0}}{\rm{.0}}\) row and the \({\rm{.00}}\) column. There's a \({\rm{0}}{\rm{.5}}\)figure there, thus

\(\begin{array}{*{20}{c}}{{\rm{f(0)}}}&{{\rm{ = 0}}{\rm{.5}}}\\{{\rm{P(0£ Z£ 1)}}}&{{\rm{ = f(1) - f(0)}}}\\{}&{{\rm{ = 0}}{\rm{.8413 - 0}}{\rm{.5}}}\\{{\rm{P(0£ Z£ 1)}}}&{{\rm{ = 0}}{\rm{.3413}}}\end{array}\)

(c) \({\rm{P( - 2}}{\rm{.5£ Z£ 0)}}\): The area beneath the standard normal curve above the interval with the left endpoint \({\rm{ - 2}}{\rm{.5}}\) and the right endpoint \({\rm{0}}\). In the diagram below, this is the shaded area. Thus

\({\rm{P( - 2}}{\rm{.5£ Z£ 0) = f(0) - f( - 2}}{\rm{.5)}}\)

Check Appendix Table A.3 at the junction of the row marked \({\rm{ - 2}}{\rm{.5}}\) and the column marked.00$ to get \({\rm{f( - 2}}{\rm{.5)}}\). \({\rm{0}}{\rm{.0062}}\) is the value there, so

\({\rm{f( - 2}}{\rm{.5) = 0}}{\rm{.0062}}\)

To find \({\rm{f(0)}}\), look in Appendix Table A.3 at the intersection of the \({\rm{0}}{\rm{.0}}\) row and the \({\rm{.00}}\) column. There's a \({\rm{0}}{\rm{.5}}\)figure there, thus

\(\begin{array}{*{20}{c}}{{\rm{f(0)}}}&{{\rm{ = 0}}{\rm{.5}}}\\{{\rm{P( - 2}}{\rm{.5£ Z£ 0)}}}&{{\rm{ = f(0) - f( - 2}}{\rm{.5)}}}\\{}&{{\rm{ = 0}}{\rm{.5 - 0}}{\rm{.0062}}}\\{{\rm{P( - 2}}{\rm{.5£ Z£ 0)}}}&{{\rm{ = 0}}{\rm{.4938}}}\end{array}\)

(d) \({\rm{P( - 2}}{\rm{.5£ Z£ 2}}{\rm{.5)}}\): The area beneath the standard normal curve above the interval with the left endpoint \({\rm{ - 2}}{\rm{.5}}\) and the right endpoint \({\rm{2}}{\rm{.5}}\). In the diagram below, this is the shaded area. Thus

\({\rm{P( - 2}}{\rm{.5£ Z£ 2}}{\rm{.5) = f(2}}{\rm{.5) - f( - 2}}{\rm{.5)}}\)

Check Appendix Table A.3 at the junction of the row marked \({\rm{ - 2}}{\rm{.5}}\) and the column marked. \({\rm{0}}{\rm{.9938}}\)to get \({\rm{f( - 2}}{\rm{.5)}}\). \({\rm{0}}{\rm{.0062}}\) is the value there, so

\({\rm{f( - 2}}{\rm{.5) = 0}}{\rm{.0062}}\)

Check Appendix Table A.3 at the junction of the row marked \({\rm{2}}{\rm{.5}}\)and the column marked. \({\rm{.00}}\)to find \({\rm{f(2}}{\rm{.5)}}\). There's a \({\rm{0}}{\rm{.9938}}\) there, thus

\(\begin{array}{*{20}{c}}{{\rm{f(0) = }}}&{{\rm{0}}{\rm{.9938}}}\\{{\rm{P( - 2}}{\rm{.5£ Z£ 2}}{\rm{.5)}}}&{{\rm{ = f(2}}{\rm{.5) - f( - 2}}{\rm{.5)}}}\\{}&{{\rm{ = 0}}{\rm{.9938 - 0}}{\rm{.0062}}}\\{{\rm{P( - 2}}{\rm{.5£ Z£ 2}}{\rm{.5)}}}&{{\rm{ = 0}}{\rm{.9876}}}\end{array}\)

(e) \({\rm{P(Z£ 1}}{\rm{.37)}}\): This is the region beneath the z curve to the left of \({\rm{1}}{\rm{.37}}\)

\({\rm{P(Z£ 1}}{\rm{.37) = f(1}}{\rm{.37)}}\)

To find \({\rm{f(1}}{\rm{.37)}}\), look in Appendix Table A.3 at the intersection of the \({\rm{1}}{\rm{.3}}\)row and the \({\rm{.07}}\)column. There's a \({\rm{0}}{\rm{.9147}}\)there, thus

\(\begin{array}{*{20}{r}}{{\rm{f(1}}{\rm{.37) = 0}}{\rm{.9147}}}\\{{\rm{P(Z£ 1}}{\rm{.37) = 0}}{\rm{.9147}}}\end{array}\)

(f) \({\rm{P( - 1}}{\rm{.75£ Z)}}\): This is the area to the right of \({\rm{ - 1}}{\rm{.75}}\)under the \({\rm{z}}\) curve.

\({\rm{P( - 1}}{\rm{.75£ Z) = 1 - f( - 1}}{\rm{.75)}}\)

Check Appendix Table A.3 at the junction of the row marked \({\rm{ - 1}}{\rm{.75}}\) and the column marked to get \({\rm{f( - 1}}{\rm{.75)}}\).

\({\rm{.05}}\)There's a \({\rm{0}}{\rm{.0401}}\) there, thus

\(\begin{array}{*{20}{c}}{{\rm{f( - 1}}{\rm{.75) = 0}}{\rm{.0401}}}\\{{\rm{P( - 1}}{\rm{.75£ Z) = 1 - 0}}{\rm{.0401}}}\\{{\rm{P( - 1}}{\rm{.75£ Z) = 0}}{\rm{.9599}}}\end{array}\)

(g) \({\rm{P( - 1}}{\rm{.5£ Z£ 2)}}\): The area under the standard normal curve above the interval with the left endpoint of \({\rm{ - 1}}{\rm{.5}}\)and the right endpoint is \({\rm{2}}\). In the diagram below, this is the shaded area. Thus

\({\rm{P( - 1}}{\rm{.5£ Z£ 2) = f(2) - f( - 1}}{\rm{.5)}}\)

Check Appendix Table A.3 at the junction of the row marked \({\rm{ - 1}}{\rm{.5}}\) and the column marked\({\rm{0}}{\rm{.0}}\)to get \({\rm{f( - 1}}{\rm{.5)}}\). There's a \({\rm{0}}{\rm{.0668}}\)figure there, thus

\({\rm{f( - 1}}{\rm{.5) = 0}}{\rm{.0668}}\)

To find \({\rm{f(2)}}\), look in Appendix Table A.3 at the intersection of the \({\rm{2}}\)row and the\({\rm{.00}}\) column. \({\rm{0}}{\rm{.9772}}\)is the figure there, therefore

\(\begin{array}{*{20}{c}}{{\rm{f(2) = 0}}{\rm{.9772}}}\\{{\rm{P( - 1}}{\rm{.5£ Z£ 2) = f(2) - f( - 1}}{\rm{.5)}}}\\{{\rm{ = 0}}{\rm{.9772 - 0}}{\rm{.0668}}}\\{{\rm{P( - 1}}{\rm{.5£ Z£ 2) = 0}}{\rm{.9104}}}\end{array}\)

(h) \({\rm{P(1}}{\rm{.37£ Z£ 2}}{\rm{.5)}}\): The area under the standard normal curve above the interval with the left endpoint of \({\rm{1}}{\rm{.37}}\)and the right end point is \({\rm{2}}{\rm{.5}}\). In the diagram below, this is the shaded area. Thus

\({\rm{P(1}}{\rm{.37£ Z£ 2}}{\rm{.5) = f(2}}{\rm{.5) - f(1}}{\rm{.37)}}\)

To find \({\rm{f(1}}{\rm{.37)}}\), look in Appendix Table A.3 at the intersection of the \({\rm{1}}{\rm{.3}}\)row and the \({\rm{.07}}\)column. There's a \({\rm{0}}{\rm{.9147}}\)there, thus

\({\rm{f(1}}{\rm{.37) = 0}}{\rm{.9147}}\)

Check Appendix Table A.3 at the junction of the row marked \({\rm{2}}{\rm{.5}}\)and the column marked. \({\rm{.00}}\)to find \({\rm{f(2}}{\rm{.5)}}\). There's a \({\rm{0}}{\rm{.9938}}\) there, thus

\(\begin{array}{*{20}{c}}{{\rm{f(0) = 0}}{\rm{.9938}}}\\{{\rm{P(1}}{\rm{.37£ Z£ 2}}{\rm{.5) = f(2}}{\rm{.5) - f(1}}{\rm{.37)}}}\\{{\rm{ = 0}}{\rm{.9938 - 0}}{\rm{.9147}}}\\{{\rm{P(1}}{\rm{.37£ Z£ 2}}{\rm{.5) = 0}}{\rm{.0791}}}\end{array}\)

i). \({\rm{P(1}}{\rm{.5£ Z)}}\): This is the region beneath the z curve to the right of \({\rm{1}}{\rm{.5}}\).

\({\rm{P(1}}{\rm{.5£ Z) = 1 - f(1}}{\rm{.5)}}\)

Check Appendix Table A.3 at the junction of the row marked \({\rm{1}}{\rm{.5}}\) and the column marked.00$ to find \({\rm{f(1}}{\rm{.5)}}\). \({\rm{0}}{\rm{.9332}}\) is the figure there, therefore

\(\begin{array}{*{20}{c}}{{\rm{f(1}}{\rm{.5)}}}&{{\rm{ = 0}}{\rm{.9332}}}\\{{\rm{P(1}}{\rm{.5£Z)}}}&{{\rm{ = 1 - 0}}{\rm{.9332}}}\\{{\rm{P(1}}{\rm{.5£ Z)}}}&{{\rm{ = 0}}{\rm{.0668}}}\end{array}\)

(j) \({\rm{P(|Z|£ 2}}{\rm{.5) = P( - 2}}{\rm{.5£ Z£ 2}}{\rm{.5)}}\)In the diagram below, this is the shaded area. As we can see, it is identical to part(d), thus we can write:

\({\rm{P(|Z|£ 2}}{\rm{.5) = 0}}{\rm{.9876}}\)