91影视

A constant is a single number or a letter such as a, b, or c that stands for a fixed number.

(a)We must find \({\rm{c}}\) in such a way that

\(\phi {\rm{(c) = 0}}{\rm{.9838}}\)

For all \({\rm{z}}\), we examine Appendix Table \({\rm{A}}{\rm{.3}}\) to see if \(\phi {\rm{(z)}}\) equals \({\rm{0}}{\rm{.9838}}\). At the junction of the marked \({\rm{2}}{\rm{.1}}\) row and the indicated \({\rm{.04}}\) column As, a result, the number there is \({\rm{0}}{\rm{.9838}}\).

\({\rm{c = 2}}{\rm{.14}}\)

Therefore, the value is \({\rm{2}}{\rm{.14}}\).

(b)It's the area above the interval with the left endpoint of \({\rm{0}}\) and the right endpoint of \({\rm{c}}\) under the standard normal curve. Thus

\({\rm{P(0}} \le {\rm{Z}} \le {\rm{c) = }}\phi {\rm{(c) - }}\phi {\rm{(0)}}\)

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

\(\phi {\rm{(0) = 0}}{\rm{.5}}\)

Then we'll be able to write:

\(\begin{array}{c}{\rm{P(0}} \le {\rm{Z}} \le {\rm{c) = }}\phi {\rm{(c) - }}\phi {\rm{(0)}}\\{\rm{0}}{\rm{.291 = }}\phi {\rm{(c) - 0}}{\rm{.5}}\\{\rm{phi(c) = 0}}{\rm{.291 + 0}}{\rm{.5}}\\\phi {\rm{(c) = 0}}{\rm{.791}}\end{array}\)

For all \({\rm{z}}\), we examine Appendix Table \({\rm{A}}{\rm{.3}}\) to see if \(\phi {\rm{(z)}}\) equals \({\rm{0}}{\rm{.791}}\). At the point where the \({\rm{0}}{\rm{.8}}\) row and the column cross \({\rm{.01}}\). As a result, the number there is \({\rm{0}}{\rm{.791}}\). Hence,

\({\rm{c = 0}}{\rm{.81}}\)

If \({\rm{Z}}\) is a continuous \({\rm{rv}}\) with cdf, then the following proposition is \(\phi {\rm{(z)}}\). Then, using \({\rm{a < b}}\), for any two numbers \({\rm{a}}\) and \({\rm{b}}\),

\({\rm{P(a}} \le {\rm{Z}} \le {\rm{b) = }}\phi {\rm{(b) - }}\phi {\rm{(a)}}\)

Therefore, the value is \({\rm{0}}{\rm{.81}}\).

(c)It is the area to the right of \({\rm{c}}\) under the \({\rm{z}}\) curve.

\(\begin{array}{c}{\rm{P(c}} \le {\rm{Z) = 1 - }}\phi {\rm{(c)}}\\{\rm{0}}{\rm{.121 = 1 - }}\phi {\rm{(c)}}\\\phi {\rm{(c) = 1 - 0}}{\rm{.121}}\\\phi {\rm{(c) = 0}}{\rm{.879}}\end{array}\)

For all \({\rm{z}}\), we examine Appendix Table \({\rm{A}}{\rm{.3}}\) to see if \(\phi {\rm{(z)}}\) equals \({\rm{0}}{\rm{.879}}\). At the point where the row marked \({\rm{1}}{\rm{.1}}\) and the column marked \({\rm{.07}}\) cross, As, a result, the number is \({\rm{0}}{\rm{.879}}\).Then,

\({\rm{c = 1}}{\rm{.17}}\)

Therefore, the value is \({\rm{1}}{\rm{.17}}\).

(d)It's the region above the interval with a left endpoint of \({\rm{ - c}}\) that falls under the standard normal curve. Thus

\({\rm{P( - c}} \le {\rm{Z}} \le {\rm{c) = }}\phi {\rm{(c) - }}\phi {\rm{( - c)}}\)

The symmetry of the normal distribution curve is used here. As a result, we can conclude that

\(\begin{array}{c}\phi {\rm{( - c) = 1 - }}\phi {\rm{(c)}}\\{\rm{P( - c}} \le {\rm{Z}} \le {\rm{c) = }}\phi {\rm{(c) - }}\phi {\rm{( - c)}}\\{\rm{P( - c}} \le {\rm{Z}} \le {\rm{c) = }}\phi {\rm{(c) - (1 - }}\phi {\rm{(c))}}\\{\rm{0}}{\rm{.668 = 2}}\phi {\rm{(c) - 1}}\\{\rm{2}}\phi {\rm{(c) = 1}}{\rm{.668}}\\\phi {\rm{(c) = }}\frac{{{\rm{1}}{\rm{.668}}}}{{\rm{2}}}\\\phi {\rm{(c) = 0}}{\rm{.834}}\end{array}\)

For all \({\rm{z}}\), we examine Appendix Table \({\rm{A}}{\rm{.3}}\) to see if \(\phi {\rm{(z)}}\) equals \({\rm{0}}{\rm{.834}}\). At the point where the row marked \({\rm{0}}{\rm{.9}}\) and the column marked \({\rm{.07}}\) cross, As, a result, the number is \({\rm{0}}{\rm{.834}}\).Then,

\({\rm{c = 0}}{\rm{.97}}\)

If \({\rm{Z}}\) is a continuous \({\rm{rv}}\) with cdf, then the following proposition is \(\phi {\rm{(z)}}\). Then, using \({\rm{a < b}}\), for any two numbers \({\rm{a}}\) and \({\rm{b}}\),

\({\rm{P(a}} \le {\rm{Z}} \le {\rm{b) = }}\phi {\rm{(b) - }}\phi {\rm{(a)}}\)

Therefore, the value is \({\rm{0}}{\rm{.97}}\).

(e)It is the region to the left of \({\rm{ - c}}\) and to the right of \({\rm{c}}\) under the standard normal curve. In the diagram below, this is illustrated by the shaded area. We can observe from the diagram that:

The area to the left is equal to \({\rm{phi( - c)}}\).

The right-hand area \({\rm{ = 1 - }}\phi {\rm{(c)}}\).

As a result, the overall probability is equal to the sum of both areas:

\({\rm{P(c}} \le {\rm{|Z|) = (}}\phi {\rm{( - c)) + (1 - }}\phi {\rm{(c))}}\)

The symmetry of the normal distribution curve is used here. As a result, we can conclude that

\(\phi {\rm{( - c) = 1 - }}\phi {\rm{(c)}}\)

Using the two equations above, we can get the following result:

\(\begin{array}{c}{\rm{P(c}} \le {\rm{|Z|) = (}}\phi {\rm{( - c)) + (1 - }}\phi {\rm{(c))}}\\{\rm{P(c}} \le {\rm{|Z|) = (1 - }}\phi {\rm{(c)) + (1 - }}\phi {\rm{(c))}}\\{\rm{0}}{\rm{.016 = 2(1 - }}\phi {\rm{(c))}}\\{\rm{0}}{\rm{.008 = 1 - }}\phi {\rm{(c)}}\\\phi {\rm{(c) = 1 - 0}}{\rm{.008}}\\\phi {\rm{(c) = 0}}{\rm{.992}}\end{array}\)

For all \({\rm{z}}\), we examine Appendix Table \({\rm{A}}{\rm{.3}}\) to see if \(\phi {\rm{(z)}}\) equals \({\rm{0}}{\rm{.992}}\). At the point where the row marked \({\rm{2}}{\rm{.4}}\) and the column marked \({\rm{.01}}\) cross, As, a result, the number is \({\rm{0}}{\rm{.992}}\).Then,

\({\rm{c = 2}}{\rm{.41}}\)

Therefore, the value is \({\rm{2}}{\rm{.41}}\).