91影视

A discrete random variable is one that can only take on a finite number of different values.

We have been granted

\({\rm{X}} \sim {\rm{Bin(20,}}{\rm{.5)}}\)

We must first determine the anticipated value and standard deviation.

The following is valid for a binomial random variable X with parameters n, p, and \({\rm{q = 1 - p}}\).

\(\begin{array}{c}{\rm{E(X) = np;}}\\{\rm{V(X) = np(1 - p)}}\\{\rm{ = npq}}\\{{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{npq}}} \end{array}\)

The anticipated outcome is:

\(\begin{aligned} E(X) &= np \\ &= 20 \times 0{\rm{.5}}\\ &= 10\end{aligned}\)

The difference is:

\(\begin{aligned}V(X) &= np(1 - p) \\ &= npq\\ &= 20 \times 0 {\rm{.5 \times 0}}{\rm{.5}}\\ &= 5\end{aligned}\)

As well as the standard deviation is:

\(\begin{aligned}{{\rm{\sigma }}_{\rm{X}}} &= \sqrt {{\rm{npq}}} \\&= \sqrt {\rm{5}} \\ &= 2 {\rm{.236}}\end{aligned}\)

Let \({\rm{k = 2}}\) first.

\(\begin{aligned}{\rm{2 \bullet }}{{\rm{\sigma }}_{\rm{X}}} &= 2 \times 2 {\rm{.236}}\\&= 4 {\rm{.472}}\end{aligned}\)

As well as we have:

\({\rm{|X - 10|}} \ge {\rm{4}}{\rm{.472}}\)

The final inequity has been met.

\({\rm{4}}{\rm{.472}} \le {\rm{X - 10 or (X - 10)}} \ge {\rm{4}}{\rm{.472}}\)

or, alternatively

\({\rm{X}} \le {\rm{5}}{\rm{.528 or X}} \ge 1{\rm{4}}{\rm{.472}}\)

We need to determine the following probability since X has a Binomial Distribution and only accepts non-negative integer values.

\(\begin{aligned}{\rm{P(X}} \le {\rm{5 or X}} \ge {\rm{15)}}\mathop = \limits^{{\rm{(1)}}} {\rm{P(X}} \le {\rm{5) + P(X}} \ge {\rm{15)}}\\\mathop = \limits^{{\rm{(2)}}} {\rm{P(X}} \le {\rm{5) + (1 - P(X < 15))}}\\\mathop = \limits^{{\rm{(3)}}} {\rm{B(5;20,0}}{\rm{.5) + 1 - B(14;20,0}}{\rm{.5)}}\\\mathop = \limits^{{\rm{(4)}}} &{\rm{0}}{\rm{.02069 + 1 - 0}}{\rm{.97931}}\\ &= 0 {\rm{.04138}}\end{aligned}\)

(1):the sequence of events is disjointed;

(2):event \({\rm{X }} \ge {\rm{ 15}}\) counterpart is event \({\rm{X < 15}}\);

(3):see Binomial Distribution cdf; \({\rm{X = 14}}\) since \({\rm{X < 15}}\) and \({\rm{X }} \le {\rm{ 14}}\) are the same event;

(4):you can get it from Appendix Table A.\({\rm{1}}\). or you can do it yourself.

Let \({\rm{k = 3}}\)

\(\begin{aligned}{\rm{3 \bullet }}{{\rm{\sigma }}_{\rm{X}}} &= 3 \times 2 {\rm{.236}}\\ &= 6 {\rm{.709}}\end{aligned}\)

As well as we have:

\({\rm{X - 10|}} \ge {\rm{6}}{\rm{.709}}\)

The final inequity has been met.

\({\rm{6}}{\rm{.709}} \le {\rm{X - 10 or (X - 10)}} \ge {\rm{6}}{\rm{.709}}\)

or, alternatively

\({\rm{X}} \le {\rm{3}}{\rm{.291 or X}} \ge {\rm{16}}{\rm{.709}}\)

We need to determine the following probability since X has a Binomial Distribution and only accepts non-negative integer values.

\(\begin{aligned}{\rm{P(X}} \le {\rm{3 or X}} \ge {\rm{17)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(X}} \le {\rm{3) + P(X}} \ge {\rm{17)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P(X}} \le {\rm{3) + (1 - P(X < 17))}}\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{B(3;20,0}}{\rm{.5) + 1 - B(16;20,0}}{\rm{.5)}}\\\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{0}}{\rm{.00129 + 1 - 0}}{\rm{.99871}}\\{\rm{ = 0}}{\rm{.00258}}\end{aligned}\)

(1):the sequence of events is disjointed;

(2):event \({\rm{X }} \ge {\rm{ 17}}\) counterpart is event \({\rm{X < 17}}\);

(3):see Binomial Distribution cdf; \({\rm{X = 16}}\) since \({\rm{X < 17}}\) and \({\rm{X }} \le {\rm{ 16}}\) are the same event;

(4):you can get it from Appendix Table A.\({\rm{1}}\). or you can do it yourself.

The cdf of a binomial random variable X with parameters n and p is the Cumulative Density Function.

\(\begin{array}{c}{\rm{B(x;n,p) = P(X}} \le {\rm{x)}}\\{\rm{ = }}\sum\limits_{{\rm{y = 0}}}^{\rm{x}} {\rm{b}} {\rm{(y;n,p),}}\quad \\{\rm{x = 0,1, \ldots ,n}}\end{array}\)

The Chebyshev鈥檚 inequality is a reminder:

\({\rm{P(|X - \mu |}} \ge {\rm{k\sigma )}} \le \frac{{\rm{1}}}{{{{\rm{k}}^{\rm{2}}}}}\)

The upper bound for \({\rm{k = 2}}\) in our scenario is

\(\begin{array}{c}\frac{{\rm{1}}}{{{{\rm{k}}^{\rm{2}}}}}{\rm{ = }}\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{2}}}}}\\{\rm{ = 0}}{\rm{.25}}\end{array}\)

This is much higher than the chance we were given

\({\rm{0}}{\rm{.25 > 0}}{\rm{.04138}}\)

For \({\rm{k = 3}}\), the upper bound in the second case is

\(\begin{array}{c}\frac{{\rm{1}}}{{{{\rm{k}}^{\rm{2}}}}}{\rm{ = }}\frac{{\rm{1}}}{{{{\rm{3}}^{\rm{2}}}}}\\{\rm{ = 0}}{\rm{.11}}\end{array}\)

which is, once again, far higher than the likelihood we estimated.

\({\rm{0}}{\rm{.11 > 0}}{\rm{.00258}}\)

We have been granted

\({\rm{X}} \sim {\rm{Bin(20,}}{\rm{.75)}}\)

We must first determine the anticipated value and standard deviation.

The anticipated outcome is:

\(\begin{aligned}E(X) &= np \\ &= 20 \times 0 {\rm{.75}}\\ &= 15 \end{aligned}\)

The difference is:

\(\begin{aligned} V(X) &= np(1 - p) \\ &= npq \\ &= 20 \times 0 {\rm{.75 \times 0}}{\rm{.75}}\\ &= 3 {\rm{.75}}\end{aligned}\)

As well as the standard deviation is:

\(\begin{aligned}{{\rm{\sigma }}_{\rm{X}}} &= \sqrt {{\rm{npq}}} \\&= \sqrt {{\rm{3}}{\rm{.75}}} \\ &= 1{\rm{.936}}\end{aligned}\)

Let \({\rm{k = 2}}\) first.

\(\begin{aligned}{\rm{2 \bullet }}{{\rm{\sigma }}_{\rm{X}}} &= 2 \times 1 {\rm{.936}}\\ &= 3 {\rm{.872}}\end{aligned}\)

As well as we have:

\({\rm{|X - 15|}} \ge {\rm{3}}{\rm{.872}}\)

The final inequity has been met.

\({\rm{3}}{\rm{.872}} \le {\rm{X - 15 or (X - 15)}} \ge {\rm{3}}{\rm{.872}}\)

or, alternatively

\({\rm{X}} \le {\rm{11}}{\rm{.128 or X}} \ge {\rm{18}}{\rm{.872}}\)

We need to determine the following probability since X has a Binomial Distribution and only accepts non-negative integer values.

\(\begin{array}{c}{\rm{P(X}} \le {\rm{11 or X}} \ge {\rm{19)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(X}} \le {\rm{11) + P(X}} \ge {\rm{19)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P(X}} \le {\rm{11) + (1 - P(X < 19))}}\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{B(11;20,0}}{\rm{.75) + 1 - B(18;20,0}}{\rm{.75)}}\\\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{0}}{\rm{.04093 + 1 - 0}}{\rm{.97569}}\\{\rm{ = 0}}{\rm{.06524}}\end{array}\)

(1):the sequence of events is disjointed;

(2):event \({\rm{X }} \ge {\rm{ 19}}\) counterpart is event \({\rm{X < 19}}\);

(3):see Binomial Distribution cdf; \({\rm{X = 18}}\) since \({\rm{X < 19}}\) and \({\rm{X }} \le {\rm{ 18}}\) are the same event;

(4):you can get it from Appendix Table A.\({\rm{1}}\). or you can do it yourself.

Let \({\rm{k = 3}}\)

\(\begin{array}{c}{\rm{3 \bullet }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ = 3 \times 1}}{\rm{.936}}\\{\rm{ = 5}}{\rm{.808}}\end{array}\)

As well as we have:

\({\rm{|X - 15|}} \ge {\rm{5}}{\rm{.808}}\)

The final inequity has been met.

\({\rm{5}}{\rm{.808}} \le {\rm{X - 15 or (X - 15)}} \ge {\rm{5}}{\rm{.808}}\)

or, alternatively

\({\rm{X}} \le {\rm{9}}{\rm{.192 or X}} \ge {\rm{20}}{\rm{.808}}\)

We need to determine the following probability since X has a Binomial Distribution and only accepts non-negative integer values.

\(\begin{array}{c}{\rm{P(X}} \le {\rm{9)}}\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{B(9;20,0}}{\rm{.75)}}\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{0}}{\rm{.00394}}\end{array}\)

(1):see Binomial Distribution of cdf

(2):you can get it from Appendix Table A.\({\rm{1}}\). or you can do it yourself.

The Chebyshev鈥檚 inequality is a reminder:

\({\rm{P(|X - \mu |}} \ge {\rm{k\sigma )}} \le \frac{{\rm{1}}}{{{{\rm{k}}^{\rm{2}}}}}\)

The upper bound for \({\rm{k = 2}}\) in our scenario is

\(\begin{array}{c}\frac{{\rm{1}}}{{{{\rm{k}}^{\rm{2}}}}}{\rm{ = }}\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{2}}}}}\\{\rm{ = 0}}{\rm{.25}}\end{array}\)

This is much higher than the chance we were given

\({\rm{0}}{\rm{.25 > 0}}{\rm{.06524}}\)

For \({\rm{k = 3}}\), the upper bound in the second case is

\(\begin{array}{c}\frac{{\rm{1}}}{{{{\rm{k}}^{\rm{2}}}}}{\rm{ = }}\frac{{\rm{1}}}{{{{\rm{3}}^{\rm{2}}}}}\\{\rm{ = 0}}{\rm{.11}}\end{array}\)

which is, once again, far higher than the likelihood we estimated.

\({\rm{0}}{\rm{.11 > 0}}{\rm{.00394}}\)

Therefore, the values are significantly bigger.