If x is a binomial random variable, use Table I in Appendix D to find the following probabilities:

a.for n = 10, p = .4

b.for n = 15, p = .6

c.for n = 5, p = .1

d.for n = 25, p = .7

e.for n = 15, p = .9

f.for n = 20, p = .2

a.

Considering the specific value of n (number of trials), p(number of successes), and x(binomial random variable), the formula for calculating $Pl(X=2)$is shown below.

$P\left(x=2\right)=P\left(x⩽2\right)-P\left(x⩽1\right)$

By subtracting $\text{P}\left(\text{x}⩽\text{1}\right)$from $\text{P}\left(\text{x}⩽\text{2}\right)$, the value of $\text{P}\left(\text{x = 2}\right)$can be found.

The calculation of is shown below.

$\begin{array}{rcl}P\left(x=2\right)& =& P\left(x⩽2\right)-P\left(x⩽1\right)\\ & =& 0.167-0.046\\ & =& 0.121\\ & & \end{array}$

The computed value of $\text{P}\left(\text{x = 2}\right)$is 0.121.

Considering the specific value of n as 15 and that of p as 0.6, the formula for calculating$\mathbf{P}\mathbf{(}\mathit{x}\mathbf{\le }\mathbf{5}\mathbf{)}$ is shown below.

$P\left(x\le 5\right)=\sum _a^{5}p\left(x\right)$

InTable 1, when n is 15 and p is 0.6,the value along the row containing k as 5 is shown below.

$\begin{array}{rcl}P\left(x⩽5\right)& =& \sum _{i=0}^{5}p\left(x\right)\\ & =& 0.034\\ & & \end{array}$

The value of$\mathbf{\text{P}}\left(\text{x}⩽\text{5}\right)$is 0.343.

By considering the specific value of n as 5 and that of p as 0.1, the method for calculating $\text{P}\left(\text{x > 1}\right)$is shown below.

$P\left(x>1\right)=\sum _{i=2}^{5}p\left(x\right)$

InTable 1, when n is 5 and p is 0.1, the value of$\text{P}\left(\text{x}⩽\text{1}\right)$ must be subtracted from 1 to get the value of $\text{P}\left(\text{x > 1}\right)$:

$\begin{array}{rcl}P\left(x>1\right)& =& \sum _{i=2}^{5}p\left(x\right)\\ & =& 1-P\left(x⩽1\right)\\ & =& 1-0.999\\ & =& 0.001\\ & & \end{array}$

The computed value of$\text{P}\left(\text{x > 1}\right)$ is 0.001.

By considering the specific value of n as 25 and that of p as 0.7, the method for calculating $p\left(x<10\right)$is shown below.

$\begin{array}{rcl}P\left(x<10\right)& =& \sum _{i=0}^{9}p\left(x\right)\\ & =& P\left(x⩽9\right)\\ & & \end{array}$

For calculating $p\left(x\le 10\right)$$\text{P}\left(\text{x}⩽\text{9}\right)$,will be considered, and the calculation is shown below.

$\begin{array}{rcl}P\left(x<10\right)& =& P\left(x⩽9\right)\\ & =& 0.000\\ & & \end{array}$

The value of$\text{P}\left(\text{x > 1}\right)$will therefore be 0.000.

By considering the specific value of n as 15 and that of p as 0.9, the method for calculating $\text{P}\left(\text{x}⩾\text{10}\right)$is shown below.

$\begin{array}{rcl}P\left(x⩾10\right)& =& \sum _{i=10}^{15}p\left(x\right)\\ & =& 1-P\left(x⩽9\right)\\ & & \end{array}$

For calculating$\text{P}\left(\text{x}⩾\text{10}\right)$$\mathbf{\text{P}}\left(\text{x}⩽\text{9}\right)$ ,must be subtracted from 1.Thecalculation is shown below.

$\begin{array}{rcl}P\left(x⩾10\right)& =& \sum _{i=10}^{15}p\left(x\right)\\ & =& 1-P\left(x⩽9\right)\\ & =& 1-0.002\\ & =& 0.998\\ & & \end{array}$

The value of $\mathbf{\text{P}}\left(\text{x}⩾\text{10}\right)$will therefore be 0.998.

Considering the specific value of n as 20 and that of p as 0.2, the formula for calculating$\text{P}\left(\text{x = 2}\right)$is shown below.

$P\left(x=2\right)=P\left(x⩽2\right)-P\left(x⩽1\right)$

By subtracting the table value $\text{P}\left(\text{x}⩽\text{1}\right)$from that of $\text{P}\left(\text{x}⩽\text{2}\right)$, the value of $\mathbf{\text{P}}\left(\text{x = 2}\right)$can be found.

The calculation of $\mathit{P}\mathit{l}\mathbf{(}\mathit{X}\mathbf{=}\mathbf{2}\mathbf{)}$with the above formula is shown below.

role="math" localid="1655726498573" $\begin{array}{rcl}P\left(x=2\right)& =& P\left(x⩽2\right)-P\left(x⩽1\right)\\ & =& 0.206-0.069\\ & =& 0.137\\ & & \end{array}$$\begin{array}{rcl}P\left(x=2\right)& =& P\left(x⩽2\right)-P\left(x⩽1\right)\\ & =& 0.206-0.069\\ & =& 0.137\\ & & \end{array}$

The computed value of$\mathbf{\text{P}}\left(\text{x = 2}\right)$is 0.137.