• Go into 2nd DISTR. The syntax for the instructions are as follows:
• To calculate ([latex]x=[/latex] value): binompdf(n, p, number) if “number” is left out, the result is the binomial probability table.
• To calculate P([latex]x\leq[/latex] value): binomcdf(n, p, number) if “number” is left out, the result is the cumulative binomial probability table.
• For this problem: After you are in 2nd DISTR, arrow down to binomcdf. Press ENTER. Enter 20,0.41,12). The result is [latex]P(x\leq12)=0.9738[/latex].

Note

If you want to find [latex]P(x=12)[/latex], use the pdf (binompdf). If you want to find [latex]P(x>12)[/latex], use [latex]1-\text{binomcdf}(20,0.41,12)[/latex].

The probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.

The graph of [latex]X\sim{B}(20,0.41)[/latex] is as follows:

The [latex]y[/latex]-axis contains the probability of [latex]x[/latex], where [latex]X=[/latex] the number of workers who have only a high school diploma.

The number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, [latex]\mu=np=(20)(0.41)=8.2[/latex].

The formula for the variance is [latex]\sigma^{2}=npq[/latex]. The standard deviation is [latex]\sqrt{{{n}{p}{q}}}[/latex].

[latex]\sigma=\sqrt{(20)(0.41)(0.59)}={2.20}[/latex]

1. [latex]x=0,1,2,3,4,5,6,7,8[/latex]
2. [latex]{X}\sim{B}{({100},\frac{{8}}{{560}})}[/latex]
   1. [latex]{P}{({x}={2})}=\text{binompdf}{({100},\frac{{8}}{{560}},{2})}={0.2466}[/latex]
   2. [latex]{P}{({x}\leq{6})}=\text{binomcdf}{({100},\frac{{8}}{{560}},{6})}={0.9994}[/latex]
   3. [latex]{P}{({x}{>}{3})}={1}-{P}{({x}\leq{3})}={1}-\text{binomcdf}{({100},\frac{{8}}{{560}},{3})}={1}-{0.9443}={0.0557}[/latex]
3. Using the formulas, calculate the (a) mean and (b) standard deviation.
   1. [latex]\text{Mean}={n}{p}={({100})}{(\frac{{8}}{{560}})}=\frac{{800}}{{560}}\approx{1.4286}[/latex]
   2. [latex]\text{Standard Deviation}=\sqrt{{{n}{p}{q}}}=\sqrt{{{({100})}{({8560})}{({552560})}}}\approx{1.1867}[/latex]

1. [latex]X\sim{B}(200,0.0128)[/latex]
2. Using the formulas, calculate the (a) mean and (b) standard deviation of [latex]X[/latex].
   1. Mean[latex]=np=200(0.0128)=2.56[/latex]
   2. [latex]\text{Standard Deviation}=\sqrt{{{n}{p}{q}}}=\sqrt{{{({200})}{({0.0128})}{({0.9872})}}}\approx{1.5897}[/latex]
3. Using the TI-83, 83+, 84 calculator:
   [latex]P(x\leq8)=\text{binomcdf}(200,0.0128,8)=0.9988[/latex]
4. [latex]P(x=5)=\text{binompdf}(200,0.0128,5)=0.0707[/latex]
   [latex]P(x=6)=\text{binompdf}(200,0.0128,6)=0.0298[/latex] So [latex]P(x=5)>P(x=6)[/latex]; it is more likely that five people will develop cancer than six.

The names of all committee members are put into a box, and two names are drawn without replacement. The first name drawn determines the chairperson and the second name the recorder. There are two trials. However, the trials are not independent because the outcome of the first trial affects the outcome of the second trial. The probability of a student on the first draw is [latex]\frac{6}{16}[/latex], when the first draw selects a staff member. The probability of drawing a student’s name changes for each of the trials and, therefore, violates the condition of independence.