Answer:
Hello here is the missing data to your question
x P(x)
0 0.167
1 0.3289
2 0.2801
3 0.149
4 0.0382
5 0.0368
The theoretical values are lower than the approximate values and the property illustrated is the mean and standard deviation
A) 1.6729
B) 1.234061
C) 1.84
D) 1.462874
Step-by-step explanation:
A) calculate the theoretical mean
theoretical mean (u) = summation of ; [tex]x_{i}p(x_{i})[/tex]
= 0(0.167)+1(0.3289)+2(0.2801)+3(0.149)+4(0.0382)+5(0.0369)
= 1.6729
B) calculate the theoretical standard deviation
theoretical standard deviation = [tex]\sqrt{variance}[/tex] = [tex]\sqrt{[summation of x_{i}^2 p(x_{i} ) ]-u^2}[/tex]
= [tex]\sqrt{[0+0.3289+1.1204+1.341+0.6112+0.92]-(2.798594)}[/tex]
= 1.234061
C) calculate the approximate mean
the generated data used for calculation is added below
mean (x) = [tex]\frac{summation of (x_{i}) }{n}[/tex] = [tex]\frac{1+2...+2+1}{25}[/tex] = 46/25 = 1.84
D) calculate the approximate standard deviation
std = [tex]\sqrt{variance}[/tex]
= [tex]\sqrt{\frac{summation of( x_{i}-x )^2}{n-1} }[/tex]
= [tex]\sqrt{2.14}[/tex] = 1.462874
