Answer to Question #141023 in Statistics and Probability for Margaret

Question #141023
Six trial are conducted in abinomial process in who the probability of success in a given trial is 0.25 .If x= the number of the following statement is incorrect
1. The mean u=15
2. The standard deviation a=1.125
3. P (X=3)
4. P (X-4)
5. P (X-2)
1
Expert's answer
2020-11-03T11:29:44-0500

XBin(n,p)X\sim Bin(n, p)

Given n=6,p=0.25n=6, p=0.25

1.


μ=np=6(0.25)=1.5\mu=np=6(0.25)=1.5

The mean u=1.5u=1.5


2.


Var(X)=σ2=np(1p)=6(0.25)(10.25)=1.125Var(X)=\sigma^2 =np(1-p)=6(0.25)(1-0.25)=1.125

standard deviation=σ=1.125=0.7521.06066\text{standard deviation}=\sigma=\sqrt{1.125}=0.75\sqrt{2}\approx1.06066

The standard deviation a=0.7521.060661.125a=0.75\sqrt{2}\approx1.06066\not=1.125

3.


P(X=3)=(63)(0.25)3(10.25)63=P(X=3)=\dbinom{6}{3}(0.25)^3(1-0.25)^{6 -3}=

=6!3!(63)!(164)(2764)=1351024=0.1318359375=\dfrac{6!}{3!(6-3)!}(\dfrac{1}{64})(\dfrac{27}{64})=\dfrac{135}{1024}=0.1318359375

P(X=3)=0.1318359375P(X=3)=0.1318359375


4.

P(X=4)=(64)(0.25)4(10.25)64=P(X=4)=\dbinom{6}{4}(0.25)^4(1-0.25)^{6 -4}=

=6!4!(64)!(1256)(916)=1354096=0.032958984375=\dfrac{6!}{4!(6-4)!}(\dfrac{1}{256})(\dfrac{9}{16})=\dfrac{135}{4096}=0.032958984375


P(X=5)=(65)(0.25)5(10.25)65=P(X=5)=\dbinom{6}{5}(0.25)^5(1-0.25)^{6 -5}=

=6!5!(65)!(11024)(34)=92048=0.00439453125=\dfrac{6!}{5!(6-5)!}(\dfrac{1}{1024})(\dfrac{3}{4})=\dfrac{9}{2048}=0.00439453125

P(X=6)=(66)(0.25)6(10.25)66=P(X=6)=\dbinom{6}{6}(0.25)^6(1-0.25)^{6 -6}=

=14096=0.000244140625=\dfrac{1}{4096}=0.000244140625


P(X>4)=P(X=5)+P(X=6)=P(X>4)=P(X=5)+P(X=6)==0.004638671875=0.004638671875

P(X4)=P(X=4)+P(X=5)+P(X=6)=P(X\geq4)=P(X=4)+P(X=5)+P(X=6)==0.03759765625=0.03759765625

P(X4)=1P(X=5)P(X=6)=P(X\leq4)=1-P(X=5)-P(X=6)==0.995361328125=0.995361328125


P(X<4)=1P(X4)=P(X<4)=1-P(X\geq4)==0.96240234375=0.96240234375

5.

P(X=2)=(62)(0.25)2(10.25)62=P(X=2)=\dbinom{6}{2}(0.25)^2(1-0.25)^{6 -2}=

=6!2!(62)!(116)(81256)=12154096=0.296630859375=\dfrac{6!}{2!(6-2)!}(\dfrac{1}{16})(\dfrac{81}{256})=\dfrac{1215}{4096}=0.296630859375




P(X=0)=(60)(0.25)0(10.25)60=P(X=0)=\dbinom{6}{0}(0.25)^0(1-0.25)^{6 -0}=

=7294096=0.177978515625=\dfrac{729}{4096}=0.177978515625




P(X=1)=(61)(0.25)1(10.25)61=P(X=1)=\dbinom{6}{1}(0.25)^1(1-0.25)^{6 -1}=

=6(14)(2431024)=7292048=0.35595703125=6(\dfrac{1}{4})(\dfrac{243}{1024})=\dfrac{729}{2048}=0.35595703125

P(X<2)=P(X=0)+P(X=1)=P(X<2)=P(X=0)+P(X=1)==0.533935546875=0.533935546875

P(X2)=P(X=0)+P(X=1)+P(X=2)=P(X\leq2)=P(X=0)+P(X=1)+P(X=2)==0.83056640625=0.83056640625

P(X>2)=1P(X2)=P(X>2)=1-P(X\leq2)==0.16943359375=0.16943359375


P(X2)=1P(X<2)=P(X\geq2)=1-P(X<2)==0.466064453125=0.466064453125


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