Answer to Question #141023 in Statistics and Probability for Margaret

Question

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)

Answer 1

$X\sim Bin(n, p)$

Given $n=6, p=0.25$

1.

$$\mu=np=6(0.25)=1.5$$

The mean $u=1.5$

2.

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

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

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

3.

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

$$=\dfrac{6!}{3!(6-3)!}(\dfrac{1}{64})(\dfrac{27}{64})=\dfrac{135}{1024}=0.1318359375$$

$P(X=3)=0.1318359375$

4.

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

$$=\dfrac{6!}{4!(6-4)!}(\dfrac{1}{256})(\dfrac{9}{16})=\dfrac{135}{4096}=0.032958984375$$

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

$$=\dfrac{6!}{5!(6-5)!}(\dfrac{1}{1024})(\dfrac{3}{4})=\dfrac{9}{2048}=0.00439453125$$

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

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

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

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

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

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

5.

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

$$=\dfrac{6!}{2!(6-2)!}(\dfrac{1}{16})(\dfrac{81}{256})=\dfrac{1215}{4096}=0.296630859375$$

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

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

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

$$=6(\dfrac{1}{4})(\dfrac{243}{1024})=\dfrac{729}{2048}=0.35595703125$$

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

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

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

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