Answer to Question #105058 in Statistics and Probability for Nicskiongi

Question

1. For two given set of data, it is known that mean x = 10 and mean of y = 4. The gradient of the regression line y on x is 0.6. Find the equation of the regression line and estimate y when x = 1 (4mks)
2. A discrete random variable has the following distribution
x 1 2 3 4
f(x) k 2k 2/3k 1/3k
i). Determine the value of k (2mks)
ii). Find the expectation of x (2mks)
iii). Find the variance of x (3mks)

3. The arithmetic mean of 10 numbers is 4. When an eleventh number x is added, the overall mean is changed to 5. When a twelfth number y is added, the mean changes to 4. Determine the values of x and y. (5mks)

PROBABILITY
4. a). A manufacturer assures his customers that the probability of having defective items is 0.005. A sample of 1000 items was inspected. Find the probabilities of having the following outcomes.
i). only one is defective
ii). at most 2 defective
iii). more than 3 defective

Answer 1

1. The line of regression of Y on X is given by $Y=A+Bx, A=\bar{y}-B\bar{x}$

Given that $\bar{x}=10, \bar{y}=4, B=0.6.$ Then

$$A=4-0.6(10)=-2$$

The line of regression of Y on X is

$$Y=-2+0.6x$$

Estimate $Y$ when $X=1$

$$Y=-2+0.6(1)=-1.4$$

2. A discrete random variable has the following distribution 

$$\def\arraystretch{1.5} \begin{array}{c:c} x & 1 & 2 & 3 & 4 \\ \hline f(x) & k & 2k & \dfrac{2}{3}k & \dfrac{1}{3}k \end{array}$$

i). Determine the value of k

$$\sum f(x_i)=1$$ $$k+2k+\dfrac{2}{3}k+\dfrac{1}{3}k=1$$ $$k=\dfrac{1}{4}$$ $$\def\arraystretch{1.5} \begin{array}{c:c} x & 1 & 2 & 3 & 4 \\ \hline f(x) & {1 \over 4} & {1 \over 2} & {1 \over 6}& {1 \over 12} \end{array}$$

ii). Find the expectation of x 

$$\mu=E(X)=\sum x_if(x_i)=$$ $$=1({1 \over 4})+2({1 \over 2})+3({1 \over 6})+4({1 \over 12})={25 \over 12}$$

iii). Find the variance of x

$$Var(X)=\sigma^2=E(X^2)-\mu^2$$

$$E(X)=\sum x_i^2f(x_i)=$$ $$=1^2 ({1 \over 4})+2^2({1 \over 2})+3^2({1 \over 6})+4^2({1 \over 12})={61 \over 12}$$

$$Var(X)=\sigma^2={61 \over 12}-({25 \over 12})^2={107 \over 144}$$

3. The arithmetic mean of 10 numbers is 4. When an eleventh number x is added, the overall mean is changed to 5. When a twelfth number y is added, the mean changes to 4. Determine the values of x and y.

Given that $\mu_{10}=4$

$$\mu_{11}={\mu_{10}(10) +x\over 11}=5$$ $${4(10)+x\over 11}=5$$ $$x=15$$

$\mu_{11}=5$

$$\mu_{12}={\mu_{11}(11) +y\over 12}=4$$ $${5(11) +y\over 12}=4$$ $$y=-7$$

4. a) Let $X=$ the number of defective items: $X\sim B(n, p).$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Given that $n=1000, p=0.005.$

The mean value and standard deviation of a binomial random variable $X$ are

$$\mu=np=1000(0.005)=5$$

$$\sigma=\sqrt{np(1-p)}=\sqrt{1000(0.005)(1-0.005)}=\sqrt{4.975}$$

Since the sample size $n$  is very large and $p$ is small we may use Approximation of Binomial Distribution by a Poisson Distribution

$$P(X=x)={e^{-\mu}\mu^x \over x!}$$

i). The probability of having only one defective item is

$$P(X=1)=\binom{1000}{1}0.005^1(1-0.005)^{1000-1}\approx0.033437$$

$$P(X=1)={e^{-5}5^1 \over 1!}=5e^{-5}\approx0.033690$$

ii). The probability of having at most 2 defective items is

$$P(X\leq2)=P(X=0)+P(X=1)+P(X=2)=$$ $$=\binom{1000}{0}0.005^0(1-0.005)^{1000-0}+\binom{1000}{1}0.005^1(1-0.005)^{1000-1}+$$ $$+\binom{1000}{2}0.005^2(1-0.005)^{1000-2}\approx0.124020$$

$$P(X\leq2)={e^{-5}5^0 \over 0!}+{e^{-5}5^1 \over 1!}+{e^{-5}5^2 \over 2!}={37 \over 2}e^{-5}\approx0.124652$$

iii). The probability of having more than 3 defective items is

$$P(X>3)=1-P(X\leq3)=$$ $$=1-P(X=0)-P(X=1)-P(X=2)-$$ $$-P(X=3)=1-\binom{1000}{0}0.005^0(1-0.005)^{1000-0}-$$ $$-\binom{1000}{1}0.005^1(1-0.005)^{1000-1}-\binom{1000}{2}0.005^2(1-0.005)^{1000-2}-$$ $$-\binom{1000}{3}0.005^2(1-0.005)^{1000-3}\approx0.735678$$

$$P(X>3)=1-{e^{-5}5^0 \over 0!}-{e^{-5}5^1 \over 1!}-{e^{-5}5^2\over 2!}-{e^{-5}5^3 \over 3!}\approx$$ $$\approx0.734974$$