Answer to Question #167095 in Statistics and Probability for Scarlett

Question

Q2(a) Suppose there are 8 multiple choices questions in a Statistics test. Each question has 5 possible

answers, and only one of them is correct. Find the probability of having at least 7 correct answers if a

student attempts to answer every question at random. (3 marks)

(b) Suppose that the number of patients admitted into Hospital K in an hour follows a Poisson distribution

with a mean of 5. What is the probability that more than 4 patients admitted into Hospital K in a given

30-minute interval? (4 marks)

(c) A batch of 2-cm nails are produced, where the lengths of 2-cm nails are normally distributed with a

mean of 2.0 cm and a standard deviation of 0.05 cm. The nails that are either shorter than 1.9 cm or

longer than 2.1 cm are unusable.

(i) Find the probability of all nails produced by this machine which are unusable. (4 marks)

(ii) Find k such that 80% of the nails will have the length more than k cm. (3 marks)

(d) In a 1000 replacement parts for a given assembly, 15% of the replacement parts are defective. Out of

these 1000 parts, 60% were made by the company itself, and the rest were bought from external

sources. Of those bought from external sources, 85% are good.

(i) Draw a well- labelled probability tree diagram with the joint probabilities calculated to

illustrate this situation. (5 marks)

(ii) If a part is randomly selected from this stock, what is the probability that

(1) the part is either bought or defective; (3 marks)

(2) the part is company made, given that it is good. (3 marks)

Answer 1

(a) $p=0.2$

$$P(X\geq7)=P(X=7)+P(X=8)$$

$$=\dbinom{8}{7}(0.2)^7(1-0.2)^1+\dbinom{8}{8}(0.2)^8$$

$$=0.00008448$$

(b) $X\sim Po(\lambda t)$

Given $\lambda t=5\cdot0.5=2.5$

$$P(X>4)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)-P(X=4)$$

$$=1-\dfrac{e^{-2.5}\cdot2.5^0}{0!}-\dfrac{e^{-2.5}\cdot2.5^1}{1!}$$

$$-\dfrac{e^{-2.5}\cdot2.5^2}{2!}-\dfrac{e^{-2.5}\cdot2.5^4}{4!}-\dfrac{e^{-2.5}\cdot2.5^3}{3!}$$

$$\approx1-0.08208-0.20521-0.25652$$

$$-0.21376-0.13360\approx0.1088$$

(c)

Given $\mu=2.0\ cm, \sigma=0.05\ cm.$

(i)

$$P(X<1.9)=P(Z<\dfrac{1.9-2}{0.05})=P(Z<-2)$$

$$\approx0.02275$$

$$P(X>2.1)=1-P(X\leq2.1)$$

$$=1-P(Z\leq\dfrac{2.1-2}{0.05})=1-P(Z<2)$$

$$\approx0.02275$$

$$P(unusable)=0.02275+0.02275=0.0455$$

(ii)

$$P(X>k)=1-P(X\leq k)$$

$$=1-P(Z\leq\dfrac{k-2}{0.05})=0.8$$

$$\dfrac{k-2}{0.05}\approx-0.841621$$

$$k=1.958\ cm$$

(d)

Let  $A=$ event that replacement part was made by the company itself,

$B=$ event that replacement part was were bought from external sources,

$D=$ event that replacement part is defective,

$D|A=$ event that replacement part made by the company itself is defective,

$D|B=$ event that replacement part bought from external sources is defective

Given $n=1000,$

$P(D)=0.15$

$P(A)=0.6, P(B)=1-P(A)=0.4,$

$P(\bar{D} | B)=0.85$

(i) $|S|=1000, |A|=600, |B|=400$

$|D|=150, |D|B|=60, |D|A|=90$

Then

$P(A)=0.6, P(B)=0.4,$

$P(D|A)=0.15, P(\bar{D}|A)=0.85,$

$P(D|B)=0.15, P(\bar{D}|B)=0.85.$

(ii)

(1)

$$P(B\cup D)=P(B)+P(D)-P(B\cap D)$$

$$=P(B)+P(D)-P(D|B)P(B)=$$

$$=0.4+0.15-0.15(0.4)=0.49$$

(2) Baye’s Rule states:

$$P(B|\bar{D})=\dfrac{P(\bar{D}|B)P(B)}{P(\bar{D}|A)P(A)+P(\bar{D}|B)P(B)}$$

$$=\dfrac{0.15(0.4)}{0.15(0.6)+0.15(0.4)}=0.4$$