Question a to c ,please help PAGE 21 oP 33 2. (a) 12 marksl A large manufacturin

Question

Question a to c ,please help

PAGE 21 oP 33 2. (a) 12 marksl A large manufacturing company produces electrical fuses. Daily produc- tion contains 5% defectivesThe quality inspector takes a random sample of 10 fuses. Find the probability that the sample contains at least 1 defective fuse. . (b) A missile protection system is set up in a partieular zone. The system consists of n radar sets operating independently. Each set has a probability of 0.95 of detecting () Suppoee that there are 6 radar sets operating in a partieular day (ie. n 6), (a) 2 marksl What is the probability that at least 5 sets detect the missile? a missile which enters the zone. and a missile enters the zone (d) [1 mark! what is the probability that none of the sets detect the missile?

Explanation / Answer

Question 2(a) Pr(Defective) = 0.05

sample size n = 10

Pr(at least one defective fuse) = 1 - Pr(no defective fuse)

Pr(no defective fuse) = 10C0 (0.95)10 = 0.5987

Pr(at least one defective fuse) = 1 - Pr(no defective fuse) =1 - 0.5987 = 0.4013

(b) Pr(Detecting a missile which enters the zone) = 0.95

(i) At least 5 sets detect the missile.

Pr(X >=5) = Pr(X = 5) + Pr(X = 6) = 6C5 (0.05) * (0.95)5 + 6C6 (0.95)6 = 0.2321 + 0.7351 = 0.9672

(ii) Pr(none of the set detect the missile) = 6C0 (0.05)6  = 0.000000054

(iii) Pr(X = 1 l X >= 1) = 6C1(0.05)5 (0.95) / [1 - 6C0 (0.05)6] = 1.78125 x 10-5  

(iv) Here let say n is the number of such missiles.

so, mathematically writing the probability statement

Pr( at least one successful radar set) >= 0.9999

Pr(all radar set shall fail) <= 1 - 0.9999

Pr(all radar set shall fail) <= 0.0001

so Pr(X = 0; n ; 0.95) <= 0.0001

nC0 (0.05)n <= 0.0001

taking log both side

n ln (0.05) <= ln (0.0001)

-2.9957 n <= -9.21

n >= 9.21/2.9957

n >= 3.07

so minimum n would be 4 . so, atleast 4 radar sets shall be there.

(c) Here median breaking strength = 25 units

(i) Here distribution of Y , which is the sum of all Y . so there will be half of the Y shall be greater than 1 and half of the Y shall be less than 1.

so the distribution would be binomial with p = 0.5 and n= number of population.

(ii) Here k = number of values greater than Y = 9

Pr(Y >= 9) = 12C9 (0.5)12 = 0.05371

(iiii) Here p - value is greater than the standard significance value 0.05 so we shall not accept the manufacture's claim that he can build ceramic tiles with higher breaking strength.

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