ID: at a speed skating manufacturer in Canada inspects and es can be used at 301

Question

ID: at a speed skating manufacturer in Canada inspects and es can be used at 3018 PyeongChang Olympic Winter Games. Based off the manufacturing process, each skate is produced independently and the probability of each one the technician needs to inspect 8 being wsed is O.8 What is the approximate probability that skates to find the second skate that cannot be used? (a)00734 (6) 0.200 (e) 0 294 () 0.250 (e) 0.0003 intersection 22. Suppose the number of green lights for a traffic light for one dire s a Pisson distribution, where the average number of green lights is once every 30 seconds. What is the probability of ten green lights in 5 minutes? (a) 0.125 (b) 0.175 (c) 0.500 (d) 0.0378 (e) 0.018i 23. Refer to the previous question. If the probability of four green lights occurring over a time peri is approximately 0.168, what is the period of time (in minutes)? (a) 2.5 minutes (b) 3 minutes (c) 4 minutes (d) 1.5 minutes (e) 2 minutes 24. Suppose medals at the 2018 Winter Olympics potentially carry defects due to subtle manufacturing flaws. On average, the flaws occur in the surface material every 88 cm2. If a medal has an area of 19.8 cm, what is the probability of the medal having two flaws? (a) 0.020 (b) 0.799 (c) 0.707 (d) 0.493 (e) 0.123 . Refer to the previous question. For 15 medals, what is the probability that ten of the medals have no flaws? (a) 0.057 (b) 0.098 c) 0.105 ) 0.707 0.085

Explanation / Answer

Question 21

Pr(Being used) = 0.80

Pr(a particular stick not being used) = 0.20

Here in initial seven skates, we discard only one and then discarded the eighth one.

So,

Pr(One discarded in initital seven and discarded the eighth one) = 7C1 (0.2)1(0.8)6 * 0.2 = 0.0734

Option A is correct.

Question 22

Expected number of green lights in 5 minutes = (1/30) * 5 * 60 = 10

so,

Pr(X = 10 ; 10) = POISSON (X = 10 ; 10) = 0.1251

Option A is correct here.

Qustion 23

Here Pr(X = 4 ; ) = 0.168

e-4/4! = 0.168

by trial and error the value of = 3

so period oftime = 3 * 0.5= 1.5 minutes

Question 24

Expected number of flaws in a medal = 19.8/88 = 0.7069

Here number of flaws if assumed x

Pr(X =2 ; 0.7069) = POISSON(X = 2; 0.7069) = 0.1232

Option E is correct

Here as probability of no flaws is

Pr(X = 0; 0.7069) = 0.4932

Now binomial distribution where n = 15 and p = 0.4932

Pr(Y = 10 ; n = 15 ; p = 0.4932) = 15C10 (0.4932)10(1 - 0.4932)5 = 0.0855 (option e is correct here)  

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