1. Superbly sun-tanned Scottish super spy Sal Sperry Stake studies the idiosyncr

Question

1. Superbly sun-tanned Scottish super spy Sal Sperry Stake studies the idiosyncracies of her enemies. At the Scottish chool of SuperSpydom, Stake learns that agents of SocetyP romotingEstremeChao ThroughRavinse weapons with a variety of magazine sizes: 20% use guns that hold 8 rounds, 30% carry 9-round weapons, 30% have 10-round weapons, and the remainder are split evenly between 12 and 15-round weapons. Let X number of rounds in a SPECTRE agent's weapon a. What proportion of agents carry: i. 10 or fewer rounds? . 15 rounds? (.6) b. Make a probability table for X; find its mean and standard deviation. c. Stake fights her way through a dozen SPECTRE agents: what is the (5,5) (6,8,6) (10) robability that at least 5 of them carry weapons of more than 10 rounds? d. Stake faces down a gauntlet of 36 SPECTRE agents. What are the chances ) (10) 3 that these agents have fewer than: i. 400 total bullets? i. 25 bullets remaining after Stake counts a fusillade of 350 bullets fired at her? (55)

Explanation / Answer

a. Proportion of agents carry

(i) 10 or fewer rounds = Pr(8 Round) + Pr( 9 - round)+ Pr( 10 round) =0.20 + 0.30 + 0.30 = 0.80

(ii) Pr(15 rounds) = 1/2 * ( 1- 0.2 - 0.3 - 0.3) = 0.10

(b) P(X) = 0.20 ; X = 8

= 0.30 ; X = 9

= 0.30 ; X = 10

= 0.10 ; X = 12

= 0.10 ; X = 15

E(X) = 0.20 * 8 + 0.30 * 9 + 0.30 * 10 + 0.10 * 12 + 0.10 * 15 = 10 rounds

Var(X) = 0.20 * (8 - 10)2 + 0.30 * (9 -10)2 + 0.30 * (10 -10)2 + 0.10 * (12 -10)2 + 0.10 * (15 -10)2 = 4

STD(X) = sqrt(4) =2 Rounds

(c) Here Pr(X > 10 rounds) = 0.2

and number of specctre agents = 12

so this is a binomial distribution where n = 12 and p = 0.2

If Y is the number of SPECTRE agents carrying more than 10 rounds

so, Pr(Y >=5 ; 12 ; 0.2) = BIN(Y >=5 ; 12; 0.2) = 0.0726

(d) Here Stake faces a gauntlet of 36 SPECTRE agents. Let say sum of rounds carried by these agents are Z.

so, E(Z) = E(36X) = 36 * 10 = 360 rounds

Var(Z) = 36 Var(X) = 36 * 4 = 144

STD(Z) = 12

so, Now as n > 36 ; by using central limit theorem and law of big numbers, we can say that sum of these rounds have normal distribution.

Pr(Z < 400) = NORMAL(Z < 400 ; 360 ; 12)

Z = (400 - 360)/12 = 3.3333

Pr(Z < 400) = NORMA(Z < 400 ; 360 ; 12) = Pr(Z < 3.3333) = 0.99957

(b) less then 25 + 350 = 375 bullets.

Pr(Z < 375) = NORMAL(Z < 375 ; 360 ; 12)

Z = (375 - 360)/12 = 1.25

Pr(Z < 375) = NORMA(Z < 375 ; 360 ; 12) = Pr(Z < 1.2) = 0.8849

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