2. Subtly suave Scot super spy Sal Sperry Stake receives an extortion ultimatum

Question

2. Subtly suave Scot super spy Sal Sperry Stake receives an extortion ultimatum from EvilEye Malocchio et scTVao)until Scotland meets their demands, SPECTRE will randomly interrupt a BBC programme daily with Punky Brewster re-runs. Let X be the time in hours past midnight at which the BBC signal is pirated b. What are the mean and standard deviation of X? c. What is the chance that the pirated signal interrupts: i. six hours of daily (6) (5,5) BBC World News broadcasts? ii. the 7-10 pm prime-time lineup? d. If Scotland refuses to bow to terrorism for 100 days, what are the chances that the proportion of times the 6 hours of daily BBC news broadcasts are interrupted is: i, more than 0.3? ii. between 0.24 and 0.28? iii. less than 0.275? (10,10,10)

Explanation / Answer

Question (2)

(a) Here the UNIFORM DISITRIUBTION will describes X. It is uniform because the SPECTRE will randomly interrrupt the BBC programme.

so Here f(x) = 1/24 ; 0 < X < 24

CDF

F(X) = x/24 ; 0 < X < 24

(b) Mean of X ; E(X) = (24 + 0)/2 = 12 HOur

Var(X) = (24 -0)2 /12 = 48

STDDEV(X) = 6.9282 HOurs

(c) (i) It will disturb six hours of daily BBC world news broadcasts.

Pr(six hours of daily BBC wold news broadcasts would be interrupted) = 6/24 = 1/4 = 0.25

(ii) Pr(7- 10 pm prime -time setup) = (10 -7)/24 = 3/24 = 1/8 = 0.125

(d) Here number of days that thing continued = 100 days

so that means as per Central LImit Theorem we can assume distribution of time that would be correupted would be as per normal distriubtion.

Expected prportion of time BBC news broadcasts are interrupted = 0.25

Standard error of that proportion = sqrt (0.25 * 0.75/100) = 0.0433

(i) Pr(p > 0.3) = BIN (p > 0.3; 0.25 ; 0.0433)

Z = ( 0.3 - 0.25)/ 0.0433 = 1.1547

Pr(p > 0.3) = 1 - Pr(Z < 1.1547) = 1 - 0.8759 = 0.1241

(b) Pr( 0.24 < p < 0.28) = BIN ( p < 0.28 ; 0.25 ; 0.0433) - BIN (p < 0.24 ; 0.25; 0.0433)

= (Z2) - (Z1 )

where is the standard normal cumulative Distriubtion.

Z2 = (0.28 - 0.25)/ 0.0433 = 0.6928

Z1 = (0.24 - 0.25)/ 0.0433 = -0.231

Pr( 0.24 < p < 0.28) = Pr(Z < 0.6928) - Pr(Z < - 0.231) = 0.7558 - 0.4087 = 0.3471

(iii) Pr( p < 0.275) =  (Z)

Z = (0.275 - 0.25)/ 0.0433 = 0.5774

Pr(p < 0.275) = Pr(Z < 0.5774) = 0.7182

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