Question

. Suppose that the lifetime of a charged electronic device is uniformly distributed in the interval [5, 6] hours. Suppose you take 20 measurements. Compute the following • The probability that the mean over the 10 measurements exceeds 5.6 hours. • The probability that the mean lies between 5.45, and 5.55 hours. • The probability that the mean exceeds 6.1 hours. Think about this last point carefully: Do it first by applying the central limit theorem, but then explain whether this answer makes sense or not.

Answer 1

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X: Lifetime of a changed electronic device x~ unif (5,6) n = 20 F(x) = 546 v(x)= (-5) F(x)= E(T EX;) V(X) = idxmono .0083 P( Meon of 10 measurements exceed 5.6 hours) 1 X-5.5 ~ N(0,1) 0083 = P(x>5.6) = PI - 5.5 5.6 5.5 .0083 -0083 = P( Z> 12.048) = 1-(19.048) 1- Coping cet) PC 8.45 < x <5.55) = 1-1=0 Ars) (same Ad before using CLT) PIX> 6.17 = 0 (Aus) Coxing CLT) Then don't make any sense as the probasility for each case wi zero