Solution :
Given that mean μ = 10 , variance σ^2 = 36
=> standard deviation σ = sqrt(variance)
= sqrt(36)
= 6
(a)
=> P(X > 5) = P((X - μ)/σ > (5 - 10)/6)
= P(Z > -0.8333)
= P(Z < 0.8333)
= 0.7967
(b)
=> P(4 < X < 16) = P((4 - 10)/6 < (X - μ)/σ < (16 -
10)/6)
= P(-1 < Z < 1)
= 0.6826
(c)
=> P(X < 8) = P((X - μ)/σ < (8 - 10)/6)
= P(Z < -0.3333)
= 1 − P(Z < 0.3333)
= 1 − 0.6293
= 0.3707
(d)
=> P(X < 20) = P((X - μ)/σ < (20 - 10)/6)
= P(Z < 1.6667)
= 0.9525
(e)
=> P(X > 16) = P((X - μ)/σ > (16 - 10)/6)
= P(Z > 1)
= 1 − P(Z < 1)
= 1 − 0.8413
= 0.1587
. (Ross 5.15) If X is a normal random variable with parameters μ-T0 and σ2-36, comput...
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