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June 2011 p61 q5
3264
(a) The random variable X is normally distributed with mean μ and standard deviation σ. It is given that 3μ = 7σ2 and that P(X > 2μ) = 0.1016. Find μ and σ.
(b) It is given that Y ~ N(33, 21). Find the value of a given that P(33 − a < Y < 33 + a) = 0.5.
Solution
(a) We start with the given equation 3μ = 7σ2 and the probability P(X > 2μ) = 0.1016. Standardizing, we have:
\(z = \frac{2μ - μ}{σ} = \frac{μ}{σ}\)
\(From the standard normal distribution table, P(Z > 1.272) = 0.1016, so:\)
\(\frac{7σ}{3} = 1.272\)
Solving for σ, we get:
\(σ = 0.545\)
Substituting back to find μ:
\(μ = \frac{7σ^2}{3} = 0.693\)
(b) Given Y ~ N(33, 21), we need to find a such that P(33 − a < Y < 33 + a) = 0.5. This implies:
\(P(X < a + 33) = 0.75\)
\(From the standard normal distribution table, z = 0.674 for 0.75 probability. Standardizing:\)
