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Nov 2013 p63 q5
3244
(a) The random variable X is normally distributed with mean 82 and standard deviation 7.4. Find the value of q such that \(P(82-q < X < 82+q) = 0.44\).
(b) The random variable Y is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is given that \(5\mu = 2\sigma^2\) and that \(P(Y < \frac{1}{2}\mu) = 0.281\). Find the values of \(\mu\) and \(\sigma\).
Solution
(a) We know that \(P(82-q < X < 82+q) = 0.44\). This implies \(P(X < 82+q) - P(X < 82-q) = 0.44\). Since the distribution is symmetric, \(P(X < 82+q) = 0.72\) and \(P(X < 82-q) = 0.28\). The standard normal variable \(Z\) corresponding to \(X\) is \(Z = \frac{X - 82}{7.4}\). For \(P(X < 82+q) = 0.72\), \(Z = 0.583\). Therefore, \(\frac{q}{7.4} = 0.583\) which gives \(q = 4.31\).
(b) We have \(5\mu = 2\sigma^2\) and \(P(Y < \frac{1}{2}\mu) = 0.281\). Standardizing, \(Z = \frac{\frac{1}{2}\mu - \mu}{\sigma} = \frac{-0.5\mu}{\sigma}\). From the standard normal table, \(Z = -0.580\). Thus, \(\frac{-0.5\mu}{\sigma} = -0.580\) gives \(\sigma = \frac{0.5\mu}{0.580}\). Substituting \(\sigma\) in \(5\mu = 2\sigma^2\), we solve to find \(\mu = 3.36\) and \(\sigma = 2.90\).
