Question

Probability of a Gaussian RV. Given a RV X ~ N (3 ,σ^2) and P{0 < X < 3} = 4.0 but σ is unknown, find P{X < 0}. Life time probability. The lifetime of a light bulb is a RV X ~ N (1000,σ^2). Find σ such that P{800 < X < 1200}= 9.0 . (Hint: use the MATLAB function "norminv").

Answer 1

a. The probability
`\(P\{X < 0\}\)` for the Gaussian random variable
`\(X \sim N(3, \sigma^2)\)` with
`\(P\{0 < X < 3\} = 0.4\) is \(0.2\).`

b. For the lifetime probability of a light bulb
`\(X \sim N(1000, \sigma^2)\)`) with
`\(P\{800 < X < 1200\} = 0.9\),` using the MATLAB function "norminv," we find
`\(\sigma \approx 109.861\).`

Step-by-step explanation:

a. The probability distribution of
`\(X\)` is symmetric around the mean
`\(\mu = 3\).` Since
`\(P\{0 < X < 3\} = 0.4\)`, we know that the probability of being below the mean is half of that, which gives
`\(P\{X < 0\} = 0.2\).`

b. To find
`\(\sigma\)` for the light bulb's lifetime distribution, we use the MATLAB function "norminv" to find the z-scores corresponding to the percentiles
`\(P\{X < 800\}\) and \(P\{X < 1200\}\)`. Solving for
`\(\sigma\)` in the equation
`\(Z = \frac{{X - \mu}}{{\sigma}}\)`, where
`\(Z\)` is the z-score, we obtain
`\(\sigma \approx 109.861\).`

In summary, the calculations involve understanding the symmetry of the normal distribution, using the properties of z-scores, and employing MATLAB functions to find the standard normal distribution's inverse.