Standardise tails of three independent normal variables

Use the completed combination foundation

This calculation combines weighted means, variances and normal tails for three mutually independent inputs. Each source is a different random quantity; repeated use of the same draw must first be collected

Keep three distinct operations

For a weighted sum of three variables, add the signed weighted means and the fixed shift. Under mutual independence, add the squared-weight variance contributions. Normal inputs make the combination exactly normal. Take the positive square root of the variance before standardising

Work an original upper tail

The three normal inputs are mutually independent. Combine them to obtain mean 8, variance 36 and standard deviation 6. Standardise the upper-tail boundary, then find its probability

For independent X∼N(2,4), Y∼N(−1,4) and Z∼N(3,16), let W=X−2Y+Z+1. Give E(W)

Which total variance matches the table?

For independent X∼N(2,4), Y∼N(−1,4) and Z∼N(3,16), let W=X−2Y+Z+1. Give the standard deviation of W

W is normal with mean 8 and standard deviation 6. Give the standardised boundary for W>11

W is normal with mean 8 and standard deviation 6. Give Pr(W>11) to four decimal places

W is normal with mean 8 and standard deviation 6. Give the standardised boundary for W<5

Justify the complete tail calculation, ungraded

For mutually independent X∼N(2,4), Y∼N(−1,4) and Z∼N(3,16), with W=X−2Y+Z+1, derive the mean and variance, state why the output is normal, standardise both events W>11 and W<5, and explain the roles of independence, signed weights and the fixed shift. Explain what would fail if Y were the same draw as X.

Model: assemble the distribution before the event

Linearity gives the mean from the three signed contributions and the fixed shift. Independence removes covariance terms, so the squared-weight contributions give the variance. Standardising the two boundaries gives opposite half-unit scores. Symmetry makes both probabilities 0.3085 to four decimals. Negative weights reverse individual effects on the mean but still contribute their squares to variance. The constant changes only location. If Y is the same draw as X, collect their terms first; the independent-source variance calculation no longer applies