Two instruments are used to measure the height, [math]h[/math], of a tower. The error made by the less accurate instrument is normally distributed with mean 0 and standard deviation 0.0056 [math]h [/math]. The error made by the more accurate instrument is normally distributed with mean 0 and standard deviation 0.0044 [math]h[/math]. The errors from the two instruments are independent of each other.

Calculate the probability that the average value of the two measurements is within 0.005 [math]h[/math] of the height of the tower.

• 0.38
• 0.47
• 0.68
• 0.84
• 0.90

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: D

Let [math]X_1[/math] and [math]X_2[/math] denote the measurement errors of the less and more accurate instruments, respectively. If [math]N(\mu, \sigma ) [/math] denotes a normal random variable with mean [math]\mu [/math] and standard deviation [math]\sigma [/math], then we are given [math]X_1[/math] is [math]N(0, 0.0056h)[/math] and [math]X_1,X_2[/math] are independent. It follows that

[[math]]Y = \frac{X_1 + X_2}{2} [[/math]]

is

[[math]] N(0, \sqrt{\frac{0.0056^2h^2 + 0.0044^2h^2}{4}}) = N(0, 0.00356h ). [[/math]]

Therefore,

[[math]] \begin{align*} \operatorname{P}[-0.005h \leq Y \leq 0.005h] &= \operatorname{P}[Y \leq 0.005h] - \operatorname{P}[Y \leq -0.005h] \\ &= \operatorname{P}[Y \leq 0.005h] - \operatorname{P}[Y \geq 0.005h] \\ &= 2\operatorname{P}[Y \leq 0.005h] - 1 \\ &= 2\operatorname{P}[Z \leq \frac{0.005h}{0.00356h}] - 1\\ &= 2\operatorname{P}[Z \leq 1.4] -1 \\ & = 2(0.9192) - 1\\ &= 0.84. \end{align*} [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.