Question

Let 𝑋₁,𝑋₂, 𝑋₃ be 𝑖.𝑖.𝑑. random variables, each having the 𝑁(2, 4) distribution. If 𝑃(2𝑋₁-3𝑋₂+6𝑋₃>17)=1-Φ(𝛽), then 𝛽 equals ________

Answer 1

Correct Answer: 0.5

We have independent identically distributed (i.i.d.) random variables X₁, X₂, X₃ each following a normal distribution N(2, 4). Our goal is to find 𝛽 such that:

𝑃(2𝐗₁−3𝐗₂+6𝐗₃>17)=1−Φ(𝛽) 

Step 1: Determine the distribution of 2𝐗₁−3𝐗₂+6𝐗₃.

Since each 𝐗_i follows 𝑁(2,4), the linear combination 2𝐗₁−3𝐗₂+6𝐗₃ is also normally distributed. The mean is calculated as:

Mean: 2(2) − 3(2) + 6(2) = 4 − 6 + 12 = 10

The variances add, so:

Variance: (2²)(4) + (−3)²(4) + (6²)(4) = 16 + 36 + 144 = 196

Thus, the standard deviation is √196 = 14. Therefore, 2𝐗₁−3𝐗₂+6𝐗₃ ~ 𝑁(10, 196).

Step 2: Standardize the inequality.

The inequality 𝑃(2𝐗₁−3𝐗₂+6𝐗₃>17) can be standardized using the normal distribution:

𝑃(𝑍 > (17−10)/14) where 𝑍 follows 𝑁(0,1)

So, 𝑃(𝑍 > 0.5) = 1−Φ(0.5)

Step 3: Conclusion.

Comparing, 𝛽 = 0.5. Ensure this meets the range provided. The required 𝛽 value of 0.5 is within the expected range: [0.5, 0.5].

Therefore, 𝛽 = 0.5.