Answer: 0.0668
Explanation:
Given the following :
Mean (m) Cost = RS385
Standard deviation (s) = RS110
Assume a normal distribution, Probability that domestic airfare is RS550 or more?
P(x > 550)
Find the z - score
Z - score = (x - m) / s
Where x = 550
Z = (550 - 385) / 110
Z = 165 / 110 = 1.5
P(Z > 1.5) = 1 - P(Z ≤1.5)
Using the z table : 1.5 = 0.9332
Therefore,
1 - P(Z ≤1.5) = 1 - 0.9332 = 0.0668
P(x > 550) = 0.0668