a) The length of a selected subcomponent is normally distributed with a mean of 112 cm and a standard deviation of 5.5 cm. We are required to find the probability that one selected subcomponent is longer than 117 cm. We will use the normal distribution formula to solve the problem as follows:Given the z-score formula, we get z=(x-μ)/σ=(117-112)/5.5 = 0.91We can look at the z-table to find out the probability that corresponds to this z-score.Z table shows that the area to the left of the z-score of 0.91 is 0.8186.Therefore, the probability that one selected subcomponent is longer than 117 cm is 0.1814. (1-0.8186)=0.1814. (Round your answer to four decimal places.)b) The lengths of subcomponents are independent and identically distributed. The sample size is greater than 30. So, we will use the central limit theorem to estimate the mean length of four subcomponents as follows: μx = μ = 112σx = σ/√n = 5.5/√4 = 2.75z = (x - μx)/σx = (117 - 112)/2.75 = 1.82Now, we can use the z-table to find the area to the right of z-score 1.82. The area is 0.0344.Therefore, the probability that the mean length of four subcomponents exceeds 117 cm is 0.0344. (Round your answer to four decimal places.)c) We will use the formula for the probability of four subcomponents with lengths exceeding 117 cm, which is: P(X>117) = P(X <117)⁴ = [1- P(X>117)]⁴From (a), we know that the probability that one selected subcomponent is longer than 117 cm is 0.1814.P(X > 117) = 1 - 0.1814 = 0.8186P(X < 117)⁴ = (0.8186)⁴ = 0.4364Therefore, the probability that if four subcomponents are randomly selected, all four have lengths that exceed 117 cm is 0.4364. (Round your answer to four decimal places.)