A company produces computer microchips, which have a life expectancy that follows a normal distribution with a mean of 90 months and a standard deviation of 3.7 months.

(a)     If a microchip is guaranteed for 84 months find the probability that it will fail before the guarantee ends.

(b)     The probability that a microchip does not fail before the end of the guarantee is required to be 99 %. For how many months should it be guaranteed?

(c)     A rival company produces microchips where the probability that they will fail after 84 months is 0.88. Given that the life expectancy also follows a normal distribution with standard deviation 3.7 months, find the mean.

(a)     \({\text{P}}(X \leqslant 84) = {\text{P}}(Z \leqslant - 1.62...) = 0.0524\)     (M1)A1     N2

Note: Accept 0.0526.

(b)     \({\text{P}}(Z \leqslant z) = 0.01 \Rightarrow z = - 2.326…\)     (M1)

\({\text{P}}(X \leqslant x) = {\text{P}}(Z \leqslant z) = 0.01 \Rightarrow z = - 2.326…\)

\(x = 81.4\,\,\,\,\,{\text{(accept 81)}}\)     A1     N2

(c)     \({\text{P}}(X \leqslant 84) = 0.12 \Rightarrow z = - 1.1749…\)     (M1)

\({\text{mean is 88.3}}\,\,\,\,\,{\text{(accept 88)}}\)     A1     N2

[6 marks]

A fair amount of students did not use their GDC directly, but used tables and more traditional methods to answer this question. Part (a) was answered correctly by most candidates using any method. A large number of candidates reversed the probabilities, i.e., failed to use a negative z value in parts (b) and (c), and hence did not obtain correct answers.