The breaking strains of cables produced by a factory are normally distributed with mean m Newtons and standard deviation s Newtons. Only 2.5% of all cables break at a strain of less than 6416 Newtons and 20% break at more than 7537 Newtons. (a) Show that m - 6416 = 1.96s. (2) (b) Obtain a second equation in m and s. (3) (c) Find the values of m and s correct to 2 significant figures. (3) (d) Find the values between which the middle 90% of breaking strains lie. (6)