We first calculate the z-score corresponding to x = 1075 kWh. Given the mean of 1050 kWh, SD of 218 kWh, and sample size of n = 50, the formula for z is:
z = (x - mean) / (SD/sqrt(n)) = (1075 - 1050) / (218/sqrt(50)) = 0.81
From a z-table, the probability that z > 0.81 is 0.2090. Therefore, the probability that the mean of the 50 households is > 1075 kWh is 0.2090.
For the first one x must not = -3 because that would make the denominator zero.
second question This simplifies to u / (u+2)