Question

Glass Houses Pty has found that the cost of fitting new sliding doors is normally distributed, with an average cost per door of R12 750, and a standard deviation of R1 095. A sample of 105 new windows is randomly selected. Assuming normal distribution: What is the probability that the average cost of fitting the doors in

the sample is less than R13 000? Interpret your answer.

Answer 1

$\bar{x} = 12750 \\ \sigma= 1095 \\ n=105 \\ P(\bar{x} <13000) = P(Z < \frac{13000-12750}{1095 / \sqrt{105}}) \\ = P(Z< 2.34) \\ = 0.99036$

The probability that the average cost of fitting the doors in the sample is less than R13000 is 0.99036.