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.
"\\bar{x} = 12750 \\\\\n\n\\sigma= 1095 \\\\\n\nn=105 \\\\\n\nP(\\bar{x} <13000) = P(Z < \\frac{13000-12750}{1095 \/ \\sqrt{105}}) \\\\\n\n= P(Z< 2.34) \\\\\n\n= 0.99036"
The probability that the average cost of fitting the doors in the sample is less than R13000 is 0.99036.
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