Answer to Question #311153 in Statistics and Probability for john

Question

The average monthly electricity bill of households in a particular subdivision is Php3 000 with standard deviation of Php800. Suppose the monthly electricity bill follows a normal distribution. What is the probability that the average monthly electricity bill of 15 households is more than Php3 200?

Answer 1

Let X be a random variable representing the average monthly electricity bill of 15 households, then X ~ N$(3000, ({\frac {800} {\sqrt{15}}})^2)$

$P(X>3200)=P(N(3000, ({\frac {800} {\sqrt{15}}})^2)>3200)=P(3000+{\frac {800} {\sqrt{15}}}N(0,1)>3200)=P(N(0,1)>0.968)=0.16652$