Question

The average electric bill in a residential area is $72 for the month of May. The standard deviation is $6. If the amounts of electric bills are normally distributed, find the probability that the mean of the bill for 15 residents will be less thatn $75

Answer 1

You need to work out the standard deviation of the mean of the bill for 15 residents. That's $6//sqrt(15), which is 1.549. So the distribution of the mean of the bill for 15 residents will be Normal, with mean $72 and standard deviation 1.549. You have to work out the probability that the mean will be less than $75, so you need work out how many standard deviations $75 is above $72, which is ($75 - $72) divided by $1.549, which is 1.936. You now need to know what the area to the left of that point (because you want the probability of the mean bill being less that $75) is, and you can get that from a set of Normal probability tables. If you were looking up 1.96 you would get an answer of 0.975, so you know it's going to be a bit less than that. I've just looked it up myself, and I get 0.9735 - so I reckon that's your answer - i.e. just over 97%.