In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. For a randomly selected home, find the probability that the September energy consumption level is between 1100 kWh and 1225 kWh.

The probability that the September energy consumption level is between 1100 kWh and 1225 kWh is 0.1972

Explanation:

The energy consumption level for a single family in September  is normally distributed, therefore to solve this problem, we are going to use the z score. Z score shows the relationship of a group of values to the mean measured in terms of standard deviation from the mean.

From the question, the mean(m) = 1050 kWh

Standard deviation(s) = 218 kWh

The formula for the z score(z) where x is the raw score is given as:

z =

Therefore to get  the probability that the September energy consumption level is between 1100 kWh and 1225 kWh, we calculate the z score for  1100 kWh and then for 1225 kWh.

For 1100 kWh,

For 1225 kWh,

The probability that the September energy consumption level is between 1100 kWh and 1225 kWh is given by:

P(1100<x<1225) = P(0.23<z<0.8) = 0.78814 - 0.59095 = 0.1972

The probability that the September energy consumption level is between 1100 kWh and 1225 kWh is 0.1972

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)