Question

According to cbs money watch, the average monthly household cellular phone bill is $100. suppose monthly household cell phone bills are normally distributed with a standard deviation of $11.35.
a. what is the probability that a randomly selected monthly cell phone bill is more than $127

Answer 1

0.0087, or 0.87% In order to solve this problem, you first need to determine how many standard deviations from the norm you are. You do this by subtracting the mean from the number you have and then divide the result by the standard deviation. 127 - 100 = 27 27 / 11.35 = 2.378854626 So you're about 2.379 standard deviations above the mean. Using a standard normal table (usually good to 2 decimal digits), or you can use an online calculator or the equivalent. Looking up 2.38 in a standard normal table, I get a value of 0.9913 which is the probability of getting a value LESS than the standard deviation I looked up. Since we're looking for the probability of getting a LARGER value, subtract the lookup value from 1, giving: 1 - 0.9913 = 0.0087 = 0.87% Using an online calculator for the standard deviation of 2.378854626, I get the return value of 0.9913167383 which is basically the same result as using a normal table except for a few extra decimal places. In any case, the probability of a randomly selected monthly cell phone bill being more than $127 is 0.0087, or 0.87%