Based on a survey conducted by Greenfield Online, 25 - 34-year-olds spend the most each week on fast food. The average weekly am
ount of $44 was reported in a May 2009 USA Today Snapshot. Assuming that weekly fast food expenditures are normally distributed with a standard deviation of $14.50, what is a probability that a person in this age group will spend more than $60 on fast food in a week?
In solving for the probability (P) that the 34-year olds
will spend more than $60 on fastfood, the mean and the standard deviation is to
be taken into account. The mean (μ) being the $44 average weekly expenditure; while
the standard deviation (σ) is $14.50. The formula goes this way P(X > 60) =
P( X−μ > 60−44 ) = P ( X−μ/σ > 60−44/14.50).
I found my answer by using desmos. It is the best app for graphing. I wrote the equation and then I looked for the point 18 on the y axis and then I looked where it was on the x axis
