tonya3498
31.01.2020 •
Mathematics
Based on a survey conducted by greenfield online, 25 - 34-year-olds spend the most each week on fast food. the average weekly amount 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?
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Ответ:
The correct answer is P = 0.1357.
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).
Ответ:
Is this geomatry?
Step-by-step explanation: