The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $430 and standard deviation $10. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05? (Round your answer to the nearest cent.)

Answer:Budgeted amount is $446.44Step-by-step explanation:Given data:mean  [tex]\mu  =  430[/tex]standard deviation is given as [tex] \sigma  = 10[/tex][tex]P(X\geq x) = 0.05[/tex][tex]1 - P(X\leq x) = 0.05[/tex][tex]P(X\leq x) = 0.95[/tex][tex]P(\frac{X - \mu}{\sigma} \leq \frac{x - \mu}{\sigma}) = 0.95[/tex][tex]P(Z\leq z) = 0.95[/tex][tex]\frac{X - \mu}{\sigma} = 1.644[/tex] { by using normal table or online calculator}[tex]\frac{x - 430}{10} = 1.644[/tex]solving for x we havex  = 446.44Budgeted amount is $446.44