Question

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 1

Answer:

Budgeted amount is $446.44

Step-by-step explanation:

Given data:

mean  $\mu = 430$

standard deviation is given as $\sigma = 10$

$P(X\geq x) = 0.05$

$1 - P(X\leq x) = 0.05$

$P(X\leq x) = 0.95$

$P(\frac{X - \mu}{\sigma} \leq \frac{x - \mu}{\sigma}) = 0.95$

$P(Z\leq z) = 0.95$

$\frac{X - \mu}{\sigma} = 1.644$ { by using normal table or online calculator}

$\frac{x - 430}{10} = 1.644$

solving for x we have

x  = 446.44

Budgeted amount is $446.44