The average do-it-yourself bedroom makeover costs $475 with a standard deviation of $86 (fictional data).Tanya redoes her daughter's room for $654.What percent of bedroom makeovers cost less than Tanya spent? (Round z score to two decimal places.)
A) 1.88%
B) 38.49%
C) 48.12%
D) 98.12%

Question

Quizplus · Accepted Answer

To find the percentage of bedroom makeovers that cost less than what Tanya spent, we first calculate the z-score using the formula: $z = \frac{X - \mu}{\sigma}$, where $X$ is Tanya's spending ($654), $\mu$ is the average spending ($475), and $\sigma$ is the standard deviation ($86). Plugging in the values, we get $z = \frac{654 - 475}{86} = \frac{179}{86} = 2.08$. Looking up a z-score of 2.08 in a standard normal distribution table or using a calculator, we find that approximately 98.12% of bedroom makeovers cost less than what Tanya spent.

The Average Do-It-Yourself Bedroom Makeover Costs $475 with a Standard