Question

paula pidcoe is doing her budget. she discovers that the average miscellaneous expense is $65.00 with a standard deviation of $18.00. what percent of her expenses in this category would she expect to fall between $54.20 and $86.60? the z for $54.20 = the percent of area associated with $54.20 = % the z for $86.60 = the percent of area associated with $86.60 = % adding the two together, the percent of his expenses between $54.20 and $86.60 is %

Explanation:

Step1: Calculate z - score for $54.20

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 65$, $\sigma=18$, and $x = 54.20$. Then $z_1=\frac{54.20 - 65}{18}=\frac{- 10.8}{18}=-0.6$.

Step2: Find the percent of area associated with $z = - 0.6$

Looking up in the standard normal distribution table, the area to the left of $z=-0.6$ is $0.2743$ or $27.43\%$.

Step3: Calculate z - score for $86.60

Using the z - score formula with $x = 86.60$, $\mu = 65$, and $\sigma = 18$. Then $z_2=\frac{86.60 - 65}{18}=\frac{21.6}{18}=1.2$.

Step4: Find the percent of area associated with $z = 1.2$

Looking up in the standard normal distribution table, the area to the left of $z = 1.2$ is $0.8849$ or $88.49\%$.

Step5: Calculate the percent of area between $z=-0.6$ and $z = 1.2$

The percent of area between the two z - scores is $0.8849-0.2743 = 0.6106$ or $61.06\%$.

Answer:

The z for $54.20=-0.6$
The percent of area associated with $54.20 = 27.43\%$
The z for $86.60 = 1.2$
The percent of area associated with $86.60=88.49\%$
Adding the two together, the percent of her expenses between $54.20$ and $86.60$ is $61.06\%$