Question

at a city pool, the mean number of patrons who come is 81 per day with a standard deviation of 16. at the same pool, the mean revenue (amount of money brought in) is $450 per day with a standard deviation of $25. find the cvar for each to decide which has more variability, number of patrons or the amount of revenue. for samples, cvar = \frac{s}{\bar{x}} \cdot 100. number of patrons has more variability. revenue has more variability.

Explanation:

Step1: Calculate CVar for number of patrons

The formula for coefficient of variation (CVar) is $CVar=\frac{s}{\bar{x}}\times100$. Given that $\bar{x}=81$ (mean number of patrons) and $s = 16$ (standard - deviation of number of patrons). Substitute the values: $CVar_{patrons}=\frac{16}{81}\times100\approx19.75$.

Step2: Calculate CVar for revenue

Given that $\bar{x}=450$ (mean revenue) and $s = 25$ (standard - deviation of revenue). Substitute into the CVar formula: $CVar_{revenue}=\frac{25}{450}\times100=\frac{2500}{450}\approx5.56$.

Step3: Compare the CVars

Since $19.75>5.56$, the number of patrons has a higher coefficient of variation.

Answer:

Number of patrons has more variability.