Question

a. One week, Park Street's revenues were $7500, which was the 15th highest revenue recorded for that restaurant. In the same wook, Bridge Road's
revenue was $7100, the 12th highest for that restaurant Use percentiles and 7-scores to compare how successful each restaurant was that wook,
relative to their typical weekly revenue

Answer 1

Answer:

The percentile rank for Park Street's revenues this week is 60th.

The percentile rank for Bridge Road's revenues this week is 73rd.

Step-by-step explanation:

The missing information are as follows:

Variable     N       Mean       SD

  Park        36       6611       3580

Bridge      40      5989       1794

A z-score (aka, a standard score) specifies the number of standard deviations an observation is from the mean.  

The formula to compute the z-score is, $Z = \frac{(X - \mu)}{\sigma}$, where X = observation, µ = mean, σ = standard deviation.

Compute the z-score for Park Street's revenues, $7500 as follows:

$Z_{p} = \frac{(X - \mu)}{\sigma}=\frac{7500-6611}{3580}=0.25$

The z-score for Park Street's revenues this week is 0.25.

Compute the percentile rank for Park Street's revenues this week as follows:

$P(Z<Z_{p})=P(Z<0.25)=0.5987\approx 0.60\ \text{or}\ 60\%$

The percentile rank for Park Street's revenues this week is 60th.

This implies that the Park Street's performed better than 60% of the revenue recorded for the restaurant.

Compute the z-score for Bridge Road's revenues, $7100 as follows:

$Z_{p} = \frac{(X - \mu)}{\sigma}=\frac{7100-5989}{1794}=0.62$

The z-score for Bridge Road's revenues this week is 0.62.

Compute the percentile rank for Bridge Road's revenues this week as follows:

$P(Z<Z_{b})=P(Z<0.62)=0.7324\approx 0.73\ \text{or}\ 73\%$

The percentile rank for Bridge Road's revenues this week is 73rd.

This implies that the Bridge Road's performed better than 73% of the revenue recorded for the restaurant.