The Candela brothers own two pizza restaurants, one on Park Street and one on Bridge Road.

The computer output below summarizes the distribution of weekly revenues at each

restaurantâ€"26 weeks for Park Street and 40 weeks for Bridge Avenue.

Answers

Answer 1

The mean, median and mode are measures of central tendency, that is they tend to indicate the location middle of the data

Required values;

(a) The performance for the week for Park Street

Revenue is Q₂ < $7,500 < Q₃The sales for the week is better than 72.91% of all sales

The performance for the week for Bridge Road

Revenue; Q₂ < $7,100 < Q₃The sale for the week is better than 59.87% of all sales

(b) The mean is $3611

The median is $3,600

The standard deviation is $3250

The Interquartile range is $6075

Reason:

The table of values that maybe used to find a solution to the question is given as follows;

[tex]\begin{array}{|l|l|l|}\mathbf{Variable} &\mathbf{Park}&\mathbf{Bridge}\\N&36&40\\Mean&6611&5989\\SE \ Mean&597&299\\StDev&3580&1794\\Minimum&800&1800\\Q_1&3600&5225\\Median&6600&6000\\Q_3&9675&7625\\Maximum&14100&8600\end{array}\right][/tex]

(a) Park Street revenue = $7,500

Bridge Road's revenue = $7,100

The two stores sold close to but below the 75th percentile

Bridge Road revenue;

The z-score is given as follows;

[tex]Z = \dfrac{x - \mu }{\sigma }[/tex]

[tex]Z = \dfrac{7100 - 5,989 }{1794 } \approx 0.6193[/tex]

From the Z-Table, we have;

The percentile= 0.7291

Therefore, the sale for the week for Park Street is better than 72.91% of all the sales

Park Street revenue;

The z-score is given as follows;

[tex]Z = \dfrac{7500 - 6611}{3580} \approx 0.25[/tex]

From the Z-Table, we have;

The percentile = 0.5987

Therefore, the sale for the week is better than 59.87 % of all the sales

(b) Given that the operating cost is $3,000, frim which we have;

The subtracted value is subtracted from the mean and median to find the new value

Profit = The revenue - Cost

New mean = 6611 - 3000 = 3611

The new mean = $3,611

The new median = 6600 - 3000 = 3600

The new median = $3,600

The standard deviation and the interquartile range remain the same, therefore, we have;

The standard deviation = $3,580

The interquartile range = 9675 - 3600 = 6075

The interquartile range = 6075

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