Respuesta :
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 < Q₃
- The sales for the week is better than 72.91% of all sales
The performance for the week for Bridge Road
- Revenue; Q₂ < $7,100 < Q₃
- The 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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