What is used to substitute inorder to verify the solution of equations?
1 answer:
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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 <u>Q₂ < $7,500 < Q₃</u>
- The sales for the week is better than <u>72.91%</u> of all sales
The performance for the week for Bridge Road
- Revenue; <u>Q₂ < $7,100 < Q₃</u>
- The sale for the week is better than <u>59.87%</u> of all sales
(b) The mean is <u>$3611</u>
The median is $<u>3,600</u>
The standard deviation is $<u>3250</u>
The Interquartile range is $<u>6075</u>
Reason:
The table of values that maybe used to find a solution to the question is given as follows;
(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;
From the Z-Table, we have;
The percentile= 0.7291
- Therefore, the sale for the week for Park Street is better than <u>72.91%</u> of all the sales
Park Street revenue;
The z-score is given as follows;
From the Z-Table, we have;
The percentile = <u>0.5987</u>
- Therefore, the sale for the week is better than <u>59.87 %</u> 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 = <u>$3,611</u>
The new median = 6600 - 3000 = 3600
- The new median = <u>$3,600</u>
The standard deviation and the interquartile range remain the same, therefore, we have;
- The standard deviation = <u>$3,580</u>
The interquartile range = 9675 - 3600 = 6075
- The interquartile range = <u>6075</u>
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