Allen is buying a video game that originally costs $49. He has a 10%-off

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

koban [17]

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;

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

(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;

$Z = \dfrac{x - \mu }{\sigma }$

$Z = \dfrac{7100 - 5,989 }{1794 } \approx 0.6193$

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;

$Z = \dfrac{7500 - 6611}{3580} \approx 0.25$

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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5 0

1 year ago

The labor force participation rate is the number of people in the labor force divided by the number of people in the country who

zepelin [54]

Answer:

The expected participation rate is 0.637.

The standard error is 0.04397

Step-by-step explanation:

For each working age people asked, there are only two possible outcomes. Either they are in the labor force, or they are not. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability:

Expected value for the participation rate: The expected value is the probability of a success. In this problem, a success is a working age people being in the labor force. 63.7% of them are. So $\pi = 0.637$ . This means that the expected participation rate is 0.637.