The Manager's Guide to Statistics Summarizing and Displaying Data

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1400 134

1500 156

1600 254

1700 351

1800 527

1900 645

Over 1950 0

The average and median of the dataset are below:

Average 1,626

Median 1,702

The histogram for this data reveals that most of the orders made are in excess of $1,900 but under $1,950. This is well above the average order of $1,626 and median order of $1,702.

Chapter 3 Exercise 15:

a) Calculating for the standard units, a return of $40 million is 1.5 standard deviations above average for investment A and 1.17 standard deviations above average for investment B. This means that investment B has the higher chance to earn a return of $40 million as such a return is less standard deviations from the average than Investment A.

(in mil) Investment A Investment B

Expect Rtn (μ) $ 25.00 $ 5.00

Std Dev (σ) $ 10.00 $ 30.00

Score (x) $ 40.00 $ 40.00

SU’s (z) 1.50 1.17

*SU’s is calculated by z = (x – μ) / σ

b) If we knew this was a skewed distribution, it would mean there were outliers affecting the average used to calculate the expected return. We could use the median in this case to see if the tail of the histogram was to the right or the left of the median to see which side occupied the most space in the histogram. If above the average of investment B was more than the median, thus occupied less than half the area on the histogram, investment A may become the better choice (depending on its own median). Likewise, if investment A had a median that was higher than its average, it may also become a better choice because then that would mean there were outliers driving the average down.

Chapter 3 Exercise 22:

a) Since the data here uses normal distribution, we can utilized the area table for areas under the normal curve. Using the calculation for standard units, we can calculate that 80 hours is 1.25 standard units from the average. This is proven by the calculation of 1.25 = (80 – 70)/8. Using the table “areas under the normal curve,” we can look up that a standard deviation of 1.25 is closest to the value 1.23. Thus with the able we can determine that we can produce the product within 80 hours 89% of the time.

b) To determine the numbers of hours needed to be promised to be able to promise the product will be produced within that number of hours 95% of the time, we need a standard deviation of 1.64 as determined by the “areas under the normal curve” table. We can adjust the formula to calculate the hours by calculating x = μ + (z * σ ) where Z = standard units, X = actual score, μ = average and σ = standard deviation.

x = 70 + ( 1.64 * 8 ) which results in x = 83.12. Therefore, we can promise that we can produce the product 95% of the time in under 83.12 hours.

c) To calculate how much the average production time needs to be reduced to achieve producing the product in under 75 hours 95% of the time, we need to use the formula μ = x - (z*σ).

Using μ = 75 - (1.64*8), μ is equal to 62 meaning the average production time needs to be reduced by 8 hours in order to produce under 75 hours 95% of the time.

References:

Pekoz, E. (n.d.). The Manager’s Guide to Statistics (2014 Edition ed., Custom Northeastern University MGSC 6200 Information Analysis).

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