Question 1086439: I am totally lost with this question on my homework.
6. The weight data from the 2008 WNBA players was collected. It was determined that the mean and standard deviation was 134.75 pounds and 10.7 pounds, respectively.
a. Find the z-score of a women spectator if her weight is 120 pounds.
b. Find and interpret the z-score of a second woman’s weight that is 165 pounds
c. Construct the graph showing the results from part (a) and (b).
d. Interpret results: are they usual or unusual?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Part A
For this part, and part B as well, we'll use the z-score formula as shown below
z = (x-mu)/sigma
where,
z = z score
x = raw score
mu = population mean
sigma = population standard deviation
In this case,
z = unknown
x = 120
mu = 134.75
sigma = 10.7
which means we can plug those values into the formula to get
z = (x-mu)/sigma
z = (120-134.75)/10.7
z = -14.75/10.7
z = -1.3785046728972
z = -1.38
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Part B
We'll follow the same basic steps as in part A. We'll use the same mu and sigma value. This time x = 165
z = (x-mu)/sigma
z = (165-134.75)/10.7
z = 30.25/10.7
z = 2.82710280373831
z = 2.83
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Part C
Graph is shown below. The red points are markings where the z score is located. Point A corresponds to the z score of part A. Point B corresponds to the z score of part B. The blue curve is the standard normal bell curve (mu = 0 and sigma = 1)
Image generated by GeoGebra (free graphing software).
As we can see, the larger the z score, the further you get away from the center (z = 0). The further you move away, the less likely to pick that value out due to random chance. Also, the height of the curve sharply drops off heading to 0 the further you move away. The height of the curve is the probability value at any given point. The middle (z = 0) is the most likely because this is where the average is located. This is shown by the fact that the curve peaks when z = 0.
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Part D
Consulting this page it states that Data beyond two standard deviations away from the mean is considered "unusual" data
Put another way, any z score that is in the interval  is considered usual. Anything outside that interval is unusual.
Using that rule, we can see that the z-score from part A is usual while the z-score from part B is unusual. It's not very likely to have a player with a weight of 165 pounds. A weight like 120 pounds is more common.
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