SOLUTION: The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1600 and the standard deviation is $50. I figure

Question 1134929: The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1600 and the standard deviation is $50.
I figured out that:
What is the approximate percentage of buyers who paid between $1550 and $1650?
68
What is the approximate percentage of buyers who paid between $1500 and $1600?
47.5
What is the approximate percentage of buyers who paid between $1600 and $1750?
49.85
but i cannot calculate :
What is the approximate percentage of buyers who paid less than $1450?
What is the approximate percentage of buyers who paid less than $1500?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
without seeing the graph, it's pretty hard for me to see if your determination is correct.

i can, however, use an online calculator that will show you the exact value and you can determine from that what your approximate value should be.

that calculator can be found at http://davidmlane.com/hyperstat/z_table.html

the answers to each of your questions in turn are shown below:

$$$

i'm not sure how you're calculating the percentages, but you were preety close for the first 3.
the fourth is so small as to be almost equal to 0.

the fifth is a little better but still very small.

generally, you would use a z-score calculator to find the area after you determined what the z-score is.

z-score = (x - m) / s

x is the raw score.
m is the meanb
s is the standard deviation.
z is the z-score.

for example, for your last problem, the formula becomes z = (1500 - 1600) / 50 = -100 / 50 = -2.

the area to the left of a z-score of -2 is equal to .022750062 which i found through the use of my TI-84 Plus calculator.

that agrees with the 0.022800000000000042 from the online calculator after taking into account rounding differences.

hope this helps.

let me know if you're ok with it.

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