The graph illustrates a normal distribution for the prices paid
for a particular model of HD television. The mean price paid is
$1000 and the standard deviation is $100.
7008009001000110012001300Distribution of Prices
What is the approximate percentage of buyers who paid between $700
and $1000?
%
What is the approximate percentage of buyers who paid between $900
and $1000?
%
What is the approximate percentage of buyers who paid more than
$1300?
%
What is the approximate percentage of buyers who paid between $900
and $1100?
%
What is the approximate percentage of buyers who paid more than
$1200?
%
What is the approximate percentage of buyers who paid between $800
and $1000?
(a) | (b) | (c) | |||||
µ | 1000 | µ | 1000 | µ | 1000 | ||
σ | 100 | σ | 100 | σ | 100 | ||
x | 1000 | x | 1000 | x | 1300 | ||
z = (x - µ)/σ | 0 | z = (x - µ)/σ | 0 | z = (x - µ)/σ | 3 | ||
p-value | 0.5000 | p-value | 0.5000 | p-value | 0.13% | ||
x | 700 | x | 900 | ||||
z = (x - µ)/σ | -3 | z = (x - µ)/σ | -1 | ||||
p-value | 0.0013 | p-value | 0.1587 | ||||
Req. | 49.87% | Req. | 34.13% | ||||
(d) | (e) | (d) | |||||
µ | 1000 | µ | 1000 | µ | 1000 | ||
σ | 100 | σ | 100 | σ | 100 | ||
x | 1100 | x | 1200 | x | 1000 | ||
z = (x - µ)/σ | 1 | z = (x - µ)/σ | 2 | z = (x - µ)/σ | 0 | ||
p-value | 0.8413 | p-value | 2.28% | p-value | 0.5000 | ||
x | 900 | x | 800 | ||||
z = (x - µ)/σ | -1 | z = (x - µ)/σ | -2 | ||||
p-value | 0.1587 | p-value | 0.0228 | ||||
Req. | 68.27% | Req. | 47.72% |
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