SOLUTION: During sleep , the reduction of a person's oxygen consumption has a normal distribution with a mean of 38.4 mL/min and a standard deviation of 4.6 mL/min.
Determine the probabili
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Question 280322: During sleep , the reduction of a person's oxygen consumption has a normal distribution with a mean of 38.4 mL/min and a standard deviation of 4.6 mL/min.
Determine the probability that during sleep a person's oxygen consumption will be reduced by:
a) more than 43.5 mL/min
b) at most 36.4 mL/min
c) anywhere from 30 to 40 mL/min
d) If 90% of people have reduction in excess of k mL/min, determine k.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
During sleep , the reduction of a person's oxygen consumption has a normal distribution with a mean of 38.4 mL/min and a standard deviation of 4.6 mL/min.
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So you can see the problem, sketch a normal curve with 38.4 at the mean
position and 4.6 under it.
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Determine the probability that during sleep a person's oxygen consumption will be reduced by:
a) more than 43.5 mL/min
Mark 43.5 on the figure you drew.
Shade the are to the right of 43.5.
Find the z-value of 43.5:
z(43.5) = (43.5-38.4)/4.6 = 1.1087
Use your z-chart or calculator to find the area to the right
of that z-value:
normalcdf(1.1087,10) = 0.1338
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b) at most 36.4 mL/min
Find the z-value of 36.4
Then find the area to the left of that z-value
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c) anywhere from 30 to 40 mL/min
Find the z-value of 40 and of 30
Then find the area between those two z-values
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d) If 90% of people have reduction in excess of k mL/min, determine k.
Mark a z-point that has 10% of the area below it.
Find that z-value with your chart of calculator:
invNorm(0.10) = -1.2816
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Use the formula x = zs+u to find "x":
x = -1.2816*4.6 + 38.4
x = 32.50mL/min
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Cheers,
Stan H.
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