The speed of nitrogen molecules in the atmosphere (at 20ºC) follows a normal distribution with a mean speed of 500 meters/second
and a standard deviation of 50 meters/second. Which conclusion does this information best support?
A. 84% of the molecules have speeds between 450 meters/second and 550 meters/second.
B. 68% of the molecules have speeds between 450 meters/second and 550 meters/second.
C. 50% of the molecules have speeds between 450 meters/second and 550 meters/second.
D. 34% of the molecules have speeds greater than 550 meters/second.
A. 84% of the molecules have speeds between 450 meters/second and 550 meters/second.
B. 68% of the molecules have speeds between 450 meters/second and 550 meters/second.
C. 50% of the molecules have speeds between 450 meters/second and 550 meters/second.
D. 34% of the molecules have speeds greater than 550 meters/second.
2 answers:
Answer:
B. 68% of the molecules have speeds between 450 meters/second and 550 meters/second.
Step-by-step explanation:
Since there is a standard deviation of 50 meters per second, and the mean speed is 500 meters per second, the speeds would range between 500 - 50 meters per second and 500 + 50 meters per second, which would be a range of 450 to 550 meters per second.
If you remember the Empirical Rule (68-95-99), things that are one standard deviation from the mean are within 68% of the data. Two standard deviations would include 95% of the data. Three standard deviations would include 99% of the data. Since this question just covers one standard deviation, 68% of the molecules have speeds between 450 meters per second and 550 meters per second.
Hope this helps!
You might be interested in
Answer:
And solving for z we have
And we can find the value for z with the following excel code:
"=NORM.INV(0.975,0,1)"
And we got z =1.96
And we can use the complement rule and we got:
And we can find the value for z with the following excel code:
"=NORM.INV(0.8686,0,1)"
And we got z =1.120
And we can find the value for z with the following excel code:
"=NORM.INV(0.67,0,1)"
And we got z =0.440
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
We want this probability:
And solving for z we have
And we can find the value for z with the following excel code:
"=NORM.INV(0.975,0,1)"
And we got z =1.96
For the next part we want to calculate:
And we can use the complement rule and we got:
And we can find the value for z with the following excel code:
"=NORM.INV(0.8686,0,1)"
And we got z =1.120
For the next part we want to calculate:
And we can find the value for z with the following excel code:
"=NORM.INV(0.67,0,1)"
And we got z =0.440