Respuesta :
Answer:
[tex]P(X>140)=1-P(X\leq 140)=1-P(\frac{X-\mu}{\sigma}<\frac{140-\mu}{\sigma})=1-P(Z<\frac{140-100}{19})=1-P(Z<2.105)=0.0176[/tex]
Number =0.0176*280000000=4928000 approximately Americans would have a IQ score more than higher than 140
Step-by-step explanation:
1) 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".
2) Solution to the problem
Let X the random variable that represent the IQ scores of the population of interest, and for this case we know the distribution for X is given by:
[tex]X \sim N(100,19)[/tex]
Where [tex]\mu=100[/tex] and [tex]\sigma=19[/tex]
We are interested on this probability
[tex]P(X>140)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula and the complement rule to our probability we got this:
[tex]P(X>140)=1-P(X\leq 140)=1-P(\frac{X-\mu}{\sigma}<\frac{140-\mu}{\sigma})=1-P(Z<\frac{140-100}{19})=1-P(Z<2.105)=0.0176[/tex]
And we can find this probability with the following excel code:
"=1-NORM.DIST(2.105,0,1,TRUE)"
This number 0.0176 represent the proportion of Americans that present a score higher than 140.
And now since we ar einterested on the approximate number of people in the United States (assuming a total population of 280,000,000) with an IQ higher than 140, we just need to do this:
Number =0.0176*280000000=4928000 approximately Americans would have a IQ score more than higher than 140
