Assume that the heights of American men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches The U.S. Marine Corps requires that men have heights between 64 and 78 inches

Assume that the heights of American men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches The U.S. Marine Corps requires that men have heights between 64 and 78 inches

Assume that the heights of American men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches The U.S. Marine Corps requires that men have heights between 64 and 78 inches Find the percentage of men meeting these height requirements 97.67% 95.88% 95.45% 96.26% 97.37%

Expert's Answer

I've already drawn us a picture of our situation. Men's heights are normally distributed with a mean of 69. And my standard deviation I have numbered to the right and two left By 2.8. Now we want to know if the Marine Corps only accepts men that are between 64 78 inches tall. What percentage of men hit these qualifications. So we're gonna need a Z score for 64 which I'm gonna do 64 minus the mean divided by the standard deviation of 2.8. So that will be negative five divided by 2.8, Which is equal to -1 79. And I'm gonna look that Z score up in the standard normal probability table to give me the area to the left of negative 1.79 which is .0367. This way .0367. And we need to do the same thing for 78. Find a Z score what? Which will essentially be negative nine divided by 2.8 Which is negative 3.2 one. And then according to the standard normal probability table, that area to the left is .9993. So all the way over here to the left This .9993. And so the area in between the red marker and the blue marker at 64 and 78 will be equal to .9993 -1367 which is equal to .9626 96 26 or 96.26%,, Which is your 4th choice?

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