Question

The heights of men in a particular country are normally distributed with a mean of 68 inches and a standard deviation of 3 inches. This means that the fraction of men with a height between a and b (with a < b) inches is given by the integral below. What percentage of American men are between 62 and 74 inches tall? Use the method of your choice and, starting with one subinterval, increase the number of subintervals by 1 until you obtain successive approximations that differ by less than $10^{-3}$. Use the last approximation as the answer.\\ $\frac{1}{3\sqrt{2\pi}} \int_{a}^{b} e^{-\frac{(x-68)^2}{2 \cdot 3^2}} dx$ \\ The percentage of American men who are between 62 and 74 inches tall is about 95.45% (Do not round until the final answer. Then round to the nearest thousandth as needed.)

          The heights of men in a particular country are normally distributed with a mean of 68 inches and a standard deviation of 3 inches. This means that the fraction of men with a height between a and b (with a < b) inches is given by the integral below. What percentage of American men are between 62 and 74 inches tall? Use the method of your choice and, starting with one subinterval, increase the number of subintervals by 1 until you obtain successive approximations that differ by less than $10^{-3}$. Use the last approximation as the answer.\\
$\frac{1}{3\sqrt{2\pi}} \int_{a}^{b} e^{-\frac{(x-68)^2}{2 \cdot 3^2}} dx$ \\
The percentage of American men who are between 62 and 74 inches tall is about 95.45% (Do not round until the final answer. Then round to the nearest thousandth as needed.)

$The heights of men in a particular country are normally distributed with a mean of 68 inches and a standard deviation of 3 inches. This means that the fraction of men with a height between a and b (with a < b) inches is given by the integral below. What percentage of American men are between 62 and 74 inches tall? Use the method of your choice and, starting with one subinterval, increase the number of subintervals by 1 until you obtain successive approximations that differ by less than 10^-3. Use the last approximation as the answer. (1)/(3√(2π))∫a^b e^-((x-68)^2)/(2 · 3^2) dx The percentage of American men who are between 62 and 74 inches tall is about 95.45% (Do not round until the final answer. Then round to the nearest thousandth as needed.)$

Added by Consuelo G.

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