63. (Use of Tech) Normal distribution of heights
The hei...

Question

63. (Use of Tech) Normal distribution of heights

The heights of U.S. men are normally distributed with a mean of 69 in and a standard deviation of 3 in. This means that the fraction of men with a height between a and b (with a < b) inches is given by the integral

(1/(3√(2π))) ∫ₐᵇ e^(-((x-69)/3)²/2) dx.

What percentage of American men are between 66 and 72 inches tall? Use the method of your choice, and experiment with the number of subintervals until you obtain successive approximations that differ by less than 10⁻³.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says the weights of adult male pandas are normally distributed with a mean of 220 pounds and a standard deviation of 15 pounds. The fraction of pandas with weights between A and B with a less than B pounds is given by the interval of 1 divided by the quantity of 15 multiplied by the square root of the quantity of 2 pi in quantity multiplied by the interval from A to B of E rates to the quantity of minus 12 multiplied by the quantity of the quantity of X minus 220 in quantity, divided by 15 in quantity squared DX. What percentage of adult male pandas weigh between 205 and 235 pounds? Use the midpoint rule with N equal to 8 subintervals, and we're given four possible choices as our answers. For choice A, we have 62.56%. For choice B, we have 68.40%. For choice C, we have 71.38%. And for choice D, we have 78.67%. So for this problem, we're asked to find the percentage of adult male pandas that weigh between 205 and 235 pounds. And we're given an interval which we can use to calculate this percentage, but we want to approximate this interval, using the midpoint rule with equal to 8 sub-intervals. So recall for your midpoint rule that the interval from A to B of F of X D X. is approximately equal to our midpoint rule with Which will denote as M of N, where in here is the number of subintervals, and recall that this midpoint rule is just the midpoint rebound sum. And so this is going to be equal to, if you recall, delta X multiplied by the sum of I equal to 1 ton of F of X of I. Where in our case here, F of X is just going to be from our integram, and if we look at the integral that we're given in the problem, we see that this is going to be equal to E raised to the quantity of minus 1/2 multiplied by the quantity of the quantity of X minus 220 in quantity divided by 15 in quantity squared. And next we need to determine our Subinterval width, which is delta X. I recall that delta X is equal to the quantity of B minus A in quantity divided by N. So here we have to use our limits of integration for A and B, and so, Delta X is going to be equal to. These is our upper limit of integration, which in our case is going to be the 235. And then we'll have minus our lower limit, which is going to be 205, and then that quantity is divided by 8, since N is equal to 8. And when we evaluate this expression, we see that this is going to be equal to 15 divided by 4, which is equal to 3.75. Next, we need to determine the midpoint of each of our sub-intervals. And so we'll denote those midpoints as Xi. And so this represents the midpoint of the I subinterval. And recall your formula for finding these midpoints is going to be equal to A plus the quantity of I minus 12 in quantity multiplied by delta X. And so this is going to be equal to 205. Plus the quantity of I minus 12 in quantity multiplied by 3.75. So, what we need to do now in order to calculate our remod sum here, is to determine the values of each of our midpoints, X of I, and then calculate our function F of X at each of those midpoints. And so to organize this, Data better, we're actually going to use a table. And so, our table is gonna have 3 columns. Our first column is going to be I. The next column is going to be XI, and the final column is going to be F of XI. And so what we're going to do is to first fill in our column for I. And then we'll evaluate X of I and F of X of by for each value of I. So, I here, since N is equal to 8, I is going to go from 1 to 8. So we'll have, under the I column, we'll have 1. 2 3 45. 6 7 and 8. So now we just need to calculate X of I and F of X of I for each of these. So, when I is equal to 1, if we use our formula or X of I, we see that X of I is going to be equal to. 206.875. And then if we use that value of X by in our function F of X to calculate F of Xabi, we see that we'll get a value of 0.68194. 0751. Next, for an X I value, we'll have 2. So, X2, if we use our formula 4 X of I, we'll see that it has a value of 210. 0625. And using that value to calculate our value of our function X of F of X of I, we see that we get a value for F of 0.822. 577. 562. When I is equal to 3, X3 is going to be equal to 214.375. And F of X3 is going to be 0.9321. 02492. When I is equal to 4, we'll have X4, and we use our firm to calculate that value, we'll have 218. 0.125. And calculating F of X4, we will get 0.99. 221. 79 38. When I is equal to 5, or X5, we will get a value of 221.875, and calculating F of X5, we'll get 0.992217938. When I is equal to 6, or X6, we will have a value of 225. Point 625. And when we evaluate F of X6, we will get a value of 0.93210. 2492. When I is equal to 7, X7 is going to be equal 2. 229. 375. And F of X7 turns out to be 0.822577. 562. And finally, when I is equal to 8, or X8, we will have a value of 233.125. And when we evaluate our function F of X at X8, we see that we get a value of 0.681940751. So, now that we have all these values, we can use them to um evaluate our midpoint rule. So, we're looking for a midpoint rule with an equal to 8 subintervals, so we need to calculate M of 8. And so this is going to be equal to Delta X, which was 3.75. And this is going to get multiplied by the quantity, and here we need to add up all of the values of F of Xi, which are in our last column in our table. And so we'll have 0.681940751. Plus 0.822. 577. 562. Plus 0 point. 932102492. Plus 0.992217938. Plus And here we'll basically get the same values twice, but in reverse order, looking at our last column here. So we'll have 0.99. 221. 7938. Plus 0.9321. 02492. Plus 0. 822-577 562. Los 0.681940751 in quantity. And so, when we evaluate this expression, we see that it's going to be equal to. 25.716 29058. So that was for a Riemann sum, but the integral that we're asked to evaluate is 1 divided by the quantity of 15, multiplied by the square root of the quantity of 2 pi in quantity, multiplied by the integral from A to B. Of E raised to the quantity of minus 1/2 multiplied by the quantity of the quantity of X minus 220, in quantity divided by 15, in quantity squared DX. And so this is going to be approximately equal to. 1, divided by the quantity of 15, multiplied by the square root of the quantity of 2 pi in quantity, multiplied by M of 8. And so substituting in the value for M of 8 that we have, we see that this integra is could be approximately equal to 1 divided by the quantity of 15 multiplied by the square root of the quantity of 2 pi in quantity multiplied by. 25.71. 629058. And so, evaluate this expression, we see that our integral is going to be approximately equal to. 0.6840. However, the final answer needs to be as a percentage, so we will multiply this by 100%. And we see that our integral is gonna be approximately equal to. 68.40%. So this here is the answer that we're looking for for this problem. And if we compare our answer here to our answer choices, We see that the correct answer choice corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Normal Distribution

Definite Integral

Numerical Integration

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