Recall that the density function of a normal distribution with mean u and standard deviation o > 0 is 1 -(-)2/(202) σν2Τ e This is an extremely important family of functions in statistics, and models the density function of many naturally occurring phenomena. As you saw in Sections 8.7 and 8.8, we often want to integrate PDFS. The problem here is that there is no elementary formula for an antiderivative of this function. Fortunately, we finally have a method for dealing with this: Taylor series! Suppose that a large number of students take an exam worth 100 points, and the distribution of student scores on this exam can be modeled by p(x), a normal distribution with mean (and median) 60 and standard deviation 15 

(a) Using the Taylor series for e centered at x = 0, e = 1 + 2 = 3! n! n-0 find the Taylor series for p(x) centered at x = 60. Write out both the first four nonzero terms and the entire series using summation notation. (You may find it useful to know that if OO Cnx" is the Taylor series for f(x) centered at x = 0, then > Cn(x - a)" is the Taylor n=0 n-0 series for f(x - a) centered at x = a.) 

(b) Use your answer to find a Taylor series for the CDF P(x) of this distribution. Again, write out both the first four nonzero terms and the entire series using summation notation. (Warning: think carefully about what the constant term should be!) 

(c) Use your formula to confirm that this Taylor series converges for all values of x p(x) dx 55

 (d) Use the degree 5 Taylor polynomial for P(x) to approximate the value of Interpret your answer in the context of the problem. 

(e) Use the degree 5 Taylor polynomial for P(x) to approximate the percentage of all scores that are within one standard deviation of the mean. 

(f) Find p100) (60), where p(100) (x) is the hundredth derivative of p(x)

Answer rating: 100% (QA)

Answer and step by step explanation a The Taylor series for e centered at x 0 is e 1 x x22 x33 x44 T... View the full answer