Solutions

 

•   Since the function $f(x)$ is the sum of the functions $a(x) = \sin (x^3)$ and $b(x) = -x$ the sixth Taylor polynomial of $f(x)$ will be the sum of the sixth Taylor polynomial of $\sin (x^3)$ and the sixth Taylor polynomial of $-x$. Since the function $b(x) = -x$ is a polynomial already it will be its own Taylor polynomial of any degree $\geq 1$. To calculate the sixth Taylor polynomial of $a(x) = \sin (x^3)$ consider the function $g(u) = \sin u$. Clearly $a(x) = g(x^{3})$ and so in order to calculate the sixth Taylor polynomial of $a(x)$ it suffices to calculate the second Taylor polynomial of $g(u)$ and then substitute $u = x^{3}$. The second Taylor polynomial of $g(u)$ is $g(0) + g'(0)u + g''(0)/2 u^{2}$,and since $g'(u) = \cos u$, $g''(u) = - \sin u$ we get $g(0) + g'(0)u + g''(0)/2 u^{2} = u$. In conclusion we have
  $$\begin{align*} p_{6}(x) & = (\text{6th Taylor poly of} \; a(x)) + (\text{6th Tay... ...\text{2nd Taylor poly of} \; g(u))_{u = x^{3}} - x \\ & = x^{3} - x.\end{align*}$$
  

• Let $X$ be the score of a person who had taken the GRE general test in 1997 and let $a$ be the lowest score in the top $20 \%$ of scores. Then $p(X \geq a) = 0.2$ or equivalently $p(X \leq a) = 1 - 0.2 = 0.8$. In terms of the standard normal variable $Z$ the latter can be rewritten as $p(Z \leq (a - 1900)/300) = 0.8$. The closest probability to $0.8$ in the table of values of standard normal areas is $p(Z \leq 0.84) = 0.7995$. Thus $(a - 1900)/300 \sim 0.84$ and so $a \sim 1900 + 300 \cdot 0.84 = 1900 + 252 = 2152$. 

•   In order for $f(x) = c\sqrt{x^{3}}$ to be a probability density function on $[0,1]$ we must have $1 = \int_{0}^{1} f(x) dx = c \int_{0}^{1} x^{3/2} dx = c \cdot (2/5) \cdot x^{5/2} \vert _{0}^{1} = 2c/5$ which implies $c = 5/2$. Now for the expected value one has $\mu = (5/2) \cdot \int_{0}^{1} x\cdot x^{3/2} dx = (5/2)\cdot (2/7) \cdot x^{7/2}\vert _{0}^{1} = 5/7$.

•   By pulling out the highest power of $n$ present in the numerator and the denominator of $a_{n}$ we get
  $$\begin{align*} \lim_{n \to \infty} a_{n} & = \lim_{n \to \infty} \frac{4 - 3n^{2... ... & = \frac{0 - 3 + 0}{0 - \sqrt[3]{0 - 27 + 0}} = \frac{-3}{3} = -1.\end{align*}$$
  

•   If $p_{n}(x)$ is the degree $n$ Taylor polynomial at zero and $R_{n}(x) = f(x) - p_{n}(x)$ is the error with which $p_{n}(x)$ approximates $f(x)$, then the Taylor remainder theorem gives

  $$\vert R_{n}(x) \vert \leq \frac{M}{(n+1)!} \vert x\vert^{n+1},$$

  on any given interval $[x_{1}, x_{2}]$ which contains zero. The constant $M$ depends on the interval and satisfies $M \geq \operatorname{max}_{x_{1} \leq x \leq x_{2}} \vert f^{(n+1)}(x)\vert$. To estimate the error $R_{n}(-1/5) = R_{n}(- 0.2)$ we will need to choose the interval $[x_{1}, x_{2}]$ so that both and $- 0.02$ belong to it. The interval of minimal length containing both and $- 0.02$ is $[ - 0.02, 0]$. The calculation of the derivatives of $f(x) = e^{2x}$ gives $f^{(n+1)}(x) = 2^{n+1} e^{2x}$ and so we have
  $$\begin{align*} M & = \operatorname{max}_{- 0.02 \leq x \leq 0 } \left\vert \frac... ...\vert x\vert^{n+1} \\ & = \frac{2^{n+1}}{(n+1)!} \vert x\vert^{n+1},\end{align*}$$
  since the function $e^{2x}$ is monotonically increasing on the interval
  $[ - 0.02, 0]$.

  Therefore $\vert R_{n}(x)\vert \leq 2^{n+1} \vert x\vert^{n+1}/(n+1)!$ on the interval $[ - 0.02, 0]$ and so $R_{n}( - 0.02) \leq (0.04)^{n+1}/(n+1)!$. By plugging in the first few values of $n$ we find that $n = 3$ is the first possible value of $n$ for which $R_{n}( - 0.01) \leq 0.01$.

•   If $X$ is the number of miles per hour by which the drivers violate the speed limit, then since $X$ is exponential has a probability density function $f(x) = k e^{-kx}$ on the interval $[0, \infty)$. Since $5 = E(X) = 1/k$ we have that $f(x) = e^{-(1/5)x}/5$. The probability in question is

  $$p(X \leq 15) = \int_{0}^{15}f(x)dx = (- e^{-(1/5)x})\vert _{0}^{15} = 1 - e^{-3}.$$

•   The probability that the sum of the two uppermost faces will be seven is $6/36 = 1/6$ and therefore we have $180$ binomial trials with probability for success $p = 1/6$ and probability for failure $q = 5/6$. The binomial random variable $x$ is thus approximated by a normal random variable $X$ having a mean $\mu = np = 180\cdot (1/6) = 30$ and a standard deviation $\sigma = \sqrt{npq} = \sqrt{180\cdot (1/6) \cdot (5/6)} = \sqrt{25} = 5$. We then have

  $$p(x \leq 62) \sim p(X \leq 62.5) = p(Z \leq \frac{62.5 - 30}{5}) = p(Z \leq 6.5).$$

•   The first Taylor polynomial of $f(x) = \sqrt{1 + x}$at $x = 3$ is $p_{1}(x) = f(3) + f'(3)(x - 3)$. Since $f'(x) = (1/2)(1 + x)^{-1/2}$ we get $p_{1}(x) = 2 + (x - 3)/4$ and hence $f(3.08) \sim p_{1}(3.08) = 2.02$.

•   According to the Chebishev inequality we have $p(\mu - a\sigma \leq X \leq \mu + a\sigma) \geq 1 - 1/a^{2}$ for any random variable $X$ and any number $a$. If $X$ is the number of computer users who chose their birth date as a password, then
  $$\begin{align*} p(276 \leq X \leq 324) &= p(300 - (3/2)\cdot 16 \leq X \leq 300 + (3/2)\cdot 16) \\ & \geq 1 - (2/3)^{2} = 5/9.\end{align*}$$
  

•   Let $X$ be the number of action movies the customer rents. Then $X$ is a finite discrete random variable with probability distribution

  $$X 1 2 3$$

  $$P(X) \frac{C(5,3)}{C(9,3)} \frac{C(5,2)C(4,1)}{C(9,3)} \frac{C(5,1)C(4,2)}{C(9,3)} \frac{C(4,3)}{C(9,3)}$$

  The expected value of $X$ is
  $$\begin{align*} E(X) & = 0\cdot \frac{C(5,3)}{C(9,3)} + 1\cdot \frac{C(5,2)C(4,1)... ...}{C(9,3)} \\ & = \frac{40 + 60 + 12}{C(9,3)} = \frac{112}{84} = 4/3 \end{align*}$$