Exam 2

  1. Question

    Given the following information:

    pineapple ++ pineapple ++ banana = 674674
    banana ++ banana ++ pineapple = 385385
    banana ++ orange ++ banana = 102102

    Compute:

    banana ++ orange ++ pineapple = ?\text{?}

    1. 321321
    2. 391391
    3. 568568
    4. 561561
    5. 385385

    Solution

    The information provided can be interpreted as the price for three fruit baskets with different combinations of the three fruits. This corresponds to a system of linear equations where the price of the three fruits is the vector of unknowns xx:

    x1=x_1 = banana x2=x_2 = orange x3=x_3 = pineapple

    The system of linear equations is then: (102201210)(x1x2x3)=(674385102) \begin{aligned} \left( \begin{array}{rrr} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 2 & 1 & 0 \end{array} \right) \cdot \left( \begin{array}{r} x_1 \\ x_2 \\ x_3 \end{array} \right) & = & \left( \begin{array}{r} 674 \\ 385 \\ 102 \end{array} \right) \end{aligned} This can be solved using any solution algorithm, e.g., elimination: x1=32,x2=38,x3=321. x_1 = 32, \, x_2 = 38, \, x_3 = 321. Based on the three prices for the different fruits it is straightforward to compute the total price of the fourth fruit basket via:

    banana ++ orange ++ pineapple =
    x1x_1 ++ x2x_2 ++ x3x_3 =
    3232 ++ 3838 ++ 321321 = 391391

    1. False
    2. True
    3. False
    4. False
    5. False

  2. Question

    What is the distance between the two points p=(1,2)p = (1, 2) and q=(5,4)q = (5, 4) in a Cartesian coordinate system?


    1. 2.9712.971
    2. 4.4724.472
    3. 1.1181.118
    4. 3.8883.888
    5. 2.4492.449

    Solution

    The distance dd of pp and qq is given by d2=(p1q1)2+(p2q2)2d^2 = (p_1 - q_1)^2 + (p_2 - q_2)^2 (Pythagorean formula).

    Hence d=(p1q1)2+(p2q2)2=(15)2+(24)2=4.472d = \sqrt{(p_1 - q_1)^2 + (p_2 - q_2)^2} = \sqrt{(1 - 5)^2 + (2 - 4)^2} = 4.472.


    1. False
    2. True
    3. False
    4. False
    5. False

  3. Question

    What is the derivative of f(x)=x8e2.7xf(x) = x^{8} e^{2.7 x}, evaluated at x=0.53x = 0.53?


    Solution

    Using the product rule for f(x)=g(x)h(x)f(x) = g(x) \cdot h(x), where g(x):=x8g(x) := x^{8} and h(x):=e2.7xh(x) := e^{2.7 x}, we obtain f(x)=[g(x)h(x)]=g(x)h(x)+g(x)h(x)=8x81e2.7x+x8e2.7x2.7=e2.7x(8x7+2.7x8)=e2.7xx7(8+2.7x). \begin{aligned} f'(x) &= [g(x) \cdot h(x)]' = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\ &= 8 x^{8 - 1} \cdot e^{2.7 x} + x^{8} \cdot e^{2.7 x} \cdot 2.7 \\ &= e^{2.7 x} \cdot(8 x^7 + 2.7 x^{8}) \\ &= e^{2.7 x} \cdot x^7 \cdot (8 + 2.7 x). \end{aligned} Evaluated at x=0.53x = 0.53, the answer is e2.70.530.537(8+2.70.53)=0.463409. e^{2.7 \cdot 0.53} \cdot 0.53^7 \cdot (8 + 2.7 \cdot 0.53) = 0.463409. Thus, rounded to two digits we have f(0.53)=0.46f'(0.53) = 0.46.


  4. Question

    The daily expenses of summer tourists in Vienna are analyzed. A survey with 112112 tourists is conducted. This shows that the tourists spend on average 126126 EUR. The sample variance sn12s^2_{n-1} is equal to 117117.

    Determine a 95%95\% confidence interval for the average daily expenses (in EUR) of a tourist.


    1. What is the lower confidence bound?
    2. What is the upper confidence bound?

    Solution

    The 95%95\% confidence interval for the average expenses μ\mu is given by: [y1.96sn12n,y+1.96sn12n]=[1261.96117112,126+1.96117112]=[123.997,128.003]. \begin{aligned} & & \left[\bar{y} \, - \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}, \; \bar{y} \, + \, 1.96\sqrt{\frac{s_{n-1}^2}{n}}\right] \\ & = & \left[ 126 \, - \, 1.96\sqrt{\frac{117}{112}}, \; 126 \, + \, 1.96\sqrt{\frac{117}{112}}\right] \\ & = & \left[123.997, \, 128.003\right]. \end{aligned}


    1. The lower confidence bound is 123.997123.997.
    2. The upper confidence bound is 128.003128.003.

  5. Question

    For 63 firms the number of employees XX and the amount of expenses for continuing education YY (in EUR) were recorded. The statistical summary of the data set is given by:

    Variable XX Variable YY
    Mean 51 202
    Variance 95 1889

    The correlation between XX and YY is equal to 0.73.

    Estimate the expected amount of money spent for continuing education by a firm with 52 employees using least squares regression.


    Solution

    First, the regression line yi=β0+β1xi+εiy_i = \beta_0 + \beta_1 x_i + \varepsilon_i is determined. The regression coefficients are given by: β^1=rsysx=0.73188995=3.2552,β^0=yβ^1x=2023.255251=35.98504.\begin{eqnarray*} && \hat \beta_1 = r \cdot \frac{s_y}{s_x} = 0.73 \cdot \sqrt{\frac{1889}{95}} = 3.2552, \\ && \hat \beta_0 = \bar y - \hat \beta_1 \cdot \bar x = 202 - 3.2552 \cdot 51 = 35.98504. \end{eqnarray*} The estimated amount of money spent by a firm with 52 employees is then given by: y^=35.98504+3.255252=205.255.\begin{eqnarray*} \hat y = 35.98504 + 3.2552 \cdot 52 = 205.255. \end{eqnarray*}