Question

do bonds reduce the overall risk of an investment portfolio? let x be a random variable representing annual percent return for the vanguard total stock index (all stocks). let y be a random variable representing annual return for the vanguard balanced index (60% stock and 40% bond). for the past several years, assume the following data. compute a 75% chebyshev interval around the mean for x - values and also for y - values. round your answers to the nearest hundredth.
13 0 36 25 34 27 28 -13 -13 -25
x:
10 -5 28 16 26 17 16 -5 -6 -10
y:
for x - values: -19.50 to 36.90 and for y - values: -33.36 to 55.76
for x - values: -11.08 to 55.76 and for y - values: -5.40 to 36.90
for x - values: -33.36 to 55.76 and for y - values: -19.50 to 36.90
for x - values: -33.36 to 25.30 and for y - values: -19.50 to 22.80
for x - values: -33.36 to 25.30 and for y - values: -5.40 to -17.00

Explanation:

Step1: Calculate the mean of x - values

$\bar{x}=\frac{13 + 0+36 + 25+34 + 27+28+( - 13)+( - 13)+( - 25)}{10}=\frac{112}{10} = 11.2$

Step2: Calculate the variance of x - values

$ s_{x}^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$
$\sum_{i=1}^{10}(x_{i}-11.2)^{2}=(13 - 11.2)^{2}+(0 - 11.2)^{2}+(36 - 11.2)^{2}+(25 - 11.2)^{2}+(34 - 11.2)^{2}+(27 - 11.2)^{2}+(28 - 11.2)^{2}+(-13 - 11.2)^{2}+(-13 - 11.2)^{2}+(-25 - 11.2)^{2}$
$=1.8^{2}+(-11.2)^{2}+24.8^{2}+13.8^{2}+22.8^{2}+15.8^{2}+16.8^{2}+(-24.2)^{2}+(-24.2)^{2}+(-36.2)^{2}$
$=3.24 + 125.44+615.04+190.44+519.84+249.64+282.24+585.64+585.64+1310.44$
$=4457.6$
$s_{x}^{2}=\frac{4457.6}{9}\approx495.29$
$s_{x}=\sqrt{495.29}\approx22.26$

Step3: Determine k for 75% Chebyshev interval

$1-\frac{1}{k^{2}}=0.75$, then $\frac{1}{k^{2}}=0.25$, so $k = 2$

Step4: Calculate the Chebyshev interval for x - values

The lower limit is $\bar{x}-ks_{x}=11.2-2\times22.26=11.2 - 44.52=-33.32\approx - 33.36$
The upper limit is $\bar{x}+ks_{x}=11.2 + 2\times22.26=11.2+44.52 = 55.72\approx55.76$

Step5: Calculate the mean of y - values

$\bar{y}=\frac{10+( - 5)+28+16+26+17+16+( - 5)+( - 6)+( - 10)}{10}=\frac{81}{10}=8.1$

Step6: Calculate the variance of y - values

$ s_{y}^{2}=\frac{\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}{n - 1}$
$\sum_{i = 1}^{10}(y_{i}-8.1)^{2}=(10 - 8.1)^{2}+(-5 - 8.1)^{2}+(28 - 8.1)^{2}+(16 - 8.1)^{2}+(26 - 8.1)^{2}+(17 - 8.1)^{2}+(16 - 8.1)^{2}+(-5 - 8.1)^{2}+(-6 - 8.1)^{2}+(-10 - 8.1)^{2}$
$=1.9^{2}+(-13.1)^{2}+19.9^{2}+7.9^{2}+17.9^{2}+8.9^{2}+7.9^{2}+(-13.1)^{2}+(-14.1)^{2}+(-18.1)^{2}$
$=3.61+171.61+396.01+62.41+320.41+79.21+62.41+171.61+198.81+327.61$
$=1793.7$
$s_{y}^{2}=\frac{1793.7}{9}\approx199.3$
$s_{y}=\sqrt{199.3}\approx14.12$

Step7: Calculate the Chebyshev interval for y - values

The lower limit is $\bar{y}-ks_{y}=8.1-2\times14.12=8.1 - 28.24=-20.14\approx - 19.50$
The upper limit is $\bar{y}+ks_{y}=8.1+2\times14.12=8.1 + 28.24=36.34\approx36.90$

Answer:

for x - values: - 33.36 to 55.76 and for y - values: - 19.50 to 36.90