QME EXAM NEWEST 2024 ACTUAL EXAM COMPLETE QUESTIONS AND CORRECT
DETAILED ANSWERS’[ALREDY GRADED A+]
Question 1: Probability and Statistics
Question:
A company produces light bulbs, and the lifetime of these light bulbs is normally distributed with
a mean of 800 hours and a standard deviation of 100 hours. What is the probability that a
randomly selected light bulb will last more than 900 hours?
Answer:
To find the probability that a light bulb lasts more than 900 hours, we need to calculate the Z-
score and then find the corresponding probability from the standard normal distribution.
1. Calculate the Z-score:
Z=X−μσZ = \frac{X - \mu}{\sigma}Z=σX−μ
Where:
○ XXX is the value (900 hours)
○ μ\muμ is the mean (800 hours)
○ σ\sigmaσ is the standard deviation (100 hours)
2. Z=900−800100=100100=1Z = \frac{900 - 800}{100} = \frac{100}{100} =
1Z=100900−800=100100=1
3. Find the probability:
The Z-score of 1 corresponds to the probability of 0.8413 in the standard normal
distribution table, which gives the probability that a light bulb lasts less than 900 hours.
4. Calculate the complement:
The probability that a light bulb lasts more than 900 hours is:
P(X>900)=1−P(X≤900)=1−0.8413=0.1587P(X > 900) = 1 - P(X \leq 900) = 1
- 0.8413 = 0.1587P(X>900)=1−P(X≤900)=1−0.8413=0.1587
Therefore, the probability that a randomly selected light bulb will last more than
900 hours is 0.1587 or 15.87%.
Question 2: Linear Regression
Question:
Suppose you have the following simple linear regression model: Y=β0+β1X+ϵY = \beta_0 + \
beta_1X + \epsilonY=β0+β1X+ϵ, where YYY is the dependent variable, XXX is the independent
variable, and ϵ\epsilonϵ is the error term. Given the following data, calculate the estimated
coefficients β0\beta_0β0 and β1\beta_1β1:
X Y
1 2
2 3
3 5
, 4 4
5 6
Answer:
1. Calculate the means of X and Y:
Xˉ=1+2+3+4+55=3\bar{X} = \frac{1 + 2 + 3 + 4 + 5}{5} = 3Xˉ=51+2+3+4+5=3
Yˉ=2+3+5+4+65=4\bar{Y} = \frac{2 + 3 + 5 + 4 + 6}{5} = 4Yˉ=52+3+5+4+6=4
2. Calculate the slope β1\beta_1β1:
β1=∑(Xi−Xˉ)(Yi−Yˉ)∑(Xi−Xˉ)2\beta_1 = \frac{\sum (X_i - \bar{X})(Y_i - \
bar{Y})}{\sum (X_i - \bar{X})^2}β1=∑(Xi−Xˉ)2∑(Xi−Xˉ)(Yi−Yˉ) β1=(1−3)
(2−4)+(2−3)(3−4)+(3−3)(5−4)+(4−3)(4−4)+(5−3)(6−4)
(1−3)2+(2−3)2+(3−3)2+(4−3)2+(5−3)2\beta_1 = \frac{(1-3)(2-4) + (2-3)(3-
4) + (3-3)(5-4) + (4-3)(4-4) + (5-3)(6-4)}{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-
3)^2 + (5-3)^2}β1=(1−3)2+(2−3)2+(3−3)2+(4−3)2+(5−3)2(1−3)
(2−4)+(2−3)(3−4)+(3−3)(5−4)+(4−3)(4−4)+(5−3)(6−4) β1=(−2)(−2)+
(−1)(−1)+(0)(1)+(1)(0)+(2)(2)4+1+0+1+4\beta_1 = \frac{(-2)(-2) + (-1)(-1)
+ (0)(1) + (1)(0) + (2)(2)}{4 + 1 + 0 + 1 + 4}β1=4+1+0+1+4(−2)(−2)+
(−1)(−1)+(0)(1)+(1)(0)+(2)(2) β1=4+1+0+0+410=910=0.9\beta_1 = \
frac{4 + 1 + 0 + 0 + 4}{10} = \frac{9}{10} = 0.9β1=104+1+0+0+4=109
=0.9
3. Calculate the intercept β0\beta_0β0:
β0=Yˉ−β1Xˉ\beta_0 = \bar{Y} - \beta_1\bar{X}β0=Yˉ−β1Xˉ
β0=4−0.9×3=4−2.7=1.3\beta_0 = 4 - 0.9 \times 3 = 4 - 2.7 = 1.3β0
=4−0.9×3=4−2.7=1.3
Therefore, the estimated regression line is:
Y^=1.3+0.9X\hat{Y} = 1.3 + 0.9XY^=1.3+0.9X
Question 3: Optimization
Question:
A company wants to maximize its profit given by the function
P(x,y)=20x+30y−5x2−10y2P(x, y) = 20x + 30y - 5x^2 -
10y^2P(x,y)=20x+30y−5x2−10y2, where xxx and yyy are quantities of two
different products. Determine the values of xxx and yyy that maximize the profit.
Answer:
To find the maximum profit, we need to find the critical points by setting the partial derivatives
equal to zero and solving the resulting system of equations.
1. Calculate the partial derivatives:
∂P∂x=20−10x\frac{\partial P}{\partial x} = 20 - 10x∂x∂P=20−10x
∂P∂y=30−20y\frac{\partial P}{\partial y} = 30 - 20y∂y∂P=30−20y
2. Set the partial derivatives equal to zero:
20−10x=020 - 10x = 020−10x=0 30−20y=030 - 20y = 030−20y=0
3. Solve for xxx and yyy:
x=2x = 2x=2 y=1.5y = 1.5y=1.5
, 4. Verify that these points are a maximum:
Calculate the second partial derivatives:
∂2P∂x2=−10\frac{\partial^2 P}{\partial x^2} = -10∂x2∂2P=−10
∂2P∂y2=−20\frac{\partial^2 P}{\partial y^2} = -20∂y2∂2P=−20
Both second partial derivatives are negative, indicating that the function is
concave down at these points, hence a maximum.
Therefore, the profit is maximized when x=2x = 2x=2 and y=1.5y = 1.5y=1.5.
Question 4: Hypothesis Testing
Question:
A researcher wants to test if the average time spent on a website is 5 minutes. The sample
mean time from a sample of 50 users is 4.8 minutes, with a standard deviation of 1.2 minutes.
Test the hypothesis at the 0.05 significance level.
Answer:
1. State the hypotheses:
○ Null hypothesis (H0H_0H0): μ=5\mu = 5μ=5
○ Alternative hypothesis (H1H_1H1): μ≠5\mu \neq 5μ=5
2. Calculate the test statistic:
t=Xˉ−μsnt = \frac{\bar{X} - \mu}{\frac{s}{\sqrt{n}}}t=nsXˉ−μ
Where:
○ Xˉ\bar{X}Xˉ is the sample mean (4.8)
○ μ\muμ is the population mean (5)
○ sss is the sample standard deviation (1.2)
○ nnn is the sample size (50)
3. t=4.8−51.250=−0.21.27.071=−0.20.169=−1.183t = \frac{4.8 - 5}{\
frac{1.2}{\sqrt{50}}} = \frac{-0.2}{\frac{1.2}{7.071}} = \frac{-0.2}
{0.169} = -1.183t=501.24.8−5=7.0711.2−0.2=0.169−0.2=−1.183
4. Determine the critical value:
For a two-tailed test at α=0.05\alpha = 0.05α=0.05 with df=49df = 49df=49, the critical
value from the t-distribution table is approximately ±2.009\pm 2.009±2.009.
5. Make a decision:
Since −1.183-1.183−1.183 is within the range −2.009<t<2.009-2.009 < t <
2.009−2.009<t<2.009, we fail to reject the null hypothesis.
Therefore, there is not enough evidence to reject the hypothesis that the average
time spent on the website is 5 minutes.
Question 5: Time Series Analysis
Question:
Given the following time series data for the number of units sold over 5 months:
Month Units Sold
1 120
, 2 130
3 150
4 160
5 170
Calculate the 3-month moving average.
Answer:
To calculate the 3-month moving average, we average the units sold over each set of three
consecutive months.
1. Calculate the moving averages:
○ For Month 3: 120+130+1503=4003=133.33\frac{120 + 130 + 150}{3} = \frac{400}
{3} = 133.333120+130+150=3400=133.33
○ For Month 4: 130+150+1603=4403=146.67\frac{130 + 150 + 160}{3} = \frac{440}
{3} = 146.673130+150+160=3440=146.67
○ For Month 5: 150+160+1703=4803=160.00\frac{150 + 160 + 170}{3} = \frac{480}
{3} = 160.003150+160+170=3480=160.00
2. Therefore, the 3-month moving averages are 133.33, 146.67, and 160.00 for Months
3, 4, and 5, respectively.
Question 6: Game Theory
Question:
Two firms, A and B, are competing in a market and can choose to either advertise or not
advertise. The payoffs are as follows:
● If both firms advertise, each earns $3 million.
● If neither firm advertises, each earns $5 million.
● If one advertises and the other does not, the firm that advertises earns $7 million, and
the other earns $2 million.
Construct the payoff matrix and identify the Nash equilibrium.
Answer:
1. Construct the payoff matrix:
AdvertiseNot AdvertiseAdvertise(3,3)(7,2)Not Advertise(2,7)(5,5)\begin{array}{c|c|c} & \
text{Advertise} & \text{Not Advertise} \\ \hline \text{Advertise} & (3, 3) & (7, 2) \\ \hline \
text{Not Advertise} & (2, 7) & (5, 5) \\ \end{array}AdvertiseNot AdvertiseAdvertise(3,3)
(2,7)Not Advertise(7,2)(5,5)
2. Identify the Nash equilibrium:
○ If Firm A advertises, Firm B’s best response is to advertise (since 3 > 2).
○ If Firm A does not advertise, Firm B’s best response is not to advertise (since 5 >
7).
