Answer Key to Econ310 Midterm Spring 2012Answer key is provided by Zichen Qiu. If you find any error, please email to [email protected]. You may also find an updated version from http://www.zichenqiu.com/teaching–spring2012-2013.html. You may also find other class materials there. Good Luck!1. According to the last census, 4 out of every 10 working women in the UnitedStates held full-time jobs in 2002. Suppose a random sample of 25 working womenis drawn. Answer the following questions:(a) [10 points] What is the probability that 8 of them held full-time jobs?Ans: From the description of the problem, it is clear it involves a Binomial randomvariable. The underlying Binomial experiment consists of 25 trials, and we candescribe “success” in the i’th trial as occurring if the i’th individual hold full-timejobs. Note that P = P r(success) =410= 0.40.Let X= number of success, i.e., the number of working women with full-time jobs.X ∼ B(25, 0.4)The question is asking for P(X=8), according to the Binomial Table,P (X = 8) = P (X ≤ 8) − P (X ≤ 7) = 0.2735 − 0.1536 = 0.1199(b) [10 points] What is the probability that no less than 10 and no morethan 15 of them held full-time jobs?Ans: This question is asking for P (10 ≤ X ≤ 15). According to binomial accumu-lative probability formulaP (10 ≤ X ≤ 15) = P (X ≤ 15) − P (X ≤ 9) = 0.9868 − 0.4246 = 0.5622(c)[15 points] Provide an interval that will include, with probability atleast 90%, the actual number of women with full-time jobs in this sam-ple?Ans:According to Chebysheff’s theorem11 −1k2= 0.9which implies thatk = 3.162According to the Binomial Formula,µx= np = 10σx=qnp(1 − p) = 2.449Interval is thus[µx−3.162·σx, µx+3.162·σx] = [10−3.162·2.449, 10+3.162·2.449] = [2.256, 17.743]2. The average number of items sold by an online company is 0.10 items per minute.Suppose we treat the number of items sold as a Poisson random variable. Answerthe following.(a) [10 points] What is the probability that at least one item will be soldin the next ten minutesAns: Let X= the number of items sold in the next 10 minutes. Then the averageµ = 0.10 · 10 = 1.0. That is,X ∼ P oisson(1.0)According to the Complement rule, and using the Poisson TableP r(X ≥ 1) = 1 − P r(X = 0) = 1 − 0.3679 = 0.6321(b) [10 points] What is the probability that 3 items will be sold in thenext fifteen minutes?Ans: Let X= the number of items sold in the next 15 minutes. Then the averageµ = 0.10 ∗ 15 = 1.5. That is,X ∼ P oisson(1.5)P r(X = 3) = P (X ≤ 3) − P (X ≤ 2) = 0.9344 − 0.8088 = 0.1256(c) [15 points] Provide an interval that will include, with probability atleast 95%, the actual number of items sold over the next 2.5 hours.Ans:2According to Chebysheff’s Theorem1 −1k2= 0.95which implies thatk = 4.472.Let X= the number of items sold in the next 2.5 hours. Then the average µ =0.10 · 60 · 2.5 = 15. That is,X ∼ P oisson(15)According to the Poisson Formula, we haveµx= σ2x= 15, thus the standard deviation isσx=√15 = 3.873So the interval is[µx−4.472·σx, µx+4.472·σx] = [15−4.472·3.873, 15+4.472·3.873] = [−2.32, 32.32]3. Using data on its most recent sales, a department store has determined a relation-ship between method of payment and price category summarized by the followingjoint probability table.Payment MethodPriceCategory Cash Credit Card Debit CardLess than $20 0.12 0.03 0.07$20-$100 0.07 0.19 0.15More than $100 0.03 0.20 0.14(a) [10 points] Suppose a purchase amount is more than $100. What isthe probability that it was paid with cash?Ans:0.030.03 + 0.20 + 0.14=0.030.37= 0.081(b) [10 points] Suppose a purchase is paid with a credit card. What isthe probability that the purchase amount is $20-$100?Ans:30.190.03 + 0.19 + 0.20=0.42= 0.452(c) [10 points] Suppose a purchase is paid with a debit card. What isthe probability that the amount purchased is $20 or more?Ans:0.15 + 0.140.07 + 0.15 + 0.14=0.290.36= 0.8054. Economic theory predicts that higher interest rates will result in less overall em-ployment because higher interest rates would make firms more reluctant to borrowand expand their business. To investigate this, an economist collects monthlydata from 1950 to 2009 for monthly prime bank rate (“interest rate”) in percent-age points an monthly unemployment rate (“unemployment”) also in percentagepoints. Suppose the data collected reveals the followingAverage(Unemployment)=5.68, Variance(Unemployment)=2.42Average(Interest Rate)=7.09, Variance(Interest Rate)=11.38Correlation(Unemployment, Interest Rate)=0.38Suppose we assume a linear relationship between unemployment and interest rate,Unemployment=b0+b1· (Interest Rate)(a) [15 points] Using the least squares method, estimate the intercept(b0) and slope (b1) coefficients. (Hint : Recall the relationship betweencorrelation and covariance.) Does the sign of b1confirm or refute thetheory described above?Ans:Let X= interest rate,¯X = 7.09, and sx=√11.38 = 3.373Let Y= unemployment, ¯y = 5.68, and sy=√2.42 = 1.555Note that sample coefficient of correlation r =sxysx·sy= 0.38then sample covariance sxy= r · sx· sy= 0.38 · 3.373 · 1.555 = 1.993ˆy = b0+ b1· xwhere b1=sxys2x, b0= ¯y − b1· ¯x.Therefore b1=sxys2x=1.99311.38= 0.175 > 0, andb0= ¯y − b1· ¯x = 5.68 − 0.175 · 7.09 = 4.439Since b1> 0, this refute the theory described above.4(b) [5 points] What would be the predicted unemployment rate for a 10percent interest rate(i.e. Interest Rate=10)?Ans:ˆy = b0+ b1· 10 = 4.439 + 0.175 · 10 = 6.189, predicted unemployment.(c) [10 points] According to the linear model, what is the level of interestrate that would product an unemployment of 5 percent (i.e., Unemploy-ment=5)?Ans:5 = b0+ b1· x = 4.439 + 0.174ximpliesx = 3.2055. Suppose the duration of long-distance calls made by employees of a certain com-pany is assumed to be normally distributed (and therefore a continuous randomvariable) with a mean of 5.5 minutes and a standard deviation of 2.5 minutes.(a) [10 points] Find the probability that a call lasts less than 2 minutes.Ans: Let X=duration of long-distance calls made by employees of a certain com-pany.X ∼ N(5.5, 2.5)According to Standard Normal Distribution Table,P (X < 2) = P (Z <2 − 5.52.5) = P (Z < −1.4) = 0.0808(b) [10 points] Find the probability that a call lasts between 6 and 10minutes.Ans:P (6 ≤ X ≤ 10) = P (X ≤ 10) − P (X ≤ 6) = P (Z ≤10−5.52.5) − P (Z ≤6−5.52.5)= P (Z ≤