Problem set 2_2024 solutions.docx

D:\\Desktop\\FDP\_SDA\\IIMA.png **Quantitative Methods 1b PGP/PGP-FABM 2024** **[Practice set 2 solutions]** *[These problems are based on Chapters 3, 4 and 5 of the lecture notes. Please work out these problems to reinforce the in-class learnings.]* **Q1.** **ITC Narmada**: A waiter at the popu...

D:\\Desktop\\FDP\_SDA\\IIMA.png **Quantitative Methods 1b PGP/PGP-FABM 2024** **[Practice set 2 solutions]** *[These problems are based on Chapters 3, 4 and 5 of the lecture notes. Please work out these problems to reinforce the in-class learnings.]* **Q1.** **ITC Narmada**: A waiter at the popular "Peshawri" restaurant ITC Narmada believes that the tips he receives from customers have a right skewed distribution with a mean of ₹100 and standard deviation of ₹25.  a\) What will be the mean and standard error of the sampling distribution of the average tip he receives from 15 customers ? Can you assume it to be normal ? Justify. *Ans: Mean will be equal to the population mean i.e 100.* *Standard error will be 25/√15 = 6.45.* *No, we cannot assume the sampling distribution to be normal because i) the population distribution is right skewed and ii) the sample size is only 15 (i.e less than 30). Hence, the CLT does not hold.* b\) What about the distribution of the average tip that he receives from 55 customers ? *Ans : Let* *be the average tip from 55 customers. Since 55 ≥ 30, we can apply the CLT and hence* *will have a normal distribution (even though the population distribution is skewed) with mean = 100 and standard error = 25/√55 = 3.37.* c\) Calculate the approximate probability that the average tip the waiter receives from 55 customers is between ₹ 110 and ₹ 140 ? *Ans : Required probability: P(110 ≤* *≤ 140) = P((110-100)/25/√55≤ Z ≤ (140-100)/25/√55)* *= P(2.97 ≤ Z ≤11.86) = P(Z ≤ 11.86) -- P(Z ≤ 2.97) = 1 -- 0.9985 =0.0015.* d\) Calculate the approximate probability that the average tip the waiter receives from 55 customers is at most ₹ 110 ? *Ans : Required probability: P(* *≤ 110) = P(Z ≤ (110-100)/25/√55)= P(Z ≤2.97) = 0.9985.* **Q2**: **Café** **Kamla**: Sometimes I visit Café Kamla, the newest restaurant on campus, to have breakfast. The probability that on any given day I would visit Café Kamla is 0.25 and the decisions to visit on successive days can be assumed to be independent. When I do visit, I always have the Masala Dosa, which costs me Rs. 70. a\) What is the distribution of my daily expenditure at Café Kamla? *Ans : Let "X" be my daily expenditure. As per the problem, X will have the following distribution* *70 with probability 0.25* *X =* *0 with probability 0.75* b\) Obtain the expectation and variance of the distribution of my daily expenditure at the café. *Ans : Expectation will be E(X) = =* 70x0.25 = 17.5. *Variance will be Var(X) =.* *Hence, standard deviation of X will be.* c\) Obtain the sampling distribution of my average daily expenditure at Café Kamla over two days. *Ans : Let denote the daily average expenditure over two days. The required sampling* *distribution of will be as follows :* *= (70+70)/2 = 70 with probability 0.25\*0.25 = 0.0625* *(0+70)/2 = 35 with probability 2\*0.75\*0.25 = 0.375* *(0 + 0)/2 = 0 with probability 0.75\*0.75 = 0.5625* d\) Obtain the expectation and variance of my average daily expenditure at the café over two days. *Ans : will have the following expectation :* *E()* = *which is nothing but the population mean,.* *Similarly,* *variance of will be* *Var() =.* *Hence, standard deviation (or standard error) of will be* se() = *which is the same as.* *Note that the standard deviation/error of is less than that of X (21.43\