Assignment-2 - Intro to BA

SIRIKI GAYATHRI R01400927 BUAN 571: Introduction to Business Analytics Assignment 2: Statistical Concepts Due in the Brightspace drop box before class on Wednesday, 9/13; work must be shown for credit . (Note: This is an individual assignment.) 1. Surgery for a painful, common back condition often results in significantly reduced back pain and better physical function than treatment with drugs and physical therapy. A researcher followed 800 patients, of whom 320 ended up getting surgery. After two years, of those who had surgery, 80% said they had a major improvement of their condition, compared with 30% among those who received nonsurgical treatment. a. What is the probability that a patient did not have surgery? b. What is the probability that a patient had surgery and experienced a major improvement in his or her condition? c. What is the probability that a patient experienced a major improvement in his or her condition? Ans: Given: Total number of patients = 800 Number of patients who had surgery = 320 Patients who had a major improvement after a surgery = 80% Number of patients who had a major improvement after surgery = 320*80% = 256 Number of patients who don’t received a surgery = 800 – 320 = 480 Patients who had a major improvement after nonsurgical treatment = 480*30% = 144 a.) Probability that a patient did not have a surgery: Probability of a patient did not have a surgery = Total patients – Patients who received surgery Total Patients = 800 – 320 800 = 0.6 b.) Probability that a patient had a surgery and experienced a major improvement: Probability = Patients who had a surgery* Patients who had a major improvement after surgery Total number of patients = 320*80%

800 = 0.32 c.) Probability that a patient experienced a major improvement in their condition: Probability = Number of patients having surgery and major improvement + Number of patients having a major improvement with nonsurgical treatment Total number of patients = 256 + 144 800 = 0.5 2. The Massachusetts State Police is trying to crack down on speeding on the Mass Pike. To aid in this effort, they employ a type of radar gun that has a 0.98 probability of detecting a speeder if the driver is actually speeding. Assume that there is a very low (1%) chance that the gun erroneously indicates a speeder even when the driver is below the speed limit. From past history, it is known that 20% of drivers exceed the speed limit on the Massachusetts Turnpike. a. What is the probability that the gun detects speeding and the driver was speeding? b. Suppose the police stop a driver because the gun detected speeding. What is the probability that the driver was actually speeding? Ans: Given: P(A) = Probability that a driver is speeding = 20% = 0.20 P(B/A) = Probability that the radar gun detects speeding when the driver is actually speeding = 0.98 P (B/A’) = Probability that there is low chance that driver exceed the speed limit = 0.01 a.) Probability that the driver was speeding if the gun detects it: P (B and A) = P (A) * P(B/A) = 0.20 * 0.98 = 0.196 b.) The probability that the driver was speeding if police stops the driver because gun detected speed: P(A/B) = P(B/A) * P(A) P(B) = 0.98 * 0.20 0.204 = 0.960784

a.) Probability that only three of these college graduates will pursue graduate degrees: P (X < 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) P (X = 0) = 10 C 0 * (0.25) ^0 * (0.75) ^10 = 0.056313515 P (X = 1) = 10 C 1 * (0.25) ^1 * (0.75) ^9 = 0.187711716 P (X = 2) = 10 C 2 * (0.25) ^2 * (0.75) ^8 = 0.281567574 P (X = 3) = 10 C 3 * (0.25) ^3 * (0.75) ^7 = 0.250282288 0.056313515 + 0.187711716 + 0.281567574 + 0.250282288 = 0.775875092 b.) Probability that exactly five of these college graduates will pursue graduate degree: P (X = 5) = 10 C 5 * (0.25) ^5 * (0.75) ^5 = 0.0583992 c.) Probability that at least six but no more than nine of these college graduates plan to pursue a graduate degree. P (6 < X < 9) = P (X = 6) + P (X = 7) + P (X = 8) + P (X = 9) = 0.016222 + 0.003089905 + 0.0003869905 + 2.86102E-05 = 0.019726753. Used Excel BINOM.DIST for rechecking n 10 p 0.25 x 0 0.056313515 1 0.187711716 2 0.281567574 P (X < or = 3) 0.77587509 2 3 0.250282288 4 0.145998001 5 0.0583992 6 0.016222 7 0.003089905 8 0.000386238 P ( 6 <= X <=9) 0.01972675 3 9 2.86102E-05 10 9.53674E-07