STA1DCT Assignment 2 Name: Dona Gayana Student Number: 21735504 This is my own work. I have not copied any of it from anyone else. Question 1: Expression 1 = B Expression 2 = A Expression 3 = C Expression 4 = E Expression 5 = D Question 2: a) 6.85 =AVERAGE(A2:A301) b) 6.99 =MEDIAN(A2:A301) c) 1.110756 =STDEV(A2:A301) d) i) 300 x 68 / 100 = 204 Assuming that a normally distributed population has 68% of its values falling within one standard deviation () of the mean (), we can apply the empirical rule to the 300 rat-pup birth weights and expect around 68% of them to fall within this range. As a result, 0.68 300 = 204 0.68 300 = 204, meaning that roughly 204 rat-pup birth weights should fall inside the range of. ii) Since I had already calculated the average and standard deviation in C2 and E2, respectively, I utilised Excel in this instance. Thus, the u+o and u-o entering G2 and H2 cells are calculated using the =C2+E2 and =C2-E2 formulas. Next, utilise the J2 form that follows and drag it down to J301, =IF(AND(A2 >= H$2, A2 <= G$2), 1, 0) After that, use =SUM(J2:J301) in cell L2 and Answer is 201.
Student Number: 21735504 STA1DCT Assignment 2 Name: Dona Gayana iii) (201 / 300) x 100% = (201/3)% = 67% iv) No, 68% of the population should fall within the range in order to roughly distribute the birth weight of rat pups normally. However, only 67% of this population falls within the range. As a result, the rat pups' birth weights are not roughly distributed normally. Question 3: b) The sample average = 6.69 c) The sample medjan =6,85 The sample average 1- 7.08 6.688625 7.96 8 53.49 5.61 48 54 6.71 48 6.91 69 64 4.13 50 48 (b) Assending order :- 4.13 5.61 6.71 6.79 6.91 7.08 7.96 8.30 median :- (8+1)th 2 6.85 11 = 9 th 2 2 13.70 = 4.5 th 12 17 16 10 6.79 6.91 + 6.85 13.70
Student Number: 21735504 STA1DCT Assignment 2 Name: Dona Gayana d) The sample standard deviation = 1.2484 (c) 6 = 1 2 (xi -1)2 Question 4: Histogram A: average = 80 Histogram B: average = 0 Histogram C: average = 50 Histogram D: average = 30 N 02= = (xi-1)2 N =(7.08-6.89)2+(7.96-6,89)2+(5.61-639)2 +(6,71-6,89)2+ (6.91-6-89)2+(4-13-6,89)2+(6.79-6.89)2+(8,30-6.89)} 8 = (0.19)2+(1.07)}(-1.28)+(0.18)+(0.02)2 +(-2.76)2+(-0.10)+(1-41)2 8 02 = 12.4679 8 o = (12.4679 8 .55845 0 = 1. 2484 standard deviation = 5 standard deviation = 40 standard deviation = 20 standard deviation = 30 Question 5: 7.96%