ISVNU-S1,09 REvision Question, Solution

# ISVNU-S1,09 REvision Question, Solution - MODULE 1 Question...

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Unformatted text preview: MODULE 1 Question 1 i A(Sample variance) = 12.627 2 = 159.44 C(Range) = 100-44 = 56 ii Mean = 60 seconds , Median = 57 seconds. and Mode = 54 seconds. Mean of 60 means that if the times taken to assemble the parts by the workers are the same, then each worker will take 60 seconds Median of 57 means that 50% of the workers took less than 57 sec. and 50% of the workers took more than 57 sec. Mode of 54 means that most workers took 54 seconds. iii Skewness is positive Reasons : From summary output, Mean(60) > median(57) From Boxplot : Whisker on right is longer than the whisker on the left iv. Outliers are 85 and 100 v. If 85 and 100 taken out – 24 times left. Mean = 24 100 85 1560 ) ( +- = 57.29 sec. Median is between 12 th and 13 th value = 56 sec. Mode remains the same i.e. 54 sec. vi. 254 25 60 627 12 2 60 2 . ) . ( _ ± = ± = ± s x = 37.746 to 85.254 25 data points are actually within the limits = ) ( 100 26 25 = 96% vii. Since the distribution is positively skewed, use Cebyshev’s Rule At least 75% of the times are within 2 standard deviation of the mean Question 2 . i. Using SD Mode, Sample Mean(A) = 55kg. Sample standard deviation(B) = 18.69 kg Sample Variance (C) = 349.18or(349.32) (D) Count(Sum) = 2750 ii. (E) - IQR = 69-41.5 = 27.5 iii. All the data are within the left and the right whiskers. Therefore there are no outliers. iv. Distribution is positively skewed, reasons: 1) mean(55) > median ( 54) 2) Whiskers on right is longer than whisker on the left 3) In the boxplot, (Q 3 – Q 2 ) > (Q 2-Q 1 ) 4) Value of skewness in the summary statistics is positive. v. 38 37 55 69 18 2 55 2 . ) . ( _ ± = ± = ± s X = 17.62 to 92.38 vi.Looking at the stem and leaf plot only 98 kg is outside the intervals. Therefore 49 boxes have weights within the intervals. 49 boxes is 98%. vii. Since the distribution is positively skewed, use Chebyshev’s theorem to predict. According to Chebyshev’s theorem, at least ) ( 2 2 1 1- = ¾(75%) of the boxes will have weights within the intervals Question 3 a. A = 47.66 B = 2383 C = 54 D = 60 and 62 E = 28.64 F = 100 – 6 =94 G = 50(50.44) = 2522 b. Mean = 47.66 marks. It means that if all the students scored the same marks, then each student scored 47.66 marks Median = 54 marks. It means that 50% of the students scored less than 47.66 marks and 50% of students scored more than 47.66 marks Mode = 60 and 62. It means that most students scored 60 and 62 marks. c. Question 1 Question 2 Negatively skewed Positively skewed i. Sk = - 0.243 Sk = 0.148 ii. Mean < Median Mean > Median iii. Me-Q1>Q3-Me Me-Q1<Me-Q1 d. 28 57 66 47 64 28 2 66 47 2 . . ) . ( . ± = ± = ±- s X = -9.62 to 104.94 Therefore all data are within this limits e. Since data is skewed use Chebyshev’s rule. At least 4 3 2 1 1 2 =- (75%) of the data are expected to be eithin this limits. So it does not contradict the answer of d....
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ISVNU-S1,09 REvision Question, Solution - MODULE 1 Question...

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