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ADM2623 (FR01A) Quiz 2 Name (print) Student ID 1. (3 marks) the following sample of time needed to review the applications of ten children at a day care: 36, 42, 18, 32, 22, 22, 25, 29, 30, 31, (a) (0.5 mark) What is the mean length of time they took to review an application? 36+ 42+ 18+32+22+22+25+29+30+31= 287/10 =28.7 (b) (0.5 mark) What is the median length of time they took to review an application? Sort: 18, 22, 22, 25, 29 , 30, 31, 32, 36, 42 Median is (29+30)/2=29.5 (c) (0.5 mark) What is the range of length of time they took to review an application? 42-18=24 (d) (1 mark) What is the variance of length of time they took to review an application? =(8703- (2872/10))/9=466.1/9=51.79 (e) (0.5 mark) What is the standard deviation of length of time they took to review an application? (51.79)0.5 = 7.20 2. (1 mark) The number of Mobil phone users in Canada grew from 3.85 million in the beginning of 2008 to 12.68 million in the end of 2015. Use the geometric mean to find the mean annual growth rate. 8 12.68 8 =√3.29= 3.85 √ 1.1607-1 =16.07% 3. (3 marks) The following table gives the frequency distribution of the number of orders received each day during the past 50 days at the office of ABC Inc. Days of orders 10 up to 13 13 up to 16 16 up to 19 19 up to 22 Frequency (f) Midpoint x CF 11.5 4 14.5 12 17.5 20 20.5 14 50 f*x 4 16 36 50 a. (1 mark) Find an approximate value of the sample mean. n=50, X =857/50= 17.14 b. (1.5 mark) Find an approximate value of the median of the data set. N CF L 2 i f Median for grouped data: (a) n/2=50/2=25. The median class is the third one [16,19). (b) Now we apply the formula: N 50 - CF -16 L+ 2 ´ i =16 + 2 ´ 3 =17.35 f 20 c. (0.5 mark) Find an approximate value of the mode of the data set. The third class [16,19) has the highest frequency. Middle point of the class [16,19) is 17.5 46 174 350 287 857