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Homework answers / question archive / The above table presents a hypothetical distribution of final scores (grouped data) obtained by students taking the statistics class
The above table presents a hypothetical distribution of final scores (grouped data) obtained by students taking the statistics class. (50 marks total) Points 40-49 5 50-59 10 60-69 10 70-79 30 80-89 S
b. Calculate i. The fiftieth (50th) percentile (2 marks) ii. The inter-quartile (Q) range (Hint: Find the 25th percentile and 75th percentile first) (5 marks - 4 marks for steps and 1 mark for the answer) c. Suppose everyone's score increased by 10, calculate i. The mode (2 marks) ii. The median (4 marks – 2 marks for steps and 2 marks for the answer) iii. The mean (4 marks - 2 marks for steps and 2 marks for the answer) iv. Variance (4 marks - 2 marks for steps and 2 marks for the answer) v. Standard Deviation (4 marks – 2 marks for steps and 2 marks for the answer) Comparing the answers of part c with part a? Do you observe any differences for the mode, median, mean, variance and standard deviation? For each of the following central tendency or dispersion measures, please use one sentence to summarize your observation. i. The mode (1 mark) ii. The median (1 mark) d. Focus h (United States) MacBook
As the given data is not continous we first make the data continuos by adding .5 to the higher limit and subtracting .5 to the lower lmit. After that we will find the cumulative frequency which is the sum of the frequency.
| points | Frequency | Cumulative Frequency |
| 40-49.5 | 5 | 5 |
| 49.5-59.5 | 10 | 5+10=15 |
| 59.5-69.5 | 10 | 5+10+10=25 |
| 69.5-79.5 | 30 | 5+10+10+30=55 |
| 79.5-89 | 5 | 5+10+10+30+5=60 |
N(Total Frequency)=60
b. i.)
50th percentile = lower limit + (50N/100 - CF (previous one))/Frequency(current one)* class limit gap
50N/100= 50*60/100=30
now lets see which Cumulative frequency is > 30. It lies in 69.5-79.5(as its CF is 55) where lower limit = 69.5 and CF (previous one) is 25 (above 55) and frequency is 30 and class gap is 10
putting this in formula we will get, 69.5+(30-25)/30*10=69.5+1.66=71.167
Therefore 50th percentile is 71.167
b. ii.) inter-quartile range
first lets find 25th percentile 25N/100= 25*60/100=15
Lets see which CF is >15 and lie in which group. The exceeding class is 59.5-69.5 where lower limit is 59.5 and CF of previous class is 15 and Frequency of current class is10
putting this in formula 59.5+ (15-15)/10*10= 59.5
so 25th percentile is 59.5
Now 75th percentile is 75N/100= 45
Exceedling class is 69.5 and 79.5 where lower limit is 69.5 and Cf previous class is 25 and current frequency is 30
putting this in formula we get 69.5+ (45-25)/30*10=76.167
Inter quartile range = 75th percentile - 25th percentile = 76.167-59.50= 16.667
