The above table presents a hypothetical distribution of final scores (grouped data) obtained by students taking the statistics class

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.

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