
b) Here mean =414.1967, SD=86.4961 so that the cut off for two deviations from mean is
mean-2SD=414.1967-2*86.4961=241.2045.
We find the Z score for the least observation 300 as (300-414.1967)/86.4961= -1.320253. Hence no observations are below 241.2045 (i.e. corresponds to a Z score of -2 or below) and the required percentage is 0%.
c) The cut off is mean+2SD=414.1967+2*86.4961= 587.1889 as no observations lie below mean-2SD=414.1967-2*86.4961=241.2045.
d) The number of runners having time more than 587.1889 is one (I.e. 599). i.e.
b) Inner fences: Q1-1.5*IQR=247.25 , Q3+1.5*IQR=467.75
c) Outer fences: Q1-3*IQR=137 , Q3+3*IQR=578
d) Observations falling outside the interval (Q1-1.5*IQR, Q3+1.5*IQR) =(247.25,467.75) are potential outliers.
Observations falling outside the interval (Q1-3*IQR, Q3+3*IQR) =(137, 578) are extreme outliers.
e) There are 15 observations below the value 360. Then percentile rank of the observation 360 is 100*15/61%=24.59%.
f) The percentile rank of 24.59% indicates that 24.59% of the running times are less than 360.
g) D5 is the fifth decile or median, which is 379.
please answer 1) and 2) PART THREE Below are the times for running a mile for 61 runners (measured in seconds). Use...