Question

The following data represent the monthly phone​ use, in​ minutes, of a customer enrolled in a fraud prevention program for the past 20 months. The phone company decides to use the upper fence as the cutoff point for the number of minutes at which the customer should be contacted. What is the cutoff​ point? 397 391 401 438 342 518 516 335 301 347 500 388 325 422 510 434 541 303 321 345 The cutoff point is minutes.

Answer 1

Upper fence = Q3 + 1.5 ( IQR )

so

The first quartile (or lower quartile or 25th percentile) is the median of the bottom half of the numbers. So, to find the first quartile, we need to place the numbers in value order and find the bottom half.

301   303   321   325   335   342   345   347   388   391   397   401   422   434   438   500   510   516   518   541   

So, the bottom half is

301   303   321   325   335   342   345   347   388   391   

The first quartile of the data set is 338.5.

The third quartile (or upper quartile or 75th percentile) is the median of the upper half of the numbers. So, to find the third quartile, we need to place the numbers in value order and find the upper half.

301   303   321   325   335   342   345   347   388   391   397   401   422   434   438   500   510   516   518   541   

So, the upper half is

397   401   422   434   438   500   510   516   518   541   

The third quartile of the data set is 469.

The interquartile range is the difference between the third and first quartiles.

The third quartile is 469.

The first quartile is 338.5.

The interquartile range = 469 - 338.5 = 130.5.

Upper fence = Q3 + 1.5 ( IQR )

= 469 + 1.5 ( 130.5 )

= 664.75

Therefore

Answer : The cutoff point is 664.75 minutes