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.

446, 363, 488, 329, 468, 304, 444, 459, 336, 415, 403, 542, 473, 448, 531, 355, 451, 547, 465, 334

What is the cutoff​ point?

Answer 1

The 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.

Arrange the given data in ascending order

304, 329, 334, 336, 355, 363, 403, 415, 444, 446, 448, 451, 459, 465, 468, 473, 488, 531, 542, 547.

The first quartile, Q₁ is the median of lower half of the number.

Lower half = 304, 329, 334, 336, 355, 363, 403, 415, 444, 446.

Q₁ = median of lower half of number

= 355 + 363 / 2

= 359

So Q₁ = 359

and

The third quartile Q₃ is the median of upper half of the number.

Upper half = 448, 451, 459, 465, 468, 473, 488, 531, 542, 547.

Q₃ = median of upper half of number

= 468 + 473 / 2

= 470.5

So Q₃ = 470.5

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

IQR = Q₃-Q₁ = 470.5 - 359 = 111.5

So IQR = 111.5

Now,

Upper fence=Q₃+(1.5×IQR) = 470.5+(1.5×111.5) = 637.75

Therefore, the cutoff point is 637.75 minutes.