A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one particular customer for the past 20 months. Use the given data to answer parts (a) and (b).
$$ \begin{array}{lllll} 328 & 346 & 371 & 455 & 325 \\ \hline 417 & 451 & 498 & 344 & 320 \\ \hline 476 & 549 & 438 & 462 & 353 \\ \hline 545 & 432 & 460 & 552 & 478 \end{array} $$
(a) Determine the standard deviation and interquartile range of the data.
(b) Suppose the month in which the customer used 328 minutes was not actually that customer's phone. That particular month the customer did not use their phone at all, so 0 minutes were used. How does changing the observation from 328 to 0 affect the standard deviation and interquartile range? What property does this illustrate?
The standard deviation _______ and the interquartile range _______
What property does this illustrate? Choose the correct answer below.
Dispersion
Empirical Rule
Resistance
Weighted Mean
Homework Answers
Mean = (328+346+371+455+325+417+451+498+344+320+476+549+438+462+353+545+432+460+552+478) / 20
= 430
Standard deviation = sqrt[[(328 - 430)2 +(346 - 430)2 +(371 - 430)2 +(455 - 430)2 +(325 - 430)2 +(417 - 430)2 +(451 - 430)2 +(498 - 430)2 +(344 - 430)2 +(320 - 430)2 +(476 - 430)2 +(549 - 430)2 +(438 - 430)2 +(462 - 430)2 +(353 - 430)2 +(545 - 430)2 +(432 - 430)2 +(460 - 430)2 +(552 - 430)2 +(478 - 430)2 ] / 19 ]
= 76.75
Data in sorted order is,
320 325 328 344 346 353 371 417 432 438 451 455 460 462 476 478 498 545 549 552
The data will be divided into two halves as
320 325 328 344 346 353 371 417 432 438
451 455 460 462 476 478 498 545 549 552
The median of these two halves are Q1 and Q3
Q1 = Mean of 5th and 6th number = (346 + 353)/2 = 349.5
Q3 = Mean of 5th and 6th number = (476 + 478) / 2 = 477
IQR = Q3 - Q1 = 477 - 349.5 = 128 (Rounding to nearest integer)
If we change any observation from 328 to 0, standard deviation will increase as the data will now be more spread out.
This property is called Dispersion
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