Question

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). 339 456 425 478 549 556 505 544 446 472 449 521 545 445 468 367 431 394 408 496 (a) Determine the standard deviation and interquartile range of the data Sul (Round to two decimal places as needed.)

Answer 1

Answer :

Given data is :

Sample size n = 20

339	456	446	472	449
425	478	521	545	445
549	556	468	367	431
505	544	394	408	496

From the data we need to find the standard deviation S :

First find mean = $\bar{X}$  $= \sum X/n$

= [339 + 456 + 446 + 472 + 449 + ............ + 408 + 496] / 20

= 9294 / 20

= 464.7

$\bar{X}$ = 464.7

Standard deviation= S = $\sqrt{\sum (X-\bar{X})^2/(n-1)}$

$=\sqrt{(339-464.7)^2+(456-464.7)^2+(446-464.7)^2+(472-464.7)^2..............+(408-464.7)^2+(496-464.7)^2/(20-1)}$$=\sqrt{(-125.7)^2+(-8.7)^2+(-18.7)^2+(7.3)^2..............+(-56.7)^2+(31.3)^2/19}$$=\sqrt{15800.49+75.69+349.69+53.29+............+3214.89+979.69/19}$$=\sqrt{71524.2/19}$

$=\sqrt{3764.432}$

$=61.355$

S = 61.355

Next we need to find interquartile range of data :

Take first 10 numbers :

339

456

446

472

449

425

478

521

545

445

Sort this data into ascending order to find median :

339

425

445

446

449

456

472

478

521

545

Median of first 10 data numbers= $Q_{1}$ = (449 + 456) / 2 = 905 / 2 = 452.5

$Q_{1}$ = 437

Now take remaining 10 numbers :

549

556

468

367

431

505

544

394

408

496

We need to sort the above data to find the median :

367

394

408

431

468

496

505

544

544

549

Median of above 10 numbers = ( 468 + 496) / 2 = 964 / 2 = 482

$Q_{3}=482$

$Q_{2}$ is nothing but the median of all complete data.

Formula to calculate interquartile range of data is :

IQR = $Q_{3}-Q_{1}$ = 482 - 437 = 45

IQR = 45