Acellular 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). 320 411 348 537 420 449 462 403 454 517 515 358 438 541 387 368 502 437 431 428 (a) determine the standard deviation and interquartile range of the data. sequals nothing (round to two decimal places as needed.) iqrequals nothing (type an integer or a decimal.) (b) suppose the month in which the customer used 320 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 320 to 0 affect the standard deviation and interquartile range

The standard deviation increased but there was no change in the interquantile range          

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

where are data points, is the mean and n is the number of observations.  

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

IQR =

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

IQR =

Thus. the standard deviation increased but there was no change in the interquantile range.

he spend 5.1 minutes chatting with his friends

step-by-step explanation:

2/5+1/2=9/10

6/1-9/10=5 1/10