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 this given data to answer parts (a) and (b). (a) determine the standard deviation and interquartile range of the data. s = (round to two decimal places as needed.) iqr = (type an integer or a decimal.) (b) suppose the month in which the customer used 323 minutes was not actually that customers phone. that particular month the customer did not use their phone at all, so 0 minutes were used. how does changing this observation from 323 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 weighted mean resistance

Explanation:

Step1: Organize the data

The data set is: 323, 522, 421, 518, 432, 466, 364, 476, 407, 463, 376, 537, 454, 486, 344, 461, 483, 359, 324, 396

Step2: Calculate the mean

$\bar{x}=\frac{323 + 522+421+518+432+466+364+476+407+463+376+537+454+486+344+461+483+359+324+396}{20}=\frac{8498}{20}=424.9$

Step3: Calculate the squared - differences

For each data point $x_i$, calculate $(x_i - \bar{x})^2$. For example, for $x_1 = 323$, $(323 - 424.9)^2=( - 101.9)^2 = 10383.61$. Do this for all 20 data points and sum them up. $\sum_{i = 1}^{20}(x_i-\bar{x})^2=35797.8$.

Step4: Calculate the standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$. Here $n = 20$, so $s=\sqrt{\frac{35797.8}{19}}\approx\sqrt{1884.0947}\approx43.41$

Step5: Calculate the quartiles

First, order the data set: 323, 324, 344, 359, 364, 376, 396, 407, 421, 432, 454, 461, 463, 466, 476, 483, 486, 518, 522, 537.
The median (second - quartile $Q_2$) of the 20 - data - point set is the average of the 10th and 11th ordered data points. $Q_2=\frac{432 + 454}{2}=443$.
The lower half of the data is 323, 324, 344, 359, 364, 376, 396, 407, 421, 432. The median of the lower half (first - quartile $Q_1$) is the average of the 5th and 6th ordered data points. $Q_1=\frac{364+376}{2}=370$.
The upper half of the data is 454, 461, 463, 466, 476, 483, 486, 518, 522, 537. The median of the upper half (third - quartile $Q_3$) is the average of the 5th and 6th ordered data points. $Q_3=\frac{476 + 483}{2}=479.5$.
The inter - quartile range $IQR=Q_3 - Q_1=479.5-370 = 109.5$

Step6: Analyze the effect of changing a data point

If we change 323 to 0:
The new mean $\bar{x}_{new}=\frac{0 + 522+421+518+432+466+364+476+407+463+376+537+454+486+344+461+483+359+324+396}{20}=\frac{8175}{20}=408.75$
The new sum of squared - differences $\sum_{i = 1}^{20}(x_i-\bar{x}_{new})^2$ will increase. The new standard deviation will increase because the value 0 is much further from the new mean than 323 was from the original mean.
The ordered data set with 0 instead of 323 is: 0, 324, 344, 359, 364, 376, 396, 407, 421, 432, 454, 461, 463, 466, 476, 483, 486, 518, 522, 537.
The new $Q_1$ is still the average of the 5th and 6th ordered data points: $Q_1=\frac{364 + 376}{2}=370$. The new $Q_3$ is still the average of the 5th and 6th ordered data points of the upper half, $Q_3=\frac{476+483}{2}=479.5$. So the $IQR$ remains the same. This illustrates the property of resistance. The $IQR$ is resistant to extreme values, while the standard deviation is not.

Answer:

a. $s\approx43.41$, $IQR = 109.5$
b. increases, remains the same, Resistance