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photoshop1234 [79]
1 year ago
10

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use of one particular cust

omer for the past 20 months. Use the given data to answerParts a and b321 397 559 454 475324 482 558 369 513385 360 459 403 498477 361 366 372 320Determine the standard deviation and interquartile range of data.
Mathematics
1 answer:
Fiesta28 [93]1 year ago
3 0

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given set of values

321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320

STEP 2: Write the formula for calculating the Standard deviation of a set of numbers

\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ where\text{ }x_i\text{ are data points,} \\ \bar{x}\text{ is the mean} \\ \text{n is the number of values in the data set} \end{gathered}

STEP 3: Calculate the mean

\begin{gathered} \bar{x}=\frac{\sum ^{}_{}x_i}{n} \\ \bar{x}=\frac{\sum ^{}_{}(321,397,559,454,475,324,482,558,369,513,385,360,459,403,498,477,361,366,372,320)}{20} \\ \bar{x}=\frac{8453}{20}=422.65 \end{gathered}

STEP 4: Calculate the Standard deviation

\begin{gathered} S\tan dard\text{ deviation=}\sqrt[]{\frac{\sum^{}_{}(x_i-\bar{x})^2}{n-1}} \\ \sum ^{}_{}(x_i-\bar{x})^2\Rightarrow\text{Sum of squares of differences} \\ \Rightarrow10332.7225+657.9225+18591.3225+982.8225+2740.52251+9731.8225+3522.4225+18319.6225+2878.3225 \\ +8163.1225+1417.5225+3925.0225+1321.3225+386.1225+5677.6225+2953.9225+3800.7225 \\ +3209.2225+2565.4225+10537.0225 \\ \text{Sum}\Rightarrow108974.0275 \\  \\ S\tan dard\text{ deviation}=\sqrt[]{\frac{111714.55}{20-1}}=\sqrt[]{\frac{111714.55}{19}} \\ \Rightarrow\sqrt[]{5879.713158}=76.67928767 \\  \\ S\tan dard\text{ deviation}\approx76.68 \end{gathered}

Hence, the standard deviation of the given set of numbers is approximately 76.68 to 2 decimal places.

STEP 5: Calculate the First and third quartile

\begin{gathered} \text{IQR}=Q_3-Q_1 \\  \\ To\text{ get }Q_1 \\ We\text{ first arrange the data in ascending order} \\ \mathrm{Arrange\: the\: terms\: in\: ascending\: order} \\ 320,\: 321,\: 324,\: 360,\: 361,\: 366,\: 369,\: 372,\: 385,\: 397,\: 403,\: 454,\: 459,\: 475,\: 477,\: 482,\: 498,\: 513,\: 558,\: 559 \\ Q_1=(\frac{n+1}{4})th \\ Q_1=(\frac{20+1}{4})th=\frac{21}{4}th=5.25th\Rightarrow\frac{361+366}{2}=\frac{727}{2}=363.5 \\  \\ To\text{ get }Q_3 \\ Q_3=(\frac{3(n+1)}{4})th=\frac{3\times21}{4}=\frac{63}{4}=15.75th\Rightarrow\frac{477+482}{2}=\frac{959}{2}=479.5 \end{gathered}

STEP 6: Find the Interquartile Range

\begin{gathered} IQR=Q_3-Q_1 \\ \text{IQR}=479.5-363.5 \\ \text{IQR}=116 \end{gathered}

Hence, the interquartile range of the data is 116

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And the best way to solve this problem is using the normal standard distribution and the z score given by:  

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