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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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The linear function f(x) = 0.9x + 79 represents the average test score in your math class, where x is the number of the test tak
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Given:

The linear function for average test score in your math class is

f(x)=0.9x+79

where x is the number of the test taken.

The linear function g(x) in the given table represents the average test score in your science class, where x is the number of the test taken.

To find:

Part A: The test average for your math class after completing test 2.

Part B: The test average for your science class after completing test 2.

Part C: Which class had a higher average after completing test 4?

Solution:

Part A:

We have,

f(x)=0.9x+79

Put x=2 in the above function, to find the test average for your math class after completing test 2.

f(2)=0.9(2)+79

f(2)=1.8+79

f(2)=80.8

Therefore, the test average for your math class after completing test 2 is 80.8.

Part B:

From the given table it is clear that g(x) =79 at x=2.

Therefore, the test average for your science class after completing test 2 is 79.

Part C:

Put x=4 in f(x), to find the test average for your math class after completing test 4.

f(4)=0.9(4)+79

f(4)=3.6+79

f(4)=82.6

From the table it is clear that the value of g(x) is increased by 1 as x is increased by 1. So, the value of function g(x) at x=4 must be 1 more than 80.

g(4)=81

It is clear that,

82.6>81

Therefore, math class had a higher average after completing test 4.

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Answer:

70 and 110

Step-by-step explanation:

First angle -- 2x

Second angle -- 3x + 5

Therefore,

2x + 3x + 5 = 180

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The formula of a distance between two points:

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We have two points (1, -5) and (-3, 6). substitute:

d=\sqrt{(-3-1)^2+(6-(-5))^2}=\sqrt{(-4)^2+11^2}=\sqrt{16+121}=\sqrt{137}

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