Solve the equation 39=3m-12

The Empirical Rule The following data represent the length of eruption for a random sample of eruptions at the Old Faithful geys

ad-work [718]

Answer:

(a) Sample Standard Deviation approximately to the nearest whole number = 6

(b) The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is.

(c) The percentage of eruptions that last between 92 and 116 seconds using the empirical rule is 95%

(d) The actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) The percentage of eruptions that last less than 98 seconds using the empirical rule is 16%

(f) The actual percentage of eruptions that last less than 98 seconds is 15.866%

Step-by-step explanation:

(a) Determine the sample standard deviation length of eruption.

Express your answer rounded to the nearest whole number.

Step 1

We find the Mean.

Mean = Sum of Terms/Number of Terms

= 90+ 90+ 92+94+ 95+99+99+100+100, 101+ 101+ 101+101+ 102+102+ 102+103+103+ 103+103+103+ 104+ 104+104+105+105+105+ 106+106+107+108+108+108 + 109+ 109+ 110+ 110+110+110+ 110+ 111+ 113+ 116+120/44

= 4582/44

= 104.1363636

Step 2

Sample Standard deviation = √(x - Mean)²/n - 1

=√( 90 - 104.1363636)²+ (90-104.1363636)² + (92 -104.1363636)² ..........)/44 - 1

= √(199.836777 + 199.836777 + 147.2913224+ 102.7458678+ 83.47314049+ 26.3822314+ 26.3822314+ 17.10950413+17.10950413+ 9.836776857+ 9.836776857, 9.836776857+9.836776857+ 4.564049585+ 4.564049585+ 4.564049585+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 0.01859504133+ 0.01859504133+ 0.01859504133+ 0.7458677685+ 0.7458677685+ 0.7458677685+ 3.473140497+ 3.473140497+ 8.200413225+ 14.92768595+ 14.92768595+ 14.92768595+ 23.65495868+ 23.65495868+ 34.38223141+ 34.38223141+34.38223141+ 34.38223141+ 34.38223141+47.10950414+ 78.56404959+ 140.7458677+ 251.6549586) /43

= √1679.181818/43

= √39.05073996

= 6.249059126

Approximately to the nearest whole number:

Mean = 104

Standard deviation = 6

(b) On the basis of the histogram drawn in Section 3.1, Problem 28, comment on the appropriateness of using the Empirical Rule to make any general statements about the length of eruptions.

The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is .

(c) Use the Empirical Rule to determine the percentage of eruptions that last between 92 and 116 seconds.

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

Mean = 104, Standard deviation = 6

For 68% μ - σ = 104 - 6 = 98, μ + σ = 104 + 6 = 110

For 95% μ – 2σ = 104 -2(6) = 104 - 12 = 92

μ + 2σ = 104 +2(6) = 104 + 12 = 116

Therefore, the percentage of eruptions that last between 92 and 116 seconds is 95%

(d) Determine the actual percentage of eruptions that last between 92 and 116 seconds, inclusive.

We solve for this using z score formula

The formula for calculating a z-score is is z = (x-μ)/σ

where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean = 104, Standard deviation = 6

For x = 92

z = 92 - 104/6

= -2

Probability value from Z-Table:

P(x = 92) = P(z = -2) = 0.02275

For x = 116

z = 92 - 116/6

= 2

Probability value from Z-Table:

P(x = 116) = P(z = 2) = 0.97725

The actual percentage of eruptions that last between 92 and 116 seconds

= P(x = 116) - P(x = 92)

= 0.97725 - 0.02275

= 0.9545

Converting to percentage = 0.9545 × 100

= 95.45%

Therefore, the actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) Use the Empirical Rule to determine the percentage of eruptions that last less than 98 seconds

The empirical rule formula:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

For 68% μ - σ = 104 - 6 = 98,

Therefore, 68% of eruptions that last for 98 seconds.

For less than 98 seconds which is the Left hand side of the distribution, it is calculated as

= 100 - 68/2

= 32/2

= 16%

Therefore, the percentage of eruptions that last less than 98 seconds is 16%

(f) Determine the actual percentage of eruptions that last less than 98 seconds.

The formula for calculating a z-score is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

For x = 98

Z score = x - μ/σ

= 98 - 104/6

= -1

Probability value from Z-Table:

P(x ≤ 98) = P(x < 98) = 0.15866

Converting to percentage =

0.15866 × 100

= 15.866%

Therefore, the actual percentage of eruptions that last less than 98 seconds is 15.866%

4 0

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What is the area of this figure?<br> Enter your answer in the box.

Aleks [24]

Step-by-step explanation:

55

the coordinates will add each other

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2 years ago

Please help ............<br>plzzzz

N76 [4]

Answer:

1774cm

Explanation:

11x14=154

20x9x9=1620

154+1620=1774

6 0

3 years ago

Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

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2 years ago

What is absolute deviation?

zlopas [31]

The average absolute deviation (or mean absolute deviation ( MAD )) about any certain point (or 'avg. absolute deviation' only) of a data set is the average of the absolute deviations or the positive difference of the given data and that certain value (generally central values). answer:

Step-by-step explanation:

7 0

3 years ago