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Tema [17]
2 years ago
5

According to the National Health Statistics Reports, the standard deviation of the weights of all one-year-old baby boys born in

the U.S. is 5.3 pounds. A random sample of 360 one-year-old baby boys born in the U.S. had a mean weight of 25.5 pounds.
a) Construct a 90% confidence interval for the mean weight of all one-year-old baby boys in the U.S. Write a sentence that interprets this interval.
b) Should this confidence interval be used to estimate the mean weights of all one-year-old babies in the U.S.? Explain.
Mathematics
1 answer:
lina2011 [118]2 years ago
3 0

Answer: A) At 90% confidence interval estimate of the population mean

is,( 25.0405 , 25.9595 )

B) YES

Step-by-step explanation:

Given that,

Point estimate = sample mean  Ж = 25.5

Population standard deviation α  = 5.3

Sample size = n =360

At 90% confidence level the z is ,

∝ = 1 - 90% = 1 - 0.90 = 0.1

∝ / 2 = 0.1 / 2 = 0.05  

Z∝/2 = Z0.05 = 1.645 ( WHEN WE USE THE Z TABLE )

Margin of error E = Z∝/2 * ( α/√n)

E = 1.645 * (5.3 / √360 )  = 0.4595

At 90% confidence interval estimate of the population mean

is

Ж - E < ц < Ж + E  

25.5 - 0.4595 <  ц < 25.5 + 0.4595  

25.0405 <  ц < 25.9595  

( 25.0405 , 25.9595 )

At 90% confidence interval estimate of the population mean

is,( 25.0405 , 25.9595 )

B) This confidence interval can be used to estimate the mean weight of all one - year old babies in the US since the mean value of 25.5 falls within the confidence values, we have sufficient evidence to

conclude that the mean weight of all one-year-old boys is 25.5

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A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

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Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

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The inner products are the same.

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