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2 years ago

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Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per po

und, with a standard deviation of 7 grams of fat per pound. A random sample of 34 farm-raised trout is selected. The mean fat content for the sample is 31.1 grams per pound. Find the probability of observing a sample mean of 31.1 grams of fat per pound or less in a random sample of 34 farm-raised trout.

1 answer:

LenaWriter [7]2 years ago

7 0

Answer:

The probability of observing a sample mean of 31.1 grams of fat per pound or less in a random sample of 34 farm-raised trout is 0.227

Step-by-step explanation:

Let X be the sample mean of fat in 34 farm-raised trout

The probability of observing a sample mean of 31.1 grams of fat per pound or less in a random sample of 34 farm-raised trout can be stated as:

P(X<31.1 grams) = P(z<z*)

where z* is the z-statistic of sample mean of 31.1 grams fat.

z* can be calculated as follows:

z*= $\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

X = 31.1 grams

M is the average grams of fat per pound (32)

s is the standard deviation (7)

N is the sample size (34)

Then z*= $\frac{31.1-32}{\frac{7}{\sqrt{34} } }$ ≈ −0.7497

and P(z<z*) ≈ 0.227

The probability of observing a sample mean of 31.1 grams of fat per pound or less in a random sample of 34 farm-raised trout is 0.227

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Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

so

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

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and

T = v × $\rho$ - v × $\rho$ × g

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solve irt we get

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Answer:

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the confidence interval is appropriate

Step-by-step explanation:

The confidence interval :

Mean ± margin of error

Sample mean = 1.27

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Sample size, n = 914

Since we are using tbe sample standard deviation, we use the T table ;

Margin of Error = Tcritical * s/√n

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