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Bingel [31]
2 years ago
9

Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and ob

serving the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met
Mathematics
1 answer:
klio [65]2 years ago
6 0

Answer:

Thirty-two percent of fish in a large lake are bass. Imagine scooping out a simple random sample of 15 fish from the lake and observing the sample proportion of bass. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.

A.  The standard deviation is 0.8795. The 10% condition is met because it is very likely there are more than 150 bass in the lake.

B. The standard deviation is 0.8795. The 10% condition is not met because there are less than 150 bass in the lake.

C. The standard deviation is 0.1204. The 10% condition is met because it is very likely there are more than 150 bass in the lake.

D. The standard deviation is 0.1204. The 10% condition is not met because there are less than 150 bass in the lake.

E. We are unable to determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 150 bass in the lake

The answer is E.

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Answer:

508.8 seconds

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A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of the sheet is decreasing at
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Answer:

dx/dt  =  0,04 m/sec

Step-by-step explanation:

Area of the circle is:

A(c) =π*x²      where   x is a radius of the circle

Applying differentiation in relation to time we get:

dA(c)/dt   =  π*2*x* dx/dt    

In this equation we know:

dA(c)/dt  = 0,5 m²/sec

And are looking for dx/dt then

0,5  = 2*π*x*dx/dt    when the area of the sheet is 12 m²  (1)

When  A(c) = 12 m²      x = ??

A(c)  =  12  =  π*x²      ⇒    12  =  3.14* x²    ⇒  12/3.14  =  x²

x²  = 3,82     ⇒   x  = √3,82    ⇒  x = 1,954 m

Finally plugging ths value in equation (1)

0,5  = 6,28*1,954*dx/dt

dx/dt  =  0,5 /12.28

dx/dt  =  0,04 m/sec

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Given f_{XY} (x,y) = c(4x + 2y +1) ; 0 < x < 40\,and\, 0 < y

a)

we know that \int\limits^\infty_{-\infty}\int\limits^\infty_{-\infty} {f(x,y)} \, dxdy=1

therefore \int\limits^{40}_{-0}\int\limits^2_{0} {c(4x+2y+1)} \, dxdy=1

on integrating we get

c=(1/6640)

b)

P(X>20, Y>=1)=\int\limits^{40}_{20}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

on doing the integration we get

                        =0.37349

c)

marginal density of X is

f(x)=\int\limits^2_{0} {\frca{1}{6640}(4x+2y+1)} \, dy

on doing integration we get

f(x)=(4x+3)/3320 ; 0<x<40

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f(y)=\int\limits^{40}_{0} {\frca{1}{6640}(4x+2y+1)} \, dx

on doing integration we get

f(y)=\frac{(y+40.5)}{83}

d)

P(01)=\int\limits^{40}_{0}\int\limits^2_{1} {\frca{1}{6640}(4x+2y+1)} \, dxdy

solve the above integration we get the answer

e)

P(X>20, 0

solve the above integration we get the answer

f)

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