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3 years ago

If 8% of all cell phones are defective, and you have a random sample group of 100 Samsung Galaxy Vills,

Mathematics

1 answer:

Sati [7]3 years ago

5 0

Answer:

Mean = 8

Variance = 7.36

Standard Deviation = 2.7129

Step-by-step explanation:

This is a binomial distribution with parameters, n and p.

Where

n is sample size (given as 100)

p is the probability of success, or probability of defective (given as 8% or 0.08)

The mean, variance, standard deviation formula for binomial distribution is shown below:

Mean = $np$

Variance = $npq$

Standard Deviation = $\sqrt{npq}$

Where q would be probability of failure, or "1 - p"

Thus,

n = 100

p = 0.08

q = 1 - 0.08 = 0.92

SO, we have:

Mean = $np=100*0.08=8$

Variance = $npq=100*0.08*0.92 = 7.36$

Standard Deviation = $\sqrt{npq}=\sqrt{7.36}=2.7129$

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3 years ago

If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

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where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

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Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

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$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

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3 years ago

A camera costs 115 Canadian Dollars. If each Canadian dollar is worth 0.952 U.S Dollars, how much would the camera cost in U.S D

PIT_PIT [208]

This is truly a simple, one-step equation if looked at in the right angle.

You're going to want to multiply 115 by .952:

115
<u> x .952
</u>    230
+   6050
<u>+ 93500
</u>  109.48

$109.48 is your final answer.

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