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erastovalidia [21]
4 years ago
7

Light bulbs of a certain type are advertised as having an averagelifetime of 750 hours. The price of these bulbs is very favorab

le,so a potential customer has decided to go ahead with a purchasearrangement unless it can be conclusively demonstrated thatthe true average lifetime is smaller than what is advertised. Arandom sample of 50 bulbs was selected, the lifetime of each bulbdetermined, and the appropriate hypothesis were tested usingMINITAB, resulting in the accompanying output.Variable N Mean St Dev SEMean ZP -Valuelifetime 50 738.44 38.20 5.40-2.14 0.016 What conclusion would be appropriate for a significance level of.05 ? .A significance level of .01 ?. What significance level wouldyou recommend ?
Mathematics
1 answer:
PilotLPTM [1.2K]4 years ago
7 0

answer 40

Step-by-step explanation:

because you added all together to make one

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What does it mean if the distance between a point p and a line l is zero? What does it mean if the distance between two lines is
-Dominant- [34]

Answer:

a) the point is on the line

b) the lines are the same line

Step-by-step explanation:

a) The distance from a point to a line is zero when the point satisfies the equation of the line. That is, the point is on the line.

b) Two lines are zero distance apart when they are on top of each other, indistinguishable. That is, they are the same line.

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3 years ago
Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I
Hatshy [7]

a. f_{X,Y} is a joint density function if its integral over the given support is 1:

\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy

=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1

so the answer is yes.

b. We should first find the density of the marginal distribution, f_Y(y):

f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy

f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}

Then

P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}

or about 0.2019.

For the other probability, we can use the joint PDF directly:

P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}

which is about 0.7326.

c. We already know the PDF for Y, so we just integrate:

E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5

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6/2 simplifyed is 3/1 or just 3

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