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Hoochie [10]
4 years ago
7

I need help doing thsi

Mathematics
1 answer:
Naily [24]4 years ago
7 0
True! Because T is the defining letter of the line, so it would be true!

Hope this helps!! :)
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The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0
valina [46]

Answer:

a) P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)

And using the complement rule, a calculator, excel or the normal standard table we have that:

P(Z>0.707)=1-P(Z

b) z=-0.674

And if we solve for a we got

a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

c) P( \bar X >1.305) = 0.05  

We can use the z score formula:

P( \bar X >1.305) = 1-P(\bar X

Then we have this:

P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

Let X the random variable that represent the amount of warpage of a population and we know

Where \mu=1.3 and \sigma=0.1

Since the sample size is large enough we can use the central limit theorem andwe know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

We can find the probability required like this:

P(\bar X >1.305)=P(Z>\frac{1.305-1.3}{\frac{0.1}{\sqrt{200}}}=0.707)

And using the complement rule, a calculator, excel or the normal standard table we have that:

P(Z>0.707)=1-P(Z

Part b

For this part we want to find a value a, such that we satisfy this condition:

P(\bar X>a)=0.75   (a)

P(\bar X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.674

And if we solve for a we got

a=1.3 -0.674* \frac{0.1}{\sqrt{200}}=

So the value of height that separates the bottom 95% of data from the top 5% is 1.295.

Part c

For this case we want this condition:

P( \bar X >1.305) = 0.05  

We can use the z score formula:

P( \bar X >1.305) = 1-P(\bar X

Then we have this:

P(z< \frac{1.305-1.3}{\frac{0.1}{\sqrt{n}}}) = 0.95

And a value that accumulates 0.95 of the area on the normal distribution z = 1.64 and we can solve for n like this:

n = (1.64*\frac{0.1}{1.305-1.3})^2= 1075.84 \approx 1076

6 0
3 years ago
Point Y (2, 1) is the midpoint of line segment XZ. Point X is located at (1, 4). Find
kolezko [41]

Answer: (3, -2) i believe

Step-by-step explanation:

3 0
3 years ago
Number 4 please I need help please I need help please help me
kotykmax [81]

Answer:

The perimeter is 112 m

Step-by-step explanation:

The perimeter is the sum of all the edges

26+ 12 + 18 +? + ?+ 30

The two question marks add to 26 so replace them with 26

26+ 12 + 18 +26+ 30

112

The perimeter is 112 m

3 0
4 years ago
The Council of American Beet Farmers estimate that 60% of Americans love beets.
harkovskaia [24]

Answer: 13% or  0.1314

Step-by-step explanation: Binomial PDF on the calculator or by hand its

22 nCr 15 (0.60)^15(1-0.6)^7 = 0.131378

5 0
3 years ago
the music hall has 20 boxes on thurs 2/5 were occupied, on friday 2/4 were occupied and o saturday 8/10 were occupied. Using sym
NikAS [45]
<span><span>1.       </span>The music hall has 20 boxes :
=> on thurs 2/5 were occupied
=> on friday 2/4 were occupied
=> and on saturday 8/10 were occupied.
Let’s find out which day has the most occupied space of the music hall.
2/5 + 2/4 + 8/10 = 6/20 + 7/20 + 10/20
Thurs 2/5 were occupied
=> 6/20 = 0.3
=> 20 * 0.3 = 6 boxes
Friday 2/4 were occupied
=> 7/20 = 0.35
=> 20 * 0.35 = 7 boxes
Saturday 8/10 were occupied.
=> 10/20 = 0.5
=> 20 * .5 = 10 boxes
Thus, during Saturday has the most number of occupied boxes.</span>



8 0
3 years ago
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