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You and some friends go miniature golfing. Your score is 3 below par. One of your friends has a score that is 12 over par. What

Leona [35]

12-(-3)= 12+3=15, so the difference is 15 points

5 0

3 years ago

Read 2 more answers

I need help with this question

Murljashka [212]

It is the third on down

4 0

3 years ago

PLS HELP 6th grade, I WILL MARK U AS BRAINLIEST PLSSS

Sergio [31]

Answer:

1.75

Other Answer:

1 3/4

7 0

3 years ago

The article "Monte Carlo Simulation—Tool for Better Understanding of LRFD" (J. Structural Engr., 1993: 1586–1599) suggests that

Sedaia [141]

Answer:

a) $P(X$

And we can find this probability using the normal standard distribution or excel and we got:

$P(z$

b) $P(X>40)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-43}{4.5})=P(z>3.78)$

And we can find this probability using the complement rule and normal standard distribution or excel and we got:

$P(z>3.78)=1-P(Z$

c) $z=0.674$

And if we solve for a we got

$a=43 +0.674*4.5=46.033$

So the value of height that separates the bottom 75% of data from the top 25% is 46.033.

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".

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(43,4.5)$

Where $\mu=43$ and $\sigma=4.5$

Part a

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard distribution or excel and we got:

$P(z$

Part b

$P(X>40)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-43}{4.5})=P(z>3.78)$

And we can find this probability using the complement rule and normal standard distribution or excel and we got:

$P(z>3.78)=1-P(Z$

Part c

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

$P(X>a)=0.25$ (a)

$P(X$ (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 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

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

$a=43 +0.674*4.5=46.033$

So the value of height that separates the bottom 75% of data from the top 25% is 46.033.

8 0

3 years ago

An article gave the accompanying data on ultimate load (kN) for two different types of beams. Assuming the underlying distributi

krok68 [10]

Answer:

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).

Step-by-step explanation:

We have to calculate a 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams.

The sample 1 (Fiberglass), of size n1=26 has a mean of 33.4 and a standard deviation of 2.2.

The sample 2 (Carbon), of size n2=26 has a mean of 42.8 and a standard deviation of 4.3.

The difference between sample means is Md=-9.4.

$M_d=M_1-M_2=33.4-42.8=-9.4$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{2.2^2}{26}+\dfrac{4.3^2}{26}}\\\\\\s_{M_d}=\sqrt{0.186+0.711}=\sqrt{0.897}=0.9473$

The critical t-value for a 99% confidence interval is t=2.678.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.678 \cdot 0.9473=2.537$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = -9.4-2.537=-11.937\\\\UL=M_d+t \cdot s_{M_d} = -9.4+2.537=-6.863$

The 99% confidence interval for the difference between the true average load for the fiberglass beams and that for the carbon beams is (-11.937, -6.863).

8 0

3 years ago