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maks197457 [2]
4 years ago
6

Find the area under the standard normal distribution curve for the following intervals. a. Between z = 0 and z = 2.0 b. To the r

ight of z = 1.5 c. To the left of z = −1.75 d. Between z = −2.78 and z = 1.66
Mathematics
1 answer:
vesna_86 [32]4 years ago
5 0

Answer:

a) P(0

And we can use the following excel code to find the probability:

"=NORM.DIST(2,0,1,TRUE)-NORM.DIST(0,0,1,TRUE)"

P(0

b) P(Z>1.5) =1-P(Z

And we can use the following code and we got:

"=1-NORM.DIST(1.5,0,1,TRUE)"

P(Z>1.5) =1-P(Z

c) P(z

And we can use the following code and we got:

"=NORM.DIST(-1.75,0,1,TRUE)"

P(z

d) P(-2.78

And we can use the following excel code to find the probability:

"=NORM.DIST(1.66,0,1,TRUE)-NORM.DIST(-2.78,0,1,TRUE)"

P(-2.78

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

Part a

We want this probability:

P(0

And we can use the following excel code to find the probability:

"=NORM.DIST(2,0,1,TRUE)-NORM.DIST(0,0,1,TRUE)"

P(0

Part b

For this case we want this probability:

P(Z>1.5)

And we can use the complement rule and we have:

P(Z>1.5) =1-P(Z

And we can use the following code and we got:

"=1-NORM.DIST(1.5,0,1,TRUE)"

P(Z>1.5) =1-P(Z

Part c

We want this probability:

P(z

And we can use the following code and we got:

"=NORM.DIST(-1.75,0,1,TRUE)"

P(z

Part d

We want this probability:

P(-2.78

And we can use the following excel code to find the probability:

"=NORM.DIST(1.66,0,1,TRUE)-NORM.DIST(-2.78,0,1,TRUE)"

P(-2.78

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Temka [501]
Integration by parts will help here. Letting u=\arctan x and \mathrm dv=\mathrm dx, you end up with \mathrm du=\dfrac{\mathrm dx}{1+x^2} and v=x. Now

\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du
\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx

For the remaining integral, setting y=1+x^2 gives \dfrac{\mathrm dy}2=x\,\mathrm dx, so

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Putting everything together, you end up with

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6 0
3 years ago
If the value of third order determinant is 11. Then what is the value of the determinant formed by its cofactors.
nordsb [41]

Answer:

146.41

Step-by-step explanation:

third order determinant = determinant of 3×3 matrix A

given ∣A∣=11

det (cofactor matrix of A) =set (transpare of cofactor amtrix of A) (transpare does not change the det)

=det(adjacent of A)

{det (cofactor matrix of A)}  ^2  = {det (adjacent of A)} ^2

 

(Using for an n×n det (cofactor matrix of A)=det (A)^n−1 )

we get

det (cofactor matrix of A)^2  = {det(A)  ^3−1 }^2

 

=(11)^2×2  = 11^4

 =146.41

5 0
3 years ago
Analyze this student's work. What did the student do correctly? What did he do incorrectly?
saw5 [17]
I don’t think he did distributive property
8 0
3 years ago
Given the graph below, which of the following statements is true?
nirvana33 [79]

Answer:

D) The graph does not represent a one-to-one function because the y-values between 0 to 2 are paired with multiple x-values.

Step-by-step explanation:

Let's find the ordered pairs.

(-4, 4), (-3, 2), (-2, 0), (-1, 0.5), (0, 1), (1, 1.5), (2, 2), (3, 0)

The graph passed through above points.

In the x-coordinates -3, 2 gives the same output. Therefore, the given function is not one-to-one.

Therefore, the answer D)

The graph does not represent a one-to-one function because the y-values between 0 to 2 are paired with multiple x-values.

Hope you will understand the concept.

Thank you.

7 0
3 years ago
Figure abcd is a parallelogram what are the measures of b and c
Over [174]

what is the question

5 0
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