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Ket [755]
3 years ago
13

The partial pressure of oxygen PaO2 is a measure ofthe amount of oxygen in the blood. Assume that the distribution ofPaO2 levels

among newborns has a mean of 38 mmHg and astandard deviation of 9 mmHg. If we take a random sample of 25newborns what is the probability that the sample meana) will be greater than 36b) will be between 36 and 41
Mathematics
1 answer:
Rus_ich [418]3 years ago
4 0

Answer: a) 0.8665

b) 0.8190

Step-by-step explanation:

Given : The partial pressure of oxygen PaO2 is a measure ofthe amount of oxygen in the blood. Assume that the distribution ofPaO2 levels among newborns has a \mu=38 mmHg and \sigma= 9 mmHg.

If we take a random sample n= 25 newborns, then  using formula z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}, we have

At x= 36

z=\dfrac{36-38}{\dfrac{9}{\sqrt{25}}}\approx-1.11

At x= 41

z=\dfrac{36-38}{\dfrac{9}{\sqrt{25}}}\approx1.67

Using table for z-values, the probability that the sample mean will be greater than 36 :

P(z>-1.11)=1-P(\leq-1.11)=1-(1-P(z\leq1.11))=P(z\leq1.11)=0.8665004\approx0.8665

The probability that the sample mean will be between 36 and 41 :-

P(-1.11

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In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t 2 , 2 + t − 5t 2 , 1 + 2t} to the standard basis of P2.
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Complete Question:

In P2, find the change-of-coordinates matrix from the basis B = {1 − 3t² , 2+t− 5t² , 1 + 2t} to the standard basis C = {1, t, t²}. Then, write t² as a linear combination of the polynomials in B.

Answer:

The change of coordinate matrix is :

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

Step-by-step explanation:

Let U =  {D, E, F} be any vector with respect to Basis B

U = D [1 − 3t²] + E [2+t− 5t²] + F[1 + 2t]..............(*)

U = [D+2E+F]+ t[E+2F] + t²[-3D-5E]...................(**)

In Matrix form;

\left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right] \left[\begin{array}{ccc}D\\E\\F\end{array}\right] = \left[\begin{array}{ccc}D+2E+F\\E+2F\\-3D-5E\end{array}\right]

The change of coordinate matrix is therefore,

M = \left[\begin{array}{ccc}1&2&1\\0&1&2\\-3&-5&0\end{array}\right]

To find D, E, F in (**) such that U = t²

D + 2E + F = 0.................(1)

E + 2F = 0.........................(2)

-3D -5E = 1........................(3)

Substituting eqn (2) into eqn (1 )

D=3F...................................(4)

Substituting equations (2) and (4) into eqn (3)

-9F+10F=1

F = 1

Put the value of F into equations (2) and (4)

E = -2(1) = -2

D = 3(1) = 3

Substituting the values of D, E, and F into (*)

U = t² = 3 [1 − 3t²] - 2 [2+t− 5t²] + [1 + 2t]

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Answer:

option A)  y > 5/4x + 10

Step-by-step explanation:

Take any point in the shaded region

( -12 , 1)

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