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3 years ago

Pls explain this, I don't understand

Mathematics

1 answer:

Rufina [12.5K]3 years ago

4 0

Answer:

AB = $\left[\begin{array}{ccc}0&-2\\4&-12\\\end{array}\right]$

Step-by-step explanation:

To obtain the first row of AB

Sum the products of corresponding elements in row 1 of A with elements in column 1 and 2 of B

To obtain the second row of AB

Sum the corresponding elements in row 2 of A with elements in column 1 and 2 of B, that is

AB = $\left[\begin{array}{ccc}0(-1)+2(0)&0(3)+2(-1)\\-4(-1)+0(0)&-4(3)+0(-1)\\\end{array}\right]$

= $\left[\begin{array}{ccc}0+0&0-2\\4+0&-12+0\\\end{array}\right]$

= $\left[\begin{array}{ccc}0&-2\\4&-12\\\end{array}\right]$

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Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Ro

puteri [66]

Answer:

(a) P(0 ≤ Z ≤ 2.87)=0.498

(b) P(0 ≤ Z ≤ 2)=0.477

(d) P(−2.20 ≤ Z ≤ 2.20)=0.972

(e) P(Z ≤ 1.01)=0.844

(f) P(−1.95 ≤ Z)=0.974

(g) P(−1.20 ≤ Z ≤ 2.00)=0.862

(h) P(1.01 ≤ Z ≤ 2.50)=0.150

(i) P(1.20 ≤ Z)=0.115

(j) P(|Z| ≤ 2.50)=0.988

Step-by-step explanation:

(a) P(0 ≤ Z ≤ 2.87)

In this case, this is equal to the difference between P(z<2.87) and P(z<0). The last term is substracting because is the area under the curve that is included in P(z<2.87) but does not correspond because the other condition is that z>0.

$P(0 \leq z \leq 2.87)= P(z$

(b) P(0 ≤ Z ≤ 2)

This is the same case as point a.

$P(0 \leq z \leq 2)= P(z$

This is the same case as point a.

$P(-2.2 \leq z \leq 0)= P(z$

(d) P(−2.20 ≤ Z ≤ 2.20)

This is the same case as point a.

$P(-2.2 \leq z \leq 2.2)= P(z$

(e) P(Z ≤ 1.01)

This can be calculated simply as the area under the curve for z from -infinity to z=1.01.

$P(z\leq1.01)=0.844$

(f) P(−1.95 ≤ Z)

This is best expressed as P(z≥-1.95), and is calculated as the area under the curve that goes from z=-1.95 to infininity.

It also can be calculated, thanks to the symmetry in z=0 of the standard normal distribution, as P(z≥-1.95)=P(z≤1.95).

$P(z\geq -1.95)=0.974$

(g) P(−1.20 ≤ Z ≤ 2.00)

This is the same case as point a.

$P(-1.20 \leq z \leq 2.00)= P(z$

(h) P(1.01 ≤ Z ≤ 2.50)

This is the same case as point a.

$P(1.01 \leq z \leq 2.50)= P(z$

(i) P(1.20 ≤ Z)

This is the same case as point f.

$P(z\geq 1.20)=0.115$

(j) P(|Z| ≤ 2.50)

In this case, the z is expressed in absolute value. If z is positive, it has to be under 2.5. If z is negative, it means it has to be over -2.5. So this probability is translated to P|Z| < 2.50)=P(-2.5<z<2.5) and then solved from there like in point a.

$P(|z|$