Given f(x, y, z) = x sin(yz), the formula for calculating the gradient of the function is expressed as ∇f(x, y, z) = fx(x, y, z)i+ fy(x, y, z)j+fz(x, y, z)k where;

fx, fy and fz are the differential of the functions with respect to x, y and z respectively.

a) ∇f(x, y, z) = sin(yz)i + xzcos(yz)j + xycos(yz)k

The gradient of f = sin(yz)i + xzcos(yz)j + xycos(yz)k

b) Directional derivative of f at (1,2,0) in the direction of v = i + 4j − k is expressed as ∇f(1, 2, 0)*v

∇f(1, 2, 0) = sin(2(0))i +1*0cos(2*0)j + 1*2cos(2*0)k

∇f(1, 2, 0) = sin0i +0cos(0)j + 2cos(0)k

∇f(1, 2, 0) = 0i +0j + 2k

Given v = i + 4j − k

∇f(1, 2, 0)*v (note that this is the dot product of the two vectors)

∇f(1, 2, 0)*v = (0i +0j + 2k)*(i + 4j − k )

Given i.i = j.j = k.k =1 and i.j=j.i=j.k=k.j=i.k = 0

∇f(1, 2, 0)*v = 0(i.i)+4*0(j.j)+2(-1)k.k

∇f(1, 2, 0)*v = 0(1)+0(1)-2(1)

∇f(1, 2, 0)*v =0+0-2

∇f(1, 2, 0)*v= -2

Hence, the directional derivative of f at (1, 2, 0) in the direction of v = i + 4j − k is -2

7 0

2 years ago

(43 points) In the US, 85% of the population has Rh positive blood. Suppose we take a random sample of 6 persons and let Y denot

VladimirAG [237]

Answer:

a) Binomial distribution with parameters p=0.85 q=0.15 n=6

b) 62.29%

c) 2.38%

d) See explanation below

Step-by-step explanation:

We could model this situation with a binomial distribution

$P(6;k)=\binom{6}{k}p^kq^{6-k}$

where P(6;k) is the probability of finding exactly k people out of 6 with Rh positive, p is the probability of finding one person with Rh positive and q=(1-p) the probability of finding a person with no Rh.

$\bf P(Y=k)=\binom{6}{k}(0.85)^k(0.15)^{6-k}$

The probability that Y is less than 6 is

P(Y=0)+P(Y=1)+...+P(Y=5)

Let's compute each of these terms

$P(Y=0)=P(6;0)=\binom{6}{0}(0.85)^0(0.15)^{6}=1.139*10^{-5}$

$P(Y=1)=P(6;1)=\binom{6}{1}(0.85)^1(0.15)^{5}=0.0000387281$

$P(Y=2)=P(6;2)=\binom{6}{2}(0.85)^2(0.15)^{4}=0.005486484$

$P(Y=3)=P(6;3)=\binom{6}{3}(0.85)^3(0.15)^{3}=0.041453438$

$P(Y=4)=P(6;4)=\binom{6}{4}(0.85)^4(0.15)^{2}=0.176177109$

$P(Y=5)=P(6;5)=\binom{6}{5}(0.85)^5(0.15)^{1}=0.399334781$

and adding up these values we have that the probability that Y is less than 6 is

$\sum_{i=1}^{5}P(Y=i)=0.622850484\approx 0.6229=62.29\%$

In this case is a binomial distribution with n=200 instead of 6.

p and q remain the same.

The mean of this sample would be 85% of 200 = 170.

In a binomial distribution, the standard deviation is

$s = \sqrt{npq}$

In this case

$\sqrt{200(0.85)(0.15)}=5.05$

<em>Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05</em>

So, the approximate probability that there are fewer than 160 persons with Rh positive blood in a sample of 200 would be the area under the normal curve to the left of 160

(see picture attached)

We can compute that area with a computer and find it is

0.0238 or 2.38%

d)<em> In order to approximate a binomial distribution with a normal distribution we need a large sample like the one taken in c).</em>

In general, we can do this if the sample of size n the following inequalities hold:

$np\geq 5 \;and\;nq \geq 5$

in our case np = 200*0.85 = 170 and nq = 200*0.15 = 30

4 0

2 years ago