Find the volume of this

(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:

a)

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.

So

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

b)

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

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

c)

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

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

Let's approximate the distribution with a normal distribution with mean 170 and standard deviation 5.05

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

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

3 years ago

How could you correctly rewrite the equation 6(11 – 3) = 4(7 455) using the distributive property?

Sergeeva-Olga [200]

Answer:

B. 66-18=28+20

Step-by-step explanation:

4 0

3 years ago

What are the values of x and y if this equation is true? 22(x+yi)+(28+4i)+72-62i

77julia77 [94]

Answer: x=2 and y=-3

Step-by-step explanation: you’re welcome

4 0

3 years ago

Monthly sales of an SUV model are expected to increase at the rate of S′1t2 = -24t 1>3 SUVs per month, where t is time in mon

uysha [10]

The company will continue to manufacture this model for 19 months.

The question is ill-formatted. The understandable format is given below.

An SUV model's monthly sales are anticipated to rise at a rate of

$S'(t)=-24^{1/3}$ SUVs each month, where "t" denotes the number of months and "S(t)" denotes the monthly sales of SUVs. When monthly sales of 300 SUVs are reached, the business intends to discontinue producing this model.

Find S if monthly sales of SUVs are 1,200 at time t=0 (t). How long will the business keep making this model?

Given $S'(t)=-24^{1/3}$ ... ... (1)

Condition (1) at t=0; s(t) =1200

We find S(t)=300 then t=?

a) We find S(t)

from $S'(t)=-24^{1/3}$

$\implies \frac{dS(t)}{dt}=-24t^{1/3} ~~[\because X'(t)=\frac{dX}{dt} \\\implies dS(t) = -24t^{1/3}dt\\$

On Integrating both sides

$S(t)=-18t^{4/3}+c ~...~...~(2)$

Now, at t=0 then S(t)=1200

So, from (2)

1200=0+c

⇒c=1200

$\therefore$ from eq (2)

$S(t)=-18t^{4/3}+1200 ~...~...~(3)$

b) Considering that the firm intends to cease production of this model once monthly sales exceed 300 SUVs.

So, take S(t)=300

from eq (2) $300=-18t^{4/3}+1200$

$t^{4/3}=\frac{1200-300}{18}\\=\frac{900}{18}=50\\\implies t=50^{3/4}\\\implies t=18.8030$

Hence the company will continue 18.8020 months.

Learn more about integration here-

brainly.com/question/18125359

#SPJ10

8 0

1 year ago

12. (2v^3 - v + 8) + (-v^3 + v - 3) 14. (4h^3 + 3h + 1) - (-5h^3 + 6h - 2)

Lana71 [14]

12. On addition, you can just combine like terms.
2v^3+(-v^3)=v^3
-v+v cancels each other out
8+(-3)=5
So you have v^3+5

14 On subtraction, you have to remember to distribute the negative sign so after you do that you have:
4h^3+3h+1+5h^3-6h+2
Then you can combine like terms
4h^3+5h^3=9h^3
3h-6h=-3h
1+2=3
So you end up with:
9h^3-3h+3

Hope that helps and feel free to ask any questions.

7 0

3 years ago