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(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

7(8i+9) (Show work please)

ivanzaharov [21]

Using the distributive property, you’ll get the equation 56i+63.

You’ll distribute (multiply) 7 to 8i, and 7 to 9

3 0

2 years ago

mrs.beluga is driving on a snow covered road with a drag factor of 0.2. she brakes suddenly for a deer. The tires leave a yaw ma

Aneli [31]

First, we are going to find the radius of the yaw mark. To do that we are going to use the formula: $r= \frac{c^2}{8m} + \frac{m}{2}$
where
$c$ is the length of the chord
$m$ is the middle ordinate
We know from our problem that the tires leave a yaw mark with a 52 foot chord and a middle ornate of 6 feet, so $c=52$ and $m=6$ . Lets replace those values in our formula:
$r= \frac{52^2}{8(6)} + \frac{6}{2}$
$r= \frac{2704}{48} +3$
$r= \frac{169}{3} +3$
$r= \frac{178}{3}$

Next, to find the minimum speed, we are going to use the formula: $s= \sqrt{15fr}$
where
$f$ is drag factor
 $r$ is the radius
We know form our problem that the drag factor is 0.2, so $f=0.2$ . We also know from our previous calculation that the radius is $\frac{178}{3}$ , so $r= \frac{178}{3}$ . Lets replace those values in our formula:
$s= \sqrt{(15)(0.2)( \frac{178}{3}) }$
$s= \sqrt{178}$
$s=13.34$ mph

We can conclude that Mrs. Beluga's minimum speed before she applied the brakes was 13.34 miles per hour.

8 0

3 years ago

Find the product: (-9)(-4)

nadya68 [22]

Answer:

36

Step-by-step explanation:

-9*-4=36

3 0

3 years ago

What is 12 divided by 15

ololo11 [35]

It would be: 12/15 = 4/5

In short, Your Answer would be 4/5

Hope this helps!

3 0

3 years ago

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