Ask question

Ask question

All categories

2 years ago

14

The student body of a large university consists of 60% female students. A random sample of 8 students is selected. Refer to Exhi

bit 5-8. What is the probability that among the students in the sample at least 6 are male?

1 answer:

jasenka [17]2 years ago

8 0

Answer:

0.0499

Step-by-step explanation:

This is a binomial probability function expressed as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\$

Given that n =8, and p(male)=1-0.6=0.4, the probability of at least 6 being male is calculated as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 6)=P(X=6)+P(X=7)+P(X=8)\\\\={8\choose 6}0.4^6(0.6)^{2}+{8\choose 7}0.4^7(0.6)^{1}+{8\choose 8}0.4^8(0.6)^{0}\\\\=0.0413+0.0079+0.0007\\\\=0.0499$

Hence, the probability of at least 6 males is 0.0499

You might be interested in

The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,

uranmaximum [27]

Answer:

Step-by-step explanation:

Assuming that the differential equation is

$\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right)$ .

We need to solve it and obtain an expression for $P(t)$ in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

$\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right)$ .

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that $k=0.04$ and $K=500$ and the initial condition $P(0)=100$ .

Notice that this equation is separable, then

$\frac{dP}{P(1-P/K)} = kdt$ .

Now, intagrating in both sides of the equation

$\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C$ .

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

$\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}$ .

So,

$\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|$ .

We have obtained that:

$-\ln\left| \frac{K-P}{P}\right| = kt +C$

which is equivalent to

$\ln\left| \frac{K-P}{P}\right|= -kt -C$

Taking exponentials in both hands:

$\left| \frac{K-P}{P}\right| = e^{-kt -C}$

Hence,

$\frac{K-P(t)}{P(t)} = Ae^{-kt}$ .

The next step is to substitute the given values in the statement of the problem:

$\frac{500-P(t)}{P(t)} = Ae^{-0.04t}$ .

We calculate the value of $A$ using the initial condition $P(0)=100$ , substituting $t=0$ :

$\frac{500-100}{100} = A}$ and $A=4$ .

So,

$\frac{500-P(t)}{P(t)} = 4e^{-0.04t}$ .

Finally, as we want the value of $t$ such that $P(t)=200$ , we substitute this last value into the above equation. Thus,

$\frac{500-200}{200} = 4e^{-0.04t}$ .

This is equivalent to $\frac{3}{8} = e^{-0.04t}$ . Taking logarithms we get $\ln\frac{3}{8} = -0.04t$ . Then,

$t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325$ .

So, the population of rats will be 200 after 25 months.

6 0

3 years ago

the ratio of boys to girls at the beach cleanup was 7:8. If there were 43 boys, how many girls were there?

faltersainse [42]

I'm pretty sure there were 48 girls at the beach clean up. 43 ÷ 7 = 6.14 (6 cause there's not gonna be floating males with missing body parts). So we multiply 6 x 8 and we get 48.

7 0

3 years ago

One hour after yolanda started walking from x to y, a distance of 45 miles, bob started walking along the same road from y to x.

Rufina [12.5K]

<span>Answer: When bob starts walking The distance between them is 45-3 = 42 miles If they meet at time t then Yolanda would have walked X/3 minutes and Bob would have walked (42-X)/4 mins X/3 = (42-X)/4 so X = 18 Bob would have walked 42-18 = 24 miles.</span>

3 0

3 years ago

1/2(8-x)=10<br> A.12<br> B.3<br> C.-12

Murljashka [212]

4-1/2x=10
-1/2x=6
3
so your answer would be B
hope that helps

7 0

3 years ago

Volanta purchase 200 beads for $48 to make necklaces if she needs to buy 25 more beads how much will she pay if she charge the s

Bogdan [553]

It's 48/200=0.24$ per a bead
She will have to pay 0.24*25=6$

4 0

3 years ago

Read 2 more answers