Find the geometric mean between 4 and 20

The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) F

cricket20 [7]

Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$n = 300, p = 0.02$

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

$E(X) = np = 300*0.02 = 6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42$

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040$

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023$

$P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143$

$P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436$

$P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485$

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

From c), we have that $P(X = 0) = 0.0023$ . So

$0.0023 + P(X \geq 1) = 1$

$P(X \geq 1) = 0.9977$

99.77% probability that we find at least 1 of the seniors are married

3 0

3 years ago

In this picture B and Fare

Igoryamba

9514 1404 393

Answer:

x = 22

Step-by-step explanation:

The midsegment BF is half length of the base segment CE, so you have ...

2x = 88/2

x = 22 . . . . . divide by 2

5 0

3 years ago

MRS.SMITH PAID#125 TO HAVE HER HAIR COLORED AND CUT. if SHE TIPS HER HAIRDRESSER 15% OF HER BILL, HOW MUCH WAS THE TIP?

Leto [7]

15% of 125 = 18.75 the tip
125 + 18.75 = 143.75 full price
Hope it helps!

5 0

4 years ago

Describe in detail how you would construct a 95% confidence interval for a set of 30 data points whose mean is 20 and population

postnew [5]

Answer:

We have been given confidence interval 95%, mean 20 , data set 30 and standard deviation 3.

We will use the formula: $mean\pm \frac{\sigma}{\sqrt{n}}\cdot (z-score)$

Here, $mean=20,\sigma=3,n=30,z-score=1.96$

Z-score value at 95% confidence interval is 1.96

On substituting the values in the formula to plug the values:

$20\pm\frac{3}{\sqrt{30}}\cdot (1.96)$

Now, we have a formula for marginal error: $z\cdot \frac{\sigma}{\sqrt{n}}$

Marginal error means your answer will be within that percentage only.

Say you have 3% marginal means your value will be within 3% real population 95% of the time.

7 0

4 years ago

Write a short paragraph discussing the reasons why someone should make a will.

LuckyWell [14K]

When someone dies everyone wonders what will happen to all of their possessions. If the person had a will then everyone will know what they have to keep. If the person did not have a will then no one will know what to keep and what not to keep. That can cause problems among family and friends. To keep the peace and sentimental value in balance a will is very important to make.

3 0

3 years ago