Which expression reflects the probability of flipping heads and landing on 25 points

USA Today reported that about 20% of all people in the United States are illiterate. Suppose you take eight people at random off

Law Incorporation [45]

Answer:

a) Figure and code attached

b) $E(X)=\mu = np = 8*0.2= 1.6$

$Var(X) =\sigma^2= np(1-p) = 8*0.2*(1-0.2) = 1.28$

$Sd(X)=\sigma= \sqrt{1.28}= 1.131$

c) $P(X\geq 7) =0.97$

And we can calculate this with the complement rule.

$P(X \geq 7) = 1-P(X$

So then we have:

$P(X \leq 6) = 0.03$

And we are interested on the valueof n who satisfy this expression.

And for this we can verify this with the following code:

"=BINOM.DIST(6,E54,0.2,TRUE)"

And as we can see on the second figure attached the value who satisfy the condition would be n = 60.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=8, p=0.2)$

The probability mass function for the Binomial distribution is given as:

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

Where (nCx) means combinatory and it's given by this formula:

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

Part a

We can use the following R code to generate the histogram for this case:

> x <- seq(0,8,by = 1)

> y <- dbinom(x,8,0.2)

> plot(x,y,type = "h",main="Histogram")

And as we can see we got the result on the figure attached. And the distribution seems to be skewed to the right.

Part b

For this case the expected value is given by:

$E(X)=\mu = np = 8*0.2= 1.6$

The variance is given by:

$Var(X) =\sigma^2= np(1-p) = 8*0.2*(1-0.2) = 1.28$

And the standard deviation would be:

$Sd(X)=\sigma= \sqrt{1.28}= 1.131$

Part c

For this case we have the following inequality:

$P(X\geq 7) =0.97$

And we can calculate this with the complement rule.

$P(X \geq 7) = 1-P(X$

So then we have:

$P(X \leq 6) = 0.03$

And we are interested on the valueof n who satisfy this expression.

And for this we can verify this with the following code:

"=BINOM.DIST(6,E54,0.2,TRUE)"

And as we can see on the second figure attached the value who satisfy the condition would be n = 60.

5 0

3 years ago

Write the slope-intercept form of the equation of each line.

yanalaym [24]

Answer:

y=mx+b

y=x+3

Step-by-step explanation:

m: 1 (rise over run by 1)

b: 3

7 0

2 years ago

Researchers believe that 24.5% of the adults in the United States are obese. A Gallup poll examined the rate of obesity in Ameri

Setler79 [48]

Answer:

The p-value of the test is 0 < 0.05(standard significance level), which means that this is enough evidence to state that more than 24.5% of all adults in the U.S. are obese

Step-by-step explanation:

Researchers believe that 24.5% of the adults in the United States are obese. Test if there is enough evidence to state that more than 24.5% of all adults in the U.S. are obese.

At the null hypothesis, we test if the proportion is of 24.5% or less, that is:

$H_0: p \leq 0.245$

At the alternative hypothesis, we test if this proportion is greater than 24.5%, that is:

$H_1: p > 0.245$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.245 is tested at the null hypothesis:

This means that $\mu = 0.245, \sigma = \sqrt{0.245*0.755}$

Survey 86,664 randomly sampled U.S. adults. Of the adults surveyed, 23,053 said that they were obese.

This means that $n = 86664, X = \frac{23053}{86664} = 0.266$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.266 - 0.245}{\frac{\sqrt{0.245*0.755}}{\sqrt{86664}}}$

$z = 14.37$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion of 0.266 or higher, which is 1 subtracted by the p-value of z = 14.37.

The p-value of z = 14.37 is 1.

1 - 1 = 0

The p-value of the test is 0 < 0.05(standard significance level), which means that this is enough evidence to state that more than 24.5% of all adults in the U.S. are obese

4 0

3 years ago

Find the volume of the cylinder.

marin [14]

Answer:

392.7

Step-by-step explanation:

5 0

3 years ago

Assortative mating is a nonrandom mating pattern where individuals with similar genotypes and/or phenotypes mate with one anothe

scZoUnD [109]

Answer:

a) P(male=blue or female=blue) = 0.71

b) P(female=blue | male=blue) = 0.68

c) P(female=blue | male=brown) = 0.35

d) P(female=blue | male=green) = 0.31

e) We can conclude that the eye colors of male respondents and their partners are not independent.

Step-by-step explanation:

We are given following information about eye colors of 204 Scandinavian men and their female partners.

Blue Brown Green Total

Blue 78 23 13 114

Brown 19 23 12 54

Green 11 9 16 36

Total 108 55 41 204

a) What is the probability that a randomly chosen male respondent or his partner has blue eyes?

Using the addition rule of probability,

∵ P(A or B) = P(A) + P(B) - P(A and B)

For the given case,

P(male=blue or female=blue) = P(male=blue) + P(female=blue) - P(male=blue and female=blue)

P(male=blue or female=blue) = 114/204 + 108/204 − 78/204

P(male=blue or female=blue) = 0.71

b) What is the probability that a randomly chosen male respondent with blue eyes has a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=blue) = 78/114

P(female=blue | male=blue) = 0.68

c) What is the probability that a randomly chosen male respondent with brown eyes has a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=brown) = 19/54

P(female=blue | male=brown) = 0.35

d) What is the probability of a randomly chosen male respondent with green eyes having a partner with blue eyes?

As per the rule of conditional probability,

P(female=blue | male=green) = 11/36

P(female=blue | male=green) = 0.31

e) Does it appear that the eye colors of male respondents and their partners are independent? Explain

If the following relation holds true then we can conclude that the eye colors of male respondents and their partners are independent.

∵ P(B | A) = P(B)

P(female=blue | male=brown) = P(female=blue)

or alternatively, you can also test

P(female=blue | male=green) = P(female=blue)

P(female=blue | male=blue) = P(female=blue)

But

P(female=blue | male=brown) ≠ P(female=blue)

19/54 ≠ 108/204

0.35 ≠ 0.53

Therefore, we can conclude that the eye colors of male respondents and their partners are not independent.

7 0

3 years ago