Answer asap! WiIl give brainliest!!!<br> Worth 11 points!

A state end-of-grade exam in American History is a multiple-choice test that has 50 questions with 4 answer choices for each que

Assoli18 [71]

Answer:

Q1) The student has a 0.01% probability of passing the test.

Q2) She has a 99.91% probability of passing in the test.

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he gets it correct, or he gets it wrong. So we solve this problem using the binomial probability distribution.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

For this problem, we have that:

Question 1.

There are 50 questions, so $n = 50$ .

The student is going to guess each question, so he has a $\pi = \frac{1}{4} = 0.25$ probability of getting it right.

He needs to get at least 25 question right.

So we need to find $P(X \geq 25)$ .

Using a binomial probability calculator, with $n = 50$ and $\pi = 0.25$ we get that $P(X \geq 25) = 0.0001$ .

This means that the student has a 0.01% probability of passing the test.

Question 2.

Now, we need to find $P(X \geq 25)$ with $\pi = 0.70$ . So $P(X \geq 25) = 0.9991$

She has a 99.91% probability of passing in the test.

7 0

4 years ago

Read 2 more answers

An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberr

Sliva [168]

Answer:

A) Option B is correct.

H₀: μ₁ = μ₂

Hₐ: μ₁ - μ₂ < 0

B) t = -2.502

p-value = 0.0112

C) Option A is correct.

Reject H₀. The data suggests that cola has a higher average compression strength than the strawberry drink.

D) Option A is correct.

The distributions of compression strengths are approximately normal.

Step-by-step explanation:

The complete Question is presented in the two attached images to this answer.

A) To perform this test we first define the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to test if the extra carbonation of cola results in a higher average compression strength. That is, that cola has a higher average compression strength than the strawberry drink.

Hence, the null hypothesis would be that there isn't significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

The alternative hypothesis is that there is significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

Mathematically, if the average compression strength of strawberry drink is μ₁, the average compression strength of cola is μ₂ and the difference in compression strengths is μ = μ₁ - μ₂

The null hypothesis is represented as

H₀: μ = 0 or μ₁ = μ₂

The alternative hypothesis is represented as

Hₐ: μ < 0 or μ₁ - μ₂ < 0

B) So, to perform this test, we need to compute the test statistic

Test statistic for 2 sample mean data is given as

Test statistic = (μ₁ - μ₂))/σ

σ = √[(s₂²/n₂) + (s₁²/n₁)]

μ₁ = average compression strength of strawberry drink = 537

n₁ = sample size of the sample of strawberry drink in cans surveyed = 10

s₁ = standard deviation of the compression strength of strawberry drink in cans surveyed= 22

μ₂ = average compression strength of cola = 559

n₂ = sample size of the sample of cola in cans surveyed = 10

s₂ = standard deviation of the compression strength of strawberry drink in cans surveyed = 17

σ = [(17²/10) + (22²/10)] = 77.5903160379 = 8.792

We will use the t-distribution as no information on population standard deviation is provided

t = (537 - 559) ÷ 8.792

= -2.502 = -2.50

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n₁ + n₂ - 2 = 10 + 10- 2 = 18

Significance level = 0.05

The hypothesis test uses a one-tailed condition because we're testing in only one direction (whether compression strength of cola in can is greater).

p-value (for t = -2.50, at 0.05 significance level, df = 18, with a one tailed condition) = 0.011154 = 0.0112 to 4 d.p.

C) The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.0112

0.0112 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis accept the alternative hypothesis & say that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.

D) The necessary conditions required before a t-test is deemed valid include.

- The samples used must be a random sample of the population distribution with each variable in the sample independent of other one.

- The distribution of the population where the samples were extracted from must be normal or approximately normal to ensure some degree of normality for the samples.

Hence, the necessary assumption for this t-test among the options is that the distributions of compression strengths are approximately normal.

Hope this Helps!!!

7 0

3 years ago

(-11r+14+9r^3) + (3-8r^5+14r^3)

murzikaleks [220]

-8r^5+23r^3-11r+17

Wasn’t Sure What It Was Asking So Hopefully That Works For You

3 0

4 years ago

A mega-discount chain store just opened a new clothing store in town emphasizing mainly women's clothing. Before opening, man

Georgia [21]

Answer:

a) Cluster sampling

Step-by-step explanation:

<em>Since here the population is divided into many groups (cluster) and one of these groups is taken as a sample. This type of sampling is called Cluster Sampling.</em>

Further,

In Cluster Sampling the population is distributed into a distinct group called cluster and clusters are chosen as a sample then it is Cluster Sampling.

Multistage Sampling is a complex form of Cluster Sampling. In this sampling population is divided into a different group then one or more groups are randomly selected then again that selected group is divided into a different small group and again one or more these small groups are randomly selected as sample. This process is done many times.

If the sampling is done by any criteria then this type of sampling method is called Systematic Sampling. Like observer is taken every 5th unit as a sample.

If the whole population is divided into many groups(strata) such that between the group units are heterogeneous and within the group units are homogeneous then units are randomly selected from these groups. This type of sampling is called Stratified Sampling.

7 0

3 years ago

A kite string is 35m long. The angle the string makes with the ground is 50o

kipiarov [429]

Answer:

22.50m

Step-by-step explanation:

With the ground, the horizontal ground distance is the adjacent side, so by SOHCAHTOA, the ratio used is cosine, i.e.

Ground distance

= hypotenuse * cos(50°)

= 35 cos(50°)

= 35*0.64279

= 22.50m

7 0

3 years ago

Answer asap! WiIl give brainliest!!! Worth 11 points!