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Nutka1998 [239]
2 years ago
10

According to the American Community Survey, 27% of residents of the United States 25 years old or older had earned a bachelor de

gree. Suppose you select 12 residents of the United States 25 years old or older and recorded the number who had earned a bachelor degree. Round probabilities to 4 decimal places.
Explain why this is a binomial experiment.

Find and interpret the probability that exactly 5 of them had a bachelor degree.

Find and interpret the probability that fewer than 5 of them had a bachelor degree.

Find and interpret the probability that at least 5 of them had a bachelor degree.

Compute the mean and standard deviation of the binomial random variable.
Mathematics
1 answer:
oksian1 [2.3K]2 years ago
3 0

a. The reason why this question is a binomial experiment is based on the fact that it is made up of an independent sample, it has a number that is fixed and a probability.

Each event is made up of two outcomes and they are random with the same success rate.

<h3>b. How to solve probability that exactly 5 had a bachelor</h3>

we have the following data n = 12, p = 0.27 and k = 5

We have to use the function to solve electronically

binompdf(n,p,k)

input the values

= binompdf(12,0.27,5)

This gives us

= 0.1255

<h3>(C) Probability that fewer than 5 have bachelor</h3>

We use the formula below

= binompdf(12,0.27,5-1)

This is = 0.7984

D. Probability of at least 5

1 - probability of fewer than 5

= 1 - 0.7984

= 0.2016

How to solve for the Mean = n*p

n = 12 , p = 0.27

Mean = 12*0.27 = 3.24

and

standard deviation = √npq

n = 12, p = 0.27 , q = 1- 0.27

= 0.73

sd = √12*.27*.73

= 1.54

Read more on binomial experiment here:

brainly.com/question/9325204

#SPJ1

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Answer:

We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

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<u><em /></u>

<u><em>Let </em></u>\mu<u><em> = average sodium content.</em></u>

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The value of z test statistics is -0.307.

<em>Since, in the question we are not given the level of significance so we assume it to be 5%. </em><em>Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.</em><em> </em>

<em>Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><em><u>we fail to reject our null hypothesis</u></em><em>.</em>

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