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valina [46]
3 years ago
7

In a class of 100 students, 20 students have hard copy and 80 students have electronic textbooks for the course. If you randomly

choose 10 students in this class, find the approximate probability using a binomial distribution that 3 of them have hard copy texts (round off to second decimal place).
Mathematics
1 answer:
Naddik [55]3 years ago
7 0

Answer:

P(X=3)

And if we use the probability mass function we got:

P(X=3)=(10C3)(0.2)^3 (1-0.2)^{10-3}=0.201 \approx 0.20

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 "number of students with hard copy texts", on this case we now that:

X \sim Binom(n=10, p=20/100=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!}

And we want to find this probability:

P(X=3)

And if we use the probability mass function we got:

P(X=3)=(10C3)(0.2)^3 (1-0.2)^{10-3}=0.201 \approx 0.20

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SOLUTION:

Step 1 :

In this question, we are told that a marketing research company needs to estimate the average total compensation of CEOs in the service industry.

We also have that: Data were randomly collected from 38 CEOs and the 98% confidence interval was calculated to be ($2,181,260, $5,836,180).

Then, we are asked to find the margin error for the confidence interval.

Step 2:

We need to recall that:

\text{Higher Confidence Interval, CI}_{H\text{ = }}X\text{ + }\frac{Z\sigma}{\sqrt[]{n}}\text{Lower Confidence Interval , CI}_{L\text{ }}=\text{ X - }\frac{Z\sigma}{\sqrt[]{n}}

It means that:

\vec{}X\text{ = }\frac{CI_{H\text{ }}+CI_L}{2}\text{Margin of error, }\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{CI_{H\text{ - }}CI_L}{2}

where,

CI_H\text{ = }$$5,836,180$\text{ }$$$CI_{L\text{ }}=\text{ }2,181,260

putting the values into the equation for the margin of error, we have that:

\text{Margin of error,}\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{5,836,180\text{  - }2,181,260\text{ }}{2}\begin{gathered} =\text{ }\frac{3654920}{2} \\ =1,\text{ 827, 460} \end{gathered}

CONCLUSION:

The margin error for the confidence interval is 1, 827, 460

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